TESTE DE FIZICÃ PENTRU BACALAUREAT

Coordonatori Elena-Mihaela Garabet middot Catildetatildelina-Valentina Stanca middot Tatiana Matilderatildendici

Liviu-Datildenuthorn Rotaru middot Victor Stoica middot Corina Dobrescu Laura-Angelica Onose middot Simona Buiu middot Ana Nithornoiu

Aurelia Daniela Florian middot Diana-Cristina Bejan middot Ion Batilderaru

TESTE DE FIZICAtilde PENTRU BACALAUREAT

BAREME DE CORECTARE

Această variantă electronică cu rezolvările testelor de fizică icircnsoţeşte lucrarea

TESTE DE FIZICĂ PENTRU BACALAUREAT (ISBN 978-606-38-0224-9)

copy Editura NICULESCU 2018

Bd Regiei 6D 060204 ndash Bucureşti Romacircnia

Telefon 021 312 97 82 Fax 021 312 97 83

E-mail edituraniculescuro

Internet wwwniculescuro

Comenzi online wwwniculescuro

Comenzi e-mail vanzariniculescuro

Comenzi telefonice 0724 505 385 021 312 97 82

Redactor Geta Vicircrtic

Tehnoredactor Carmen Birta Şerban-Alexandru Popină

ISBN 978-606-38-0224-9

Toate drepturile rezervate Nicio parte a acestei catilderthorni nu poate fi reprodusatilde sau transmisatilde sub nicio formatilde ordmi prin niciun mijloc electronic sau mecanic inclusiv prin fotocopiere icircnregistrare sau prin orice sistem de stocare ordmi accesare a datelor fatilderatilde permisiunea Editurii NICULESCU Orice nerespectare a acestor prevederi conduce icircn mod automat la ratildespunderea penalatilde fathornatilde de legile nathornionale ordmi internathornionale privind proprietatea intelectualatilde ____________________________________________________________________________________________________________________________________________________________________________________________

Editura NICULESCU este partener ordmi distribuitor oficial OXFORD UNIVERSITY PRESS icircn Romacircnia E-mail oxfordniculescuro Internet wwwoxford-niculescuro

TESTE DE NIVEL MINIMAL

Teste de fizicatilde pentru bacalaureat

4

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 d 34 c 35 a 3

TOTAL pentru Subiectul I 15p

Subiectul al II-lea Parţial Punctaj

a

Greutatea lăzii este G = m middot g 2p

4pDeci masa lăzii este Gmg

= 1p

Rezultă m = 2 kg 1p

b

Forţa de frecare la alunecare dintre ladă şi suprafaţa orizontală se determină din relaţia Ff = μ sdot N 2p

4pUnde N = G 1pRezultă Ff = 4 N 1p

c Lada este trasă pe suprafaţă orizontală cu viteză constantă de către forţa F

deci 0fF F =- 2p3p

Rezultă F = 4 N 1p

d

Sub acţiunea forţei 1 2F F=

lada se mişcă uniform accelerat cu acceleraţia 1a Icircn acest caz avem relaţia 1 1fF F m aminus = sdot

2p

4pObţinem 1

2 fF F Fam m

= =- 1p

Rezultă a1 = 2 ms2 1p

TOTAL pentru Subiectul al II-lea 15p

Subiectul al III-lea Parţial Punctaj

aEnergia potenţială maximă a sistemului bilăndashPămacircnt p maxE m g h= sdot sdot 2p

3Rezultă p max 2 JE = 1p

b

Energia totală a bilei icircn momentul icircn care aceasta este lăsată sa cadă liber este 1 maxpE E= 1p

4Energia totală a bilei icircn momentul icircn care aceasta atinge solul este 2 maxcE E= 1p

Aplicăm legea de conservare a energiei mecanice 1 2E E= 1p

Rezultă Ec max = 2 J 1p

Bareme teste nivel minimal

5

c

Din condiţiile problemei cA A pE E= deci energia totală a bilei este 2 2A cA pA pA AE E E E m g h= + = = sdot sdot 1p

4pAplicăm legea de conservare a energiei mecanice 1 AE E= 1p

Obţinem 2Ahh = 1p

Rezultă 10 mAh = 1p

d

Aplicăm teorema de variaţie a energiei potenţiale (1 2)conservativpE Lrarr∆ = minus 1p

4pDeci 0 Gm g h Lminus sdot sdot = minus 1p

Lucrul mecanic efectuat de greutatea corpului din momentul icircn care bila cade liber pacircnă cacircnd aceasta atinge

solul este GL m g h= sdot sdot 1p

Rezultă 2 JGL = 1p

TOTAL pentru Subiectul al III-lea 15p


6

B TERMODINAMICAtilde

Subiectul I Punctaj

1 b 32 b 33 a 34 c 35 d 3



aNumărul de moli de azot este 1

11

mν =

micro2p

3p

Rezultă 1 0035 moliν cong 1p

b

Numărul de moli de oxigen poate fi exprimat atacirct din relaţia 22

A

NN

ν = 1p

4pcacirct şi din relaţia 2

22

mν =

micro 1p

Din relaţiile precedente obţinem numărul de molecule de oxigen 22 A

2

mN N= sdotmicro

1p

Rezultă 222 301 10N = sdot 1p

c

Presiunea azotului este 11

1 1

m R TpVsdot

= sdotmicro 1p

4pRezultă 5 2

1 0445 10 N mp cong sdot 1p

Presiunea oxigenului este 22

2 2

m R TpVsdot

= sdotmicro

1p

Rezultă 5 22 0311 10 N mp cong sdot 1p

d

Numărul de moli din cele două baloane este 1 2 ν = ν + ν 1p

4p

unde 1 2m m+ν =

micro 1p

Icircn final obţinem 1 2 1 2 1 2

1 2 1 2 2 1

1 2

( )m m m mm m m m

+ + sdotmicro sdotmicromicro = =

sdotmicro + sdotmicro+micro micro

1p

Rezultă 3033 g molmicro cong 1p



7


a

Pentru starea de echilibru termodinamic 1 13

1 0

311

1

300 K

831 831 10 Pa

= 3 dm

Tp p

R TVp

=

= = sdotν sdot sdot

=

1p

5pPentru starea de echilibru termodinamic 2 2 1

32 1

32 1

2 600 K

831 10 Pa

=2 6 dm

T Tp pV V

= =

= = sdot

=

2p

Pentru starea de echilibru termodinamic 3 3 2

313

33 2

600 K

4155 10 Pa2

=2 12 dm

T Tpp

V V

= =

= = sdot

=

2p

b

Căldura primită de gaz icircn procesul 1rarr2 este

12 2 1 V 2 1 17( ) ( ) ( )2

= ν sdot sdot minus = ν sdot + sdot minus = ν sdot sdotpQ C T T C R T T R T 1p

4p

Căldura primită de gaz icircn procesul 2rarr3 este3

23 2 12

ln 2 ln 2VQ R T R TV

= ν sdot sdot = ν sdot sdot 1p

Căldura totală primită de gaz este

12 23 17 2ln 22

= + = ν sdot sdot sdot +

Q Q Q R T 1p

Rezultă 121808 JQ cong 1p

c

Variaţia energiei interne icircn procesul 1rarr2 este

12 V 2 1 15( )2

U C T T R T∆ = ν sdot sdot minus = ν sdot sdot 1p2p

Rezultă 12 62325 JU∆ = 1p

d

Lucrul mecanic efectuat de gaz asupra mediului exterior icircn procesul 1rarr2 este12 2 1 1( )L R T T R T= ν sdot sdot minus = ν sdot sdot 1p

4p

Lucrul mecanic efectuat de gaz asupra mediului exterior icircn procesul 2rarr3 este3

23 2 12

ln 2 ln 2VL R T R TV


Lucrul mecanic total efectuat de gaz asupra mediului exterior este12 23 1 (1 2ln 2)L L L R T= + = ν sdot sdot sdot + 1p

Rezultă 59483 JL cong 1p



8

C PRODUCEREA ordfI UTILIZAREA CURENTULUI CONTINUU

Subiectul I Punctaj

1 c 32 a 33 d 34 d 35 b 3



aRezistenţa internă a grupării celor două surse o determinăm din relaţia 1 2Sr r r= + 2p

3p

Rezultă 2 Sr = Ω 1p

b

Tensiunea electromotoare echivalentă a grupării celor două surse este 1 2SE E E= + 1p

4pIntensitatea curentului prin rezistorul R cacircnd icircntrerupătorul k se află icircn poziţia deschis este S

S

EIR r

=+

2p

Rezultă 1 AI = 1p

c

Dacă icircntrerupătorul 1k este deschis şi icircntrerupătorul 2k este icircnchis atunci intensitatea curentului prin rezis-

torul R este 1

1

EIR r

prime =+

2p

4pCăderea de tensiune pe rezistenţa R este U I Rprime= sdot 1p

Rezultă 431 VU cong 1p

dDacă ambele icircntrerupătoare sunt icircnchise atunci 1

1SC

EIr

= 3p4p

Rezultă 9 ASCI = 1p



9


a

Pentru gruparea serie a rezistoarelor rezistenţa echivalentă SR o calculăm din relaţia 1 2SR R R= + 1p

5p

Pentru gruparea paralel a rezistoarelor rezistenţa echivalentă PR o calculăm din relaţia 1 2

1 1 1

PR R R= + 1p

Deoarece S P P P= avem relaţia 2 2

S PS P

E ER RR r R r

sdot = sdot

+ + 1p

După efectuarea calculelor obţinem S Pr R R= sdot 1p

Rezultă 4 r = Ω 1p

b

Pentru gruparea serie a rezistoarelor randamentul circuitului este SS

S

RR r

η =+

1p

4pRezultă 7142 Sη cong 1p

Pentru gruparea paralel a rezistoarelor randamentul circuitului este PP

P

RR r

η =+

1p

Rezultă 2857 Pη cong 1p

c

Icircn acest caz puterea P1 furnizată de sursă rezistorului R1 este 2

1 11

EP RR r

= sdot

+ 1p

3pDeci 1

11

( ) PE R rR

= + sdot 1p

Rezultă 6 VE = 1p

dIntensitatea curentului de scurtcircuit al sursei este SC

EIr

= 2p3p




10

DOPTICAtilde

Subiectul I Punctaj

1 b 32 a 33 d 34 a 35 c 3



aConvergenţa lentilei este 1C

f= 2p

3p

Rezultă 15 mC minus= 1p

b

Din prima formulă fundamentală a lentilelor 2 1

1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+1p

Rezultă 2 30 cm 03 mx = = 1p

c

Mărirea liniară transversală este 2

1

xx

β = 1p

5pRezultă 12

β = minus 1p

Imagine reală 1pImagine răsturnată 1p

Imagine de două ori mai mică decacirct obiectul 1p

dDistanţa dintre obiectul luminos şi imaginea acestuia prin lentilă este 1 2d x x= minus + 2p

3p

Rezultă 90 cm 09 md = = 1p



11


aLungimea de undă a radiaţiei monocromatice de frecvenţă 1ν este 1

1

cλ =

ν2p

3p

Rezultă 1 600 nmλ = 1p

b

Pentru radiaţia de frecvenţă 1ν avem 1 1ex Sh L e Usdot ν = + sdot 1p

4pPentru radiaţia de frecvenţă 2ν avem 2 2ex Sh L e Usdot ν = + sdot 1p

Din relaţiile precedente obţinem 2 1

2 1

( )

S S

heU Usdot ν minus ν

=minus

1p

Rezultă 1916 10 Ce minus= sdot 1p

c

Lucrul mecanic de extracţie este 0exL h= sdot ν 1p

4pDar 1 1ex SL h e U= sdot ν minus sdot sau 2 2ex SL h e U= sdot ν minus sdot 1p

Obţinem 10 1

Se Uh

sdotν = ν minus sau 2

0 2Se U

hsdot

ν = ν minus 1p

Rezultă 140 46972 10 Hzν = sdot 1p

d

Pentru radiaţia de frecvenţă 1ν energia cinetică a fotoelectronilor emişi de către catodul celulei fotoelectrice

este 1 1c SE e U= sdot1p

4pRezultă 22

1 01249 eV = 19984 10 JcE minus= sdot 1p

Pentru radiaţia de frecvenţă 2 ν energia cinetică a fotoelectronilor emişi de către catodul celulei fotoelec-

trice este 2 2c SE e U= sdot1p

Rezultă 222 05374 eV = 85984 10 JcE minus= sdot 1p



12

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1a [ ] J Nm= =SIL 3

2 c 2

2 2 00 2 270 m

2vv v ad d

a= + rArr = = 3

3b Din legea vitezei 0

2

10 6

6 ms

v v at v t

a

= + = +

rArr =

854 kgFma

= = 3

4 b 2

2

2

2

C

C

mvE p mv

pEm

= =

rArr =

2

2

2

2

C

C

mvE p mv

pEm

= =

rArr = 3

5 b 2

0max

0

20 m 10 m2

5 JC C p C

vh hg

E E E E

= = rArr =

= + rArr =

3



a

Reprezentăm forţele care acţionează asupra corpului şi aplicăm principiul al doilea al dinamicii la mişcarea corpului pe planul icircnclinat

1 1

1

t f

n

G F ma

N G

minus =

= 2p

4p

1 1(sin cos )a g= α minus micro α 1p

21 25 msa = 1p

b

Din ecuaţia lui Galilei 2 20 1 02 0v v a l v= + =

2 20 1 02 0v v a l v= + = 1p

3p1 12v a l= 1p

1 10 316 msv = = 1p

c

Reprezentăm forţele care acţionează asupra corpului şi aplicăm principiul al doilea al dinamicii la mişcarea corpului pe planul orizontal

2 2

2

fF ma

G N

minus =

=

1p

4p2 2a g= minusmicro 1p

2 2 21 2 22 1 msv v a d a= + rArr = minus 1p

22 01a

gmicro = minus =

1p


13

d


1t f

n

G F FN G

= +

=

2p

4p

(sin cos )F mg= α minus micro α 1p

5 NF = 1p



a

1 1f fL F d= minus

1p

4p1f

y

F N

N G F

= micro

= minus

1p

1( sin )fF mg F= micro minus α

1p

1

75 JfL = minus 1p

bcosL Fd= α 2p

3p

255 JL = 1p

c

1C fE L L∆ = + 1p

4p

2

2CmvE =

1p

2 CEvm

=

1p

2 3 34 msv =

1p

d

2C fE L∆ =

2 2f fL F D= minus

1p

4p2f

F N

G N

= micro

=1p

C

f

EDF

=

1p

D = 6 m 1p



14


Subiectul I Punctaj

1b mpV RT=

micro

30 kPamp RT

V= =

micro

3

2 b Din teorie 0 0L Qne = 3

3 c VL C T= minusν ∆ 3

4c

3127 kgmpRT

microρ = = 3

5

b 1 1 2 23 32 2

U RT U RT= ν = ν

2 1 2 12 2U U T T= rArr =

11

2 200 J3UL R T RT= ν ∆ = ν = =

3



a

Din ecuaţiile de stare 1 21 2

1 2

p V p Vm mRT RTmicro micro

= = 1p

4p

1 2m m m∆ = minus 1p

1 2

1 2

V p pmR T T

micro∆ = minus

1p

1 2

1 2 2 1( )mRTTV

p T p T∆

=micro minus

33324 mV =

1p

b

11

1

p VRT

ν =

2p3p

1 026 kmolν = 1p

c

22 2

mp V RT=micro

2p

4p22

2

p VmRT

micro=

1p

2 512 kgm = 1p

d

11

1

pRT

microρ =

3p

4p

31 25 kgmρ = 1p



15


a

Reprezentarea grafică a transformărilor icircn coordonate (p V)

3p 3p

b

Din legea procesului izobar 1 22 1

1 2

3V V V VT T

= rArr = 1p

4p12 1 2 1( )L p V V= minus 1p

121

2 1( )Lp

V V=

minus 1p

5 21 2 10 Nmp = sdot

1p

c

Gazul cedează căldură icircn procesele 2rarr3 şi 3rarr1

23 31cedQ Q Q= + 1p

4p323 2 1 1( ) 3VQ C T T p V= ν minus = minus

1p

131 1 1 1

3

ln ln3VQ RT p VV

= ν = minus

1p

cedat 1640 JQ minus 1p

d

2 1( )abs pQ C T T= ν minus 1p

4pp VC C R= + 1p

1 1 15 5absQ RT p V= ν = 1p

absorbit 2000 JQ = 1p



16


Subiectul I Punctaj

1b 1 Wh = 3600 J 6720 10 JW = sdot

3

2 c 1 1 2

2

E I EIR r R R r

= =+ + +

1 1 2

2

E I EIR r R R r

= =+ + +

1 22

1

2 8R R r RR r+ +

rArr = rArr = Ω+

3

3 b 23ABRR = = Ω 3

4 c V1sc

EIr

= =Ω

3

5c

2

bec 90n

n

URP

= = Ω

bec 0 (1 ) 2000 CR R t t= + α rArr = deg

3



a

1e pR R R= + 1p

3p2 3

2 3p

R RRR R

=+

1p

44eR = Ω 1p

b

1e

EIR r

=+

1p

4pV eU E Ir IR= minus = 2p

88VU V= 1p

c

1 2 3I I I= + 1p

4p2 2 3 3I R I R= 1p

1 23

2 3

I RIR R

=+

1p

3 08I A= 1p


17

d

Prin schimbarea locurilor icircntre ampermetrul ideal şi sursă ampermetrul va măsura intensitatatea prin R1

1 23

1 2e

R RR RR R

= ++

1p

4p125 A

e

EIR r

=+

1p

1 2

1 1 2 2

I I I I R I R

= +=

1p

12 083 A3II = =

1p



a

Aplicacircnd legile lui Kirchhoff rezultă

1 3 3E Ir U I R= + +

2p

4p

13

3

E Ir UIR

minus minus=

1p

3 15 AI = 1p

b

2 3 05 AI I I= minus =

2 2 3 3I R I R= 1p

4p3 3

22

75I RRI

= = Ω

1p

22 2 2P R I= 1p

2 1875 WP = 1p

c

1e pR R R= + 1p

4p

11 30UR

I= = Ω

1p

2 3

2 3p

R RRR R

=+

1p

4875eR = Ω 1p

d

1 1

s

P UP E

η = =

2p3p

05 50η = = 1p



18

DOPTICAtilde

Subiectul I Punctaj

1 c 3

2 a 2

1

sin 3sin 60sin 2

i n r rr n

= rArr = rArr = 3

3 b 1 2 3 3 4C C C C C= + + rArr = δ 3

4 c ndash19026 10 Jex C Chc L E E= + rArr = sdotλ

3

5b 2 2

1 1

2 1

3

3 150 cm

y xy x

x x

β = = = minus

rArr = minus =

2 1

1 1 1 375 cmff x x

= minus rArr =

3



a

2 21

1 1

3y xy x

β = = = minus

1p

4p2 13x x= minus 1p

11

43fx = minus

1p

1 40 cmx = minus 1p

b

32

1

3xx

β = =

1p

4p3 13x x= 1p

1 32

1 3

x xfx x

=minus

1p

2 60 cmf = 1p

c 3p 3p


19

d

1 2

1 1 1F f f

= + 1p

4p

20 cmF = 1p

14

1

FxxF x

=+

1p

43

1

1xx

β = = minus 1p



aDefinirea efectului fotoelectric 1p

4pEnunţarea legilor efectului fotoelectric 3p

b

0L h= ν 1p

4p

150 055 10 Hzν = sdot

1p

151

1

044 10 Hzcν = = sdot

λ 1p

1 0 ν lt ν nu se produce efect fotoelectric folosind radiaţia monocromatică cu λ1 1p

c2

2

hcε =

λ 2p

3p

19495 10 JCE minus= sdot 1p

d

2 ex CL Eε = + 2p

4pCS

EUe

=

1p

079 VSU = 1p



20

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 a 34 a 35 d 3



a

Reprezentăm forţele care acţionează asupra corpului şi aplicăm principiul al doilea al dinamicii la mişcarea

0eF G+ =

1p

3p1 1middot middotk l m g∆ =

11

m glk

∆ = 1p

Rezultat final 40 Nm=k 1p

b

fF F= 1p

4pN G= 1p

fF N= micro

Rezultat final 22 micro

∆ =m glk 2 125 cm∆ =l 1p

c

e fF F ma+ =

1p

4peF F= 1p

2

2

minus micro=

F m gam 1p

Rezultat final 23 ms=a 1p

d

22

2pk lE ∆

=

2p

4p2

Flk

∆ =

1p

Rezultat final -25∙10pE J= 1p



21


a

F G ma+ =

1p

4pminus =F G ma rArr ( )F m a g= + 2p

Rezultat final 36 NF = 1p

b

vat

∆=

∆1p

3p0 = +v v at 0 0v = 1p

Rezultat final 10 ms=v 1p

c

total∆ =cE L 2p

4p2

2cmvE∆ = 1p

Rezultat final total 150=L J 1p

d

Conservarea energiei 1 2E E= 1p

4p

2

1 2

= +mvE mgh

2

2 2PmvE =

1p

2( )2∆

=a th

1p

2 2v v gh= + Rezultat final 10 6 msPv =

1p



22


Subiectul I Punctaj

1 a 32 d 33 b 34 a 35 c 3



a A

N mN

=micro

2p4p

Rezultat final 231125∙10N = 2p

b

0∆ρ = ρ minus ρ 1p

4p0

0

mV

ρ =

mV

ρ = 1p

0

0middotmVV

V m=

∆ρ +1p

Rezultat final 4V l= 1p

cV V 2 1( )∆ = ν ∆ = ν minusU C T C T T 1 2T T= 2p

3p

Rezultat final 0U∆ = 1p

d

0middot p V RT= ν 2p

4p0 p VRT

ν =

Rezultat final 016 molν =

2p



a

Din grafic procesul 1 2rarr este reprezentată printr-o dreaptă ce trece prin originea axelor de coordonate deci presiunea şi volumul sunt proporţionale Astfel rapoartele volumelor şi ale presiunilor sunt egale

2 2

1 1

=p Vp V

2p

3p

Rezultat final 2

1

2pp

= 1p


23

b

12 V 12 V 2 1( )∆ = ν ∆ = ν minusU C T C T T 1p

4p1 1 1sdot = νp V RT 2 2 2sdot = νp V RT 1p

2 14T T= 1p

Rezultat final 12 192

U RT∆ = ν rArr 12 10800U J∆ = 1p

c

Lucrul mecanic total reprezintă aria figurii formate de ciclul termodinamic reprezentat icircn coordonate (pV)

( )=L A p V 2p

4p1 1 1

2 2p V RTL ν

= =

Rezultat final 1200JL =

2p

d

ced 2 3 3 1minus minus= +Q Q Q 1p

4p1 2 3 2( )minus = ν minusvQ C T T 3 1 1 3( )minus = ν minuspQ C T T 1p

2 14 =T T 3 12T T= 1p

Rezultat final ced 111 2

= minus νQ RT ced 13200= minusQ J 1p



24


Subiectul I Punctaj

1 d 32 b 33 a 34 c 35 c 3



aRS= R1+ R2 2p

3pRezultat final RS= 30 Ω 1p

b

UAB = R2 I 1p

4ps

EIR r

=+ 2p

Rezultat final UAB = 25 V 1p

c s A

EIR R r

=+ +

3p4p

I primecong 12 A 1p

d 1

EIR r

=+

3p4p

I primeprimeprimecong 33 A 1p



a

Din grafic E = 40 V 1p

4pISC = 10 A 1p

ISC = scEIr

= 1p

Rezultat final E = 40 V r = 4 Ω 1p

b

2middot middotW R I t= ∆ 2p

4pEIR r

=+

1p

Rezultat final W = 5760 J 1p

c

echivalent

echivalent

RR r

η =+

2p

4pechivalent 10 bR R= 1p

91η 1p

dPuterile consumate de către doi rezistori diferiţi conectaţi pe racircnd la bornele unei surse electrice sunt egale dacă R1R2= r2 2p

3pRezultat final R2= 2 Ω 1p



25

DOPTICAtildeSubiectul I Punctaj

1 a 32 b 33 b 34 d 35 a 3



a Construcţia imaginii 4p 4p

b

2 1

1 1 1x x f

minus = 1p

4p1fC

= 1p

Rezultate finale 1 100 cm= minusx 2p

c

2 2

1 1

β = =x yx y 2p

3p

Rezultat final 2 2 cm= minusy 1p

d

1 1 60 cm= + = minusx d x 1p

4p12

1

x fx x f

=+

1p

Rezultat final 2 200 cm=x 2 2 2ndash 100 cmx x x∆ = = 2p



aDin grafic 1= minusU V 2p

3p

Rezultat final stopare 1 VU = 1p

bfW h= ν 2p

4p

Rezultat final 1946 10 Jminus= sdotfW 2p

c max stopareW eU= 2p4p

Rezultat final 19max 16 10 Jminus= sdotW 2p

d extractie maxndashfL W W= 2p4p

Rezultat final ndash19extractie 3 10 J= sdotL 2p



26

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1c

5

8

km = 3 10 uas = 012

1ua min1 km = 15 10

cc

sdot rArrsdot

3

2b Pentru determinarea distanţei folosim metoda grafică

m

m

km = 200 m 48 h

= 15 s

dvt

d vt

= ∆ rArr =∆

3

3c

25

3

= 75 10 J2 = 208 kW h

11 J = 1 W s = 10 kW h3600

c c

c

m vE EE

minus

sdot= rArr sdot rArr sdot

sdot sdot

3

4

c Aplicacircnd principiul fundamental al dinamicii sau teorema variaţiei impulsului corpului asupra căruia se exercită forţa şi

proiectacircnd relaţia vectorială pe axa mişcării obţinem( )1 2

= 75 Nm m

m v vF F

tsdot +

= rArr∆

m mF v Fuarruarr ∆ rArr

este orientată orizontal spre stacircnga 3

5

b Scriem legea lui Hooke 0

0 0 02

0 0 2

0

4 1 3828 cm4

F l F lE lS l E S

d FS l l ld E

l l l

∆ sdot = sdot rArr ∆ = sdot π sdot sdot = rArr = + rArr = π sdot sdot

∆ = minus

3



a

[ ]05 s 10 N 0x ft R Fisin = rArr = 1p

4p2 = 2 m sxx

RR m a a am

minus= sdot rArr = rArr sdot 1p

10 10 m sv v a t v minus= + sdot ∆ rArr = sdot 2p

b 10 s 0 N 10 Nx f ft R F F F= = rArr = rArr = 3p 3p

cmax max 02fF m g= micro sdot sdot rArr micro = 3p 3p


27

d

[ ]1020 s 0 N constxt R visin = rArr = 1p

5p

Aplicacircnd teorema variaţiei impulsului pentru [ ]010 st isin p R t∆ = sdot ∆

1p

şi folosind interpretarea grafică

obţinem că din momentul 10 st = 1 15 m sv minus= sdot şi se menţine constantă

1p

Aplicacircnd teorema variaţiei energiei cinetice pentru [ ]020 st isin

5625 Jif if ifc R RE L L∆ = rArr =

2p



a

Reprezentăm printr-un desen situaţia din problemă punacircnd icircn evidenţă poziţiile avionului la momentele 1t

şi 2t Folosind datele problemei observăm că triunghiul care are ca vacircrfuri staţia radar şi poziţile avionului este dreptunghic

= 11314 mdrArr

2p

5p2p

2 1 2

= 14 s

= 12 h 24 min 34 s

dv tt

t t t t

= rArr ∆∆

∆ = minus rArr1p

b

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr =

2p3p

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr = 1p

c

Aplicăm teorema variaţiei energiei cinetice icircntre punctele A şi B c RE L∆ = (1) 1p

4p

Viteza avionului menţinacircndu-se constantă variaţia energiei cinetice este egală cu zero Motorul fiind oprit singurele forţe care se exercită asupra avionului sunt greutatea şi forţele de rezistenţă Icircnlocuind aceste observaţii icircn ecuaţia (1) şi exprimacircnd lucrul mecanic al greutăţii avionului icircntre A şi B obţinem

2p

( )1 20

879 MJr

r

F

F

m g h h L

L

= sdot sdot minus + rArr

rArr = minus

1p

d

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

2p

3p

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

1p



28


Subiectul I Punctaj

1 c Realizacircnd corespondenţa celor două scări de temperatură cu scara Celsius obţinem că0 degX corespund temperaturii 25 degC şi 0 degY corespund temperaturii 50 degC

3

2a Scriind ecuaţia termică de stare pentru gazele din cele două incinte icircn stările iniţială respectiv finală ţinacircnd cont de

datele problemei şi prelucracircnd ecuaţiile obţinute rezultă 1 23 2m msdot = sdot 3

3

b 1 1

2 2

1 1 11

22 2 2

5 cu 80 K 52 1673 3 cu 80 K2

V V

V V

U C T C R T UUU C T C R T

∆ = sdot sdot∆ = sdot ∆ = ∆ rArr = minus = minus ∆∆ = sdot sdot∆ = sdot ∆ = minus

ν

ν3

4

d 23 23 23

2323

40 J0

Q U LU

L= ∆ +

rArr ∆ = minus=

31 31 31 31 31 31 31 70 J Q U L U Q L U= ∆ + rArr ∆ = minus rArr ∆ = minus

123112

1231 12 23 31

0110 J

UU

U U U U∆ =

rArr ∆ =∆ = ∆ + ∆ + ∆

12 12 12 12 21190 J 190 JQ U L L L= ∆ + rArr = rArr = minus

3

5

c 22 72 10

VA

VA

Q C TQ NN NNC T

N

= ν sdot sdot ∆ sdot rArr = rArr = sdotν = sdot ∆

3



a

1 2 m m m= = 1p

3p1 2

2 1

125ν micro= =

ν micro2p

b1 2

1 2

2 g 356

molm m m

ν = ν + νsdot

= + rArr micro =micro micro micro

1p

4p1 2

1 2

2 g 356

molm m m

ν = ν + νsdot

= + rArr micro =micro micro micro 3p

c

3

kg 157 m

mp V R TpR Tm

V

sdot = sdot sdot sdotmicromicro rArr ρ = rArrsdotρ =

rArr ρ =

3p

4p

3

kg 157 m

mp V R TpR Tm

V


rArr ρ = 1p


29

d5

5 10 Pa

mp V R T

m R TpV

p

sdot = sdot sdot rArrmicro

sdot sdot= rArr

micro sdot

rArr = sdot

1p

4p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot 2p



a 11966 J L R T L= ν sdot sdot ∆ rArr = 3p 3p

b 20034 JQ U L U Q L U= ∆ + rArr ∆ = minus rArr ∆ = 3p 3p

c

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆

∆ = ν sdot sdot ∆

γ = = rArr γ =∆

1p

4p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

2p

d

5 1 3

pp v

v v

p v

CC C R RC C

C C R

γ = rArr = γ sdot sdot rArr = =

γ minus= +

1p

5p

Notăm cu f fracţiunea din cantitatea totală de substanţă care o constituie heliul şi exprimăm icircn două moduri energia internă a amestecului de gaze

( )( )

1 2 1 2

1 2

1 2

1 2

1

1v v v v v v

U U UC T C T C T C f C f C

f f

= +ν sdot sdot = ν sdot sdot + ν sdot sdot rArr = sdot + minus sdot

ν = sdot ν rArr ν = minus sdot ν

2p

2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = =

1p2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = = 1p



30


Subiectul I Punctaj

1a 2

2

514 nC

U S qI U r tl t ql

S rq

sdot = = sdot π sdot sdot ∆ρ sdot ∆ rArr =ρ sdot= π sdot

=

3

2

a ( )( )

1 0 1 2 1

1 2 2 12 0 2

3 1

1 - 1

4 10

R R t R RR t R tR R t

Kminus minus

= sdot + α sdot minus rArr α =sdot sdot= sdot + α sdot

α = sdot

10 0

1

9 1

RR Rt

= rArr = Ω+ α sdot

3

3

a Aplicacircnd teorema I a lui Kirchhoff pentru nodul A intensitatea curentului prin ramura AB va fi

15 AABI = minus deci acest curent are intensitatea 15 AI = şi circulă de la A la B

Aplicacircnd teorema a II-a a lui Kirchhoff ( ) 25 VAB ABU E I R r U= + sdot + rArr =

3

4b

=12800 C

PIP tU q q

q UIt

= sdot ∆ rArr = rArr=∆

3

5

c Folosind seturile complete de date din tabel ((2) şi (3)) 44 UR RI

= rArr = Ω

Pentru determinarea (4) 2 A 2000 mAUI IR

= rArr = =

Pentru determinarea (5) 132 VU R I U= sdot rArr =

3



aLa momentul iniţial cursorul fiind icircn A 0R = 1p

3p Intensitatea curentului icircn circuit va avea expresia 1 1

1

9 E EI R r R

R r I= rArr = minus rArr = Ω

+ 2p

b

Din grafic deducem că icircn momentul icircn care cursorul ajunge icircn B intensitatea curentului icircn circuit are valoa-

rea 071 AI prime = 1p

5p

iar expresia ei este ( )11

1817 18

E EI R R rR R r I

R

prime = rArr = minus + rArrprime+ +

rArr = Ω cong Ω1p

Cursorul se mişcă cu viteză constantă parcurgacircnd icircn intervalul de timp 20 st∆ = 1p

determinat din grafic icircntreaga lungime a reostatului 20 cmlv l v t lt

= rArr = sdot ∆ rArr =∆

1p

2 05 mml lR S SS R

ρ sdot= ρ sdot rArr = rArr = 1p


31

c

9 2 2l Rx Rprime= rArr = = Ω

1p

4p1

1

105 A 11 s

EIR R r

I t

primeprime =prime + +

primeprime = rArr =

2p

1

1

105 A 11 s

EIR R r

I t


primeprime = rArr = 1p

dI = 0 1p

3pU = E 1pU = 20 V 1p



a = 2400 mA h = 8640 C q sdot 2p

4p10368 J W q U W= sdot rArr = 2p

b1

06 AUI IR

= rArr = 3p 3p

c 14400 s = 4 hq qI t tt I

= rArr ∆ = rArr ∆ =∆

3p 3p

d

După o oră de funcţionare a circuitului de la punctul b s-a consumat sarcina electrică 1 1 1 2160 Cq I t q= sdot ∆ rArr =Sarcina electrică rămasă va fi 2 1 2 6480 Cq q q q= minus rArr =

1p

5p

Noua intensitate a curentului electric prin ramura principală a circuitului devine 1 Ap

UI IR

prime prime= rArr = 1p

Iar durata funcţionării acestuia va fi2 2

2 22

6480 s = 18 hq qI t tt I

prime = rArr ∆ = rArr ∆ =prime∆

1p

2 22

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =

sdot sdot ∆= rArr = rArr = sdot

∆

1p2 2

2

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆ 1p



32


1

b Din desen se observă că doar punctul 2 poate fi unit prin reflexie cu O

3

2 c 3

3

d Pentru a nu produce efect fotoelectric pentru orice radiaţie din domeniul vizibil trebuie ca pentru materialul respectiv

extrL ge ε ε fiind energia fotonului incident h csdotε =

λPentru domeniul vizibil [ ]380760 nmλ isin [ ] [ ]19 262 523 10 J sau 163 327 eVminusε isin sdot ε isin Folosind datele din tabelul cu valorile lucrului de extracţie pentru diferite materiale se observă că doar cuprul nu produce efect fotoelectric

3

4c Prin introducerea dispozitivului icircn apă fără a schimba caracteristicile geometrice ale acestuia şi folosind aceeaşi radia-

ţie interfranja devine 2

D iil n n

λ sdotprime = =sdot

3

5 b Scriind ecuaţia lui Einstein pentru situaţiile descrise icircn problemă obţinem 0

0

2 4

s

s

h c h c e U

h c h c e U

sdot sdot = + sdot λ λ sdot sdot sdot = + sdot λ λ

Rezolvacircnd sistemul de ecuaţii obţinem 0 3λ = sdot λ

3



a

22 1

12

1

2 4 cm

2 2 cm

xx x

xx

x

β = = sdot rArr = minus β = rArr = minus

2p3p

rArr x1 = minus 2 cm 1p

b

2 1

1 1 1

1

25 4 cm

x x f

Cf

Cf

minus = =

rArr = δrArr =

2p4p

rArr C = 25 δ 1prArr f = 4 cm 1p

c

2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx

minus = prime prime primerArr = minus = minus sdotprime prime= minus

prime = minus

2p

4p2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus 1p

Lupa trebuie icircndepărtată de obiect cu 6 cm 1p


33

d

Imagine dreaptă 1 20 0x xrArr β gt rArr lt rArr imaginea este virtuală 1p

4p

2

1 2 1

1 1

2 1

22

8 cm 1875 1 1

xx x x

x x C

C Cx x

primeprime β = primeprime primeprime primeprimerArr = sdot β = primeprime prime prime= = minus rArr = minus δ

primeminus = +

primeprime primeprime

lentila este divergentă

3p



a ( )0 0 k x= rArr = forall λ rArr maximul central este alb şi situat pe axa de simetrie a dispozitivului4p 4p

b

1

1

22

kDxD li

l

= λ sdot rArr =λ sdot

sdot= sdot

1p

4pLărgimea maximului de ordinul I va fi( )

2

R V Dx

lλ minus λ sdot

∆ =sdot

2p

076 mmx∆ = 1p

c2

2

2 2 192 mm

2

k x ixDi

l

= rArr = sdot rArr =λ sdot

= sdot 3p 3p

d

1 1 2 2k i k isdot = sdot 1p

4p

1 21 22 2

D Dk kl l

λ sdot λ sdotsdot = sdot

sdot sdot1p

1 2 1 2500 750 2 3 k k k ksdot = sdot rArr sdot = sdot 1p

Soluţia 1

2

32

kk

= =

1p


0Cprime lt rArr


34

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj

1 a 32 b 3

3

a Aplicacircnd principiul I al dinamicii obţinem 0Gt Ff N Gn+ + + =

0 sin cos 0sincos

Gt Ff mg mg

tg

minus = =gt α minus micro α =α

micro = = αα

3

4c L = F sdot d sdot cos α

a = 0 rArr L = F sdotd = 40 J 3

5

c Conform principiului II al dinamicii

N = G + m sdot a rArr 1 1600 NG aN G a Gg g

= + = + =

3



a

Forţele care acţionează asupra corpurilor sunt reprezentate icircn figura alăturată

4p 4p

b

Acceleraţiile corpurilor 1a

şi 2a

transmise prin fir sunt egale icircn modul şi de sensuri opuse 2 1a a= minus

1p

4pproiectăm ecuaţiile pe axa verticală Oy ţinacircnd cont că 1 2a a a= = 1 1

2 2

T m g m aT m g m a

minus sdot = sdot

minus sdot = minus sdot 2p

rezolvacircnd sistemul obţinem 22 1

1 2

( ) 111 m sm m gam m

minusminus sdot= = sdot

+ 1p

c T = m1(a + g) = 2222 N 3p 3p

d

Forţa măsurată de dinamometru

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p

4p

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p



35


a

2p

3p

cos5640 J

L F dL

= sdot α=

1p

b

NFfL Ff d

Ff= minus sdot

= micro

1p

4pN

FfL Ff dFf

= minus sdot

= micro 1p

sinN G F= minus α 1p

( )sin 1036 JFfL mg F d= minusmicro minus α = minus 1p

cAplicăm teorema variaţiei energiei cinetice f i F FfEc Ec L Lminus = + 2p

4p4504 JfEc = 2p

d564 W

med

med

LPt

P

=∆

=

2p4p

564 W

med

med

LPt

P

=∆

= 2p



36


Subiectul I Punctaj

1 c 3

2 a 33 b 34 a 35 c 3



a3096 kgm

pRT

microρ =

ρ =

2p3p

3096 kgm

pRT

microρ =

ρ = 1p

b

a

a

N m N mNN

sdot= rArr =

micro micro1p

4p

Masa gazului care părăseşte butelia m2 = f middot m =gt N2 = N middot f 1p

Din ecuaţia de stare 11

a

a

N p VNp V RT NN RT

= rArr = 1p

2312 139 10 moleculeap VNN f

RT= = sdot 1p

c

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus=

1p

4p

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus= 1p

1

2

11

pp f

=minus

1p

p2 = p1(1 ndash f ) = 96 ∙ 105 Pa 1p

d

Aplicăm conservarea numărului de moli ν1 + ν2 = ν TOT 1p

4p

ν2 = reprezintă numărul de moli din prima butelie icircn starea finală ν1 + ν2 = Totmmicro

1p

21 O 2

1 2

He 3391 10 kgmolminusmicromicro = = sdot

ν + micro ν

ν + ν2p



37


a Procesul 1rarr2 icircncălzire izocorăProcesul 2rarr3 destindere izotermă

Procesul 3rarr1 răcire izobară

3p 3p

b

Aplicacircnd legea procesului izocor pe procesul 1rarr2 obţinem

2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= =

2p

4p2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= = 2p

c9kJ

p

p

LO

LO

η = rArr

= =η

2p

4p

9kJ

p

p

LO

LO

η = rArr

= =η

2p

d

ΔU = νCν(T3 minus T2) 2p

4pT2 = T3 1p

ΔU = 0 1p



38


Subiectul I Punctaj

1 a 32 c 33 c 34 a 35 b 3



a

Pentru gruparea paralel a rezistoarelor rezistenţa echivalentă RP o calculăm din relaţia 1 2

1 1 1

PR R R= + 1p

4pPentru gruparea serie a rezistoarelor rezistenţa echivalentă RS o calculăm din relaţia Rs = R3 + R4 1p

Rezistenţa circuitului exterior R = Rs + Rp 1p

R = 11 Ω 1p

b

Curentul care circulă prin sursa I icircl putem calcula aplicacircnd legea lui Ohm pentru un circuit simplu 1p

4pPentru a calcula I1 I2 aplicăm legile lui Kirchhoff 2p

Rezolvacircnd sistemul de ecuaţii obţinem I1 = 12 A I2 = 08 A şi I3 = I = 2 A 1p

cE = I3r + UAB 2p

3pUAB = 22 V 1p

d

Aplicăm legea lui Ohm pentru o porţiune de circuit 11

1

UIR

= 1p

4pRezultă U1 = I1 R1 1p

Aplicăm legea lui Ohm pentru o porţiune de circuit I2 = 2

2

UR

Rezultă U2 = I2 R2 1p

U1 = U2 = 96 V 1p



aPuterea la bornele becului P = Ub ∙ I 2p

3pUb = 120 V 1p

b

Aplicăm legea lui Ohm pentru o porţiune de circuit (fir) I = Uf r 2p

4pRezultă Uf = I middot r 1p

Uf = 4 V 1p

cExpresia puterii P = I 2 R 2p

4pR = P I 2 1pRezultă R = 12 Ω 1p

d Expresia energiei debitată de o sursă icircn icircntreg circuitul W = I 2(R + r) Δt 3p4p

W = 4464 kJ 1p



39


1 c 32 a 33 d 34 c 35 c 3



aDin definiţia indicelui de refacţie cn

v= unde c este viteza de propagare a luminii icircn vid 2p

3p

Rezultă cvn

= v = 212 ∙108 ms 1p

b

Reprezentarea corectă a razelor de lumină 1p

5pAplicăm legea refracţiei icircn punctul I n sin i = naer sin r 2p

r = 450 1p090 45rθ = minus = 1p

c

Icircn cazul icircn care lumina trece dintr-un mediu optic mai dens icircn unul mai puţin dens unghiul de refracţie este totdeauna mai mare decacirct unghiul de incidenţă şi de aceea el poate atinge valoarea 2

π Valoarea l a unghiu-lui de incidenţă pentru care r =

2π poartă numele de unghi limită

1p

4p1sin sin2

n l n π= 1p

1

2

sin nln

= 1p

l = 450 1p

dPentru orice valoare a unghiului de incidenţă mai mare decacirct unghiul limită lumina nu mai trece icircn mediul al doilea ci se reflectă icircn punctul de incidenţă icircntorcacircndu-se icircn primul mediu conform legilor reflexiei suprafaţa de separare comportacircndu-se ca o oglindă

3p 3p



aConvergenţa unei lentile

1Cf

= 2p3p

C = 5 m-1 1pb Realizarea corectă a desenului 4p 4p

c

2 1

1 1 1x x f

minus = 2p

5px2 = 40 cmx2 gt 0 imagine reală 1p

Expresia măririi liniare 2 2

1 1

x yx y

β = = 1p

y2= y1=10 cm 1p

dPentru un sistem optic centrat cu cele două lentile alipite convergenţa sistemuluiCs = C1 + C2

2p3p

Rezultă Cs=15m-1 1p



40

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj

1 c 32 a 33 c 34 c 35 b 3



a

2p

4p

2p

b

fM a T Fsdot = minus

2m a G Tsdot = minus2p

4pPrin adunare rezultă ( )( )

a M m mg Nmg Mg g m Ma

M m M m

+ = minus microminus micro minus micro

= =+ +

1p

21msa = 1p

c 2 ( )T G ma mg ma m g a= minus = minus = minus 2p3p

09 NT = 1p

d

Precizarea noului sens de acţiune a forţei de frecare sau desenarea ei 1p

5p( )F mg Mg g m M= + micro = + micro 2p

17 NF = 1p



a

Lucrul mecanic consumat este lucrul mecanic minim necesar ridicării corpului pe planul icircnclinat 1p

4pfc u FL L L= + 2p

2500JcL = 1p


41

b

u

c

LL

η = 2p

3p2000 08 802500

η = = = 1p

c

cossin sinfF f

h hL F l N mg= minus sdot = minusmicro sdot sdot = minusmicro α sdotα α

1p

4p

u GL L mgh= = 1p

f

f

FF u

u

LL ctg L

L ctg

minus= minusmicro α sdot rArr micro =

α1p

1 0254

micro = = 1p

d

2 60 40 s3

t∆ = sdot = 1p

4p

u cu c

L LP Pt t

= =∆ ∆

1p

2000 50 W40

2500 625 W40

u

c

P

P

= =

= =2p



42


Subiectul I Punctaj

1 d 32 d 33 b 34 c 35 c 3



a

1 1 1 1 1

2 2 2 2 2

( ) ( 2 )2 2 2

( - ) ( - 2 )2 2 - 2

l l p lpV p V p S p d S p l p l d pl d


sdot= hArr sdot sdot = + hArr sdot = + rArr =

+sdot

= hArr sdot sdot = hArr sdot = rArr =2p

4p5

51

55

2

10 1 5 10 Pa18 9

10 1 5 10 Pa02

p

p

sdot= = sdot

sdot= = sdot

2p

b

F

trebuie să acţioneze spre compartimentul cu presiunea mai mare 1p

3p1 2 2 1( )p S F p S F S p p+ = hArr = minus 1 2 2 1( )p S F p S F S p p+ = hArr = minus 1p

133(3) NF = 1p

c

1 1 1

2 2 2

( 2 ) ( 2 )

pV p V p l p l hpV p V p l p l h

= hArr sdot = += hArr sdot = minus

1 2 2 1mgp S mg p S p pS

prime prime prime prime+ = hArr = +

2p

4p

51

52

510 Pa6625= 10 Pa

6

p

p

prime =

prime2p

d

1

2 p ll hpsdot

+ = 1p

4p1

1 12

p lhp

sdot= minus prime

1p

01m = 10 cmh = 1p

Pistonul se va stabili la 40cm de capătul inferior al cilindrului 1p



43


a Reprezentarea corectă a graficului

3p4p

Notarea corectă a stărilor 1p

b ΔU = 0 pentru că procesul este ciclic 3p 3p

c

212 1 1 1

1

ln ln 600 27 1620 JVQ RT p V eV

= ν = = sdot =

23 3 2 3 2 3 1 1 2 15 5 5( ) ( ) ( ) ( )2 2 2pQ C T T RT RT RT RT V p p= ν minus = ν minus ν = ν minus ν = minus

1p

5p

5 51 11 1 2 2 2 1

2 1

3 10 3 10 PaV Vp V p V p p eV eV

= rArr = = sdot = sdot

3 523

5 2 10 3 10 (1 27) 1500 17 2550J2

Q minus= sdot sdot sdot minus = minus sdot = minus2p

31 1 3 1 3 1 1 2 1 1 1 23 3 3( ) ( ) ( ) ( )2 2 2VQ C T T RT RT p V p V V p p= ν minus = ν minus ν = minus = minus 31 1530 JQ = 1p

1620 2550 1530 600 JQ = minus + = 1p

d

U Q L∆ = minus 1p

3pL Q= 1p

600JL = 1p



44


Subiectul I Punctaj

1 a 32 a 33 d 34 b 35 d 3



a

1 1 1 1 1AU I R I= rArr =

2 2 2 2 075AU I R I= rArr =2p

4p( )1 1 2 1 1 2

2 2 1 2

( )6V

( )E I R r R R I I

EE I R r I I

= + minusrArr = = = + minus

2p

b

1 11 1 2 2

2 2

( )( ) ( )

( )E I R r

I R r I R rE I R r

= +rArr + = + = +

2p

4p2 2 1 1 2 1

1 2 1 2

I R I R U UrI I I I

minus minus= =

minus minus 1p

2r = Ω 1p

c

66

15012 10 603 10fir

lRS

minusminus= ρ sdot = sdot = Ω

sdot2p

4p6 075A 075 60 45C8fir

EI q I tR r

= = = = sdot ∆ = sdot =+ 2p

d0

00 0

1(1 ) 1 1

fir

fir firfir

RR R RR R t t t tR R

minus= + α rArr + α = α = minus rArr =

α2p

3p

40 Ct = 1p



a

ss

s

EIR r

=+ 1p

4p10sE E= 1p

10sr r= 1p

4 133 A3sI = = 1p


45

b

pp

p

EI

R r=

+ 1p

4p10prr = 1p

10

110p

ErE E

r

= = 1p

05ApI = 1p

c

1 sU E I r= minus 1p

4p2 pU I R= 1p

1 1 VU

1p

2 35VU = 1p

d1

s

RR r

η =+

2p

RR r

η =+

1p

3p

17 259227

η = 235 972236

η =

2p



46


1 d 32 c 33 c 34 c 35 a 3



a 4p 4p

b

sin sini n r= 1p

4psin sin n r i= 1p

alterne interner r= 1p

Din relaţiile anterioare rezultă i = iprime = 60deg 1p

c

Deviaţia δ = BC 1p

4p

sin sin

3sin 12sin

23

i n r

irn

=

= = =1p

30r = deg 1p

sin( ) sin( ) 058 cmcosADBC AB i r i r

r= sdot minus = sdot minus 1p

d cosADAB

r= 2p

3p

115cmAB 1p



47


a

maxc stoparech L E h L eUν = + hArr = +λ

1

2

1

2

s

s

ch L eU

ch L eU

= +λ

= +λ

2p

5p

Prin scăderea celor două ecuaţii rezultă 2 1

2

2 1

1 1 2

1 2

Prin scatildederea celor douatilde ecuathornii rezultatilde1 1( ) ( )

( )( )

s s

s s

hc e U U

e U Uh

c

minus = minusλ λ

minus λ λ=

λ minus λ

2p

3466 10 J sh minus= sdot sdot 1p

b1

1s

cL h eU= minusλ 2p

3p1961 10 J 38 eVL minus= sdot = 1p

c0 0

Lh Lh

ν = hArr ν = 2p3p

150 0924 10 Hzν = sdot 1p

d

2max

2smveU = 1p

4pmax2 seUv

m= 1p

61 max 048 10 msv = sdot 1p




48

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj

1 b Reprezentăm grafic forţa F = f (x)

0 m 8 N4 m 0 N

x Fx F

= rArr == rArr =

Lucrul mecanic este egal cu aria trapezului format de graficul forţei icircntre 1x şi 2x

(6 2) 2 8 J

2L + sdot

= =

3

2 c Aplicăm legea conservării energiei mecanice2 2 2

22 2i f

kx mv kxE E vm

= rArr = rArr = = 3

3

a Viteza medie mdvt

= (1)

Notăm cu 2v v= şi 1v nv=

Avem 1 1 12 2d dv t nv t= sdot rArr = sdot şi 2 2 22 2

d dv t v t= sdot rArr = sdot

Rezultă timpii de mişcare 1 2dtnv

= şi 2 2dtv

=Iar timpul total 1 2t t t= +

Icircnlocuind icircn (1) rezultă ( )12 36

1 22 2

mm m

n vd nvv v vd d n nnv v

+= rArr = rArr = =

++km h

Rezultă 1 2 72v v= = km h şi 2 36v v= = km h

3

4 a 35 d 3



a

Distanţa parcursă este egală cu aria de sub graficul vitezei d = d1 +d2

1km20h

v =

2km40h

v =

1130min h2

t = =

2115min h4

t = = 2p4p

d1 = 10 km d2 = 10 km 1pRezultă d = 20 km 1p

b

Viteza medie mdvt

= 2p

4pUnde d = 20 km 345min h4

t = = 1p

Rezultă vm = 266 kmh 1p

ct1 = 30 min d1 = 10 km t2 = 15 min d2 = 10 km 2p

4pReprezentarea graficului mişcării 2p

d

Energiile cinetice corespunzătoare celor două viteze sunt 1

21

2cmvE =

2

22

2cmvE =

Rezultă 2

1

222

1

c

c

E vE v

=2p

3p

2

1

4c

c

EE

= 1p



49


a

Reprezentarea corectă a forţelor 1p

4pTeorema variaţiei energiei cinetice pe planul icircnclinat cE L∆ = rArr 11fCB co G F NE E L L Lminus = + +

Lungimea planului icircnclinat sin

hl =α

iar 1

0NL =

Rezultă ECB = mgh minus μmgl cos α = mgh (1 minus ctg α)

2p

ECB = 32 J 1p

bLFf

= LFf1 + LFf2

=μmg(h ctg α + d ) 2p3p

Rezultă LFf = minus 20 J 1p

c

Teorema variaţiei energiei cinetice pe planul orizontal22fCC CB G F NE E L L Lminus = + +

1p

4p2

0NL = şi 0GL = rezultă 20 JCC CBE E mgd= minus micro = 1p

Din legea conservării energiei mecanice rezultă maxCCE mgh= 1p

Rezultă max 1 mCCEhmg

= = 1p

d

Aplicăm legea conservării energiei cinetice icircntre punctele C şi D CC PC CD PDE E E E+ = + 1p

4pdar EPC = 0 iar 4PD

CDEE = 1 1

15

4 4CCmgh mghE mgh= + = 2p

Rezultă h1 = 08 m 1p



50


Subiectul I Punctaj

1 b 32 c 33 b 3

4b Scriem formula variaţiei energiei interne 2 1 2 1 2 1

3 3( ) ( ) ( )2 2VU C T T R T T RT RT∆ = ν minus = ν minus = ν minus ν

Din ecuaţiile de stare p1V1 = νRT1 şi p2V2 = νRT2 rezultă 2 2 1 13 ( ) 600 J2

U p V p V∆ = minus =

3

5 c 3



a

Relaţia dintre temperaturi este 2 1 1T T fT= + 2 1 1(1 ) 12 360KT T f TrArr = + = = 2p

4pDin legea procesului izocor 1 22 1

1 2

12p p p pT T

= rArr = 1p

2p = 5

2

N12 10m

sdot 1p

bLegea procesului izocor 1 max max 1

max1 max 1

p p p TTT T p

= rArr = 2p4p

max 12 600KT T= = 2p

c

Relaţia dintre densitatea gazului icircn starea (1) şi densitatea gazului icircn condiţii normale0 1 0

1 00 1

91p Tp T

ρρ = = ρ 2p

3p

31 0819kgmρ = 1p

d

Din ecuaţiile de stare 11 1

mp V RT=micro

şi 21 2

mp V RT=micro 1 1 2 2m RT m RTrArr = 1p

4p2 1m m m= minus ∆ rezultă 1 1 1 2( )m T m m T= minus ∆ 1p

12 112

6mT T m= rArr ∆ = 1p

3 31 1 819 10 kg 1365 10 kgm V mminus minus= ρ = sdot rArr ∆ = sdot 1p



a

1rarr2 proces izoterm 2rarr3 proces izobar 3rarr1 proces izocor 1p

3p

2p


51

b

Legea procesului izoterm 1 1 2 2p V p V= dar 2 12p p= 12 2

VVrArr = 1p

4pLegea procesului izobar 2 3

2 3

V VT T

= cum 2 1T T= şi 3 1V V= 3 12T TrArr = 1p

Ecuaţia de stare pentru starea iniţială 1 1 1 1 1444Kp V RT T= υ rArr = 1p

2 12 2888KT T= = 1p

c1 3 3 1( )vU C T Trarr∆ = υ minus

3

1 2p

vv

C RC RC

γ = rArr = =γ minus 2p

4p

1 3 1 1 1 1 13 3 3(2 ) 1800 J2 2 2

U R T T RT p Vrarr∆ = υ minus = υ = = 2p

d

2

1

| |1 QQ

η = minus 1p

4p

1 2 3 3 2( )5

1 2p

Q Q Cp T TRC R

rarr= = υ minusγ

= =γ minus

Rezultă 1 1 1 15 252

Q RT p V= υ =

1p

12 1 2 3 1 1 3 1 1 1 1 1 1 1

2

3| | | | ln ( ) ln 2 (ln 2 15) 21932v

VQ Q Q RT C T T RT RT p V p VV

υ υ υ υrarr rarr= + = + minus = + = + = 1p

122η = 1p



52


Subiectul I Punctaj

1 c 32 c 33 d 34 d 3

5 c Avem relaţia dintre tensiuni E = U + IrDin grafic I = 0 A rezultă E = U = 6 V U = 0 V I = 4 A rezultă r = 075 Ω 3



a

Din grafic P1= 80 W I1 = 4A 1p

4pUtilizacircnd formula P1= I12 R1 rezultă 1

1 21

PRI

= 2p

R1 = 5 Ω 1p

b

Prin legarea celor două rezistenţe icircn serie rezultă rezistenţa echivalentă a circuitului exterior Re = R1 + R2= 8 Ω 2p

4pAplicacircnd legea lui Ohm pe icircntregul circuit determinăm intensitatea electrică 2Ae

EIR r

= =+

1p

Puterea electrică pe R2 este P2= I2 R2 = 12 W 1p

cRandamentul circuitului este e

e

RR r

η =+

2p3p

80η = 1p

d

Condiţia ca sursa să debiteze o putere maximă este Re = rprime rArr rprime = 8 Ω 2p

4pPuterea maximă

2

max 125 W4 EPr

= = 2p



a

Reprezentarea circuitul electric echivalent 1p

5p

Rezistenţa echivalentă a circuitului exterior este 2 1

2

V

p

V

RRR RR

= = Ω+

32e pRR R= + = Ω 2p

Intensitatea curentului electric prin ramura principală a circuitului este 5Ae

EIR r

= =+

1p

Tensiunea indicată de voltmetru UV = IRp UV = 5 V 1p

b Puterea electrică pe circuitul exterior P = I2Re 2p

3pP = 75 W 1p

cPuterea totală a sursei PE = EI 2p

3pPE = 100 W 1p

dRandamentul circuitului e

e

RR r

η =+ 2p

4pη = 75 2p



53


1 c Relaţia dintre energia cinetică maximă a electronilor emişi şi tensiunea de stopare este

2max

max max2

2S

C S SmV eUE eU eU V

m= rArr = rArr = 3

2 b 33 a 34 c 3

5

a Determinăm convergenţa lentilei subţiri biconcave ţinacircnd cont de semnele razelor de curbură

1

2

R RR R

= minus= +

1

1 2

1 1 1 1 2( 1)( 1)( ) ( 1)( ) 5mnC n nR R R R R

minusminusrArr = minus minus = minus minus minus = minus = minus 3



aDeoarece avem o imagine reală rezultă că avem o lentilă convergentă 1p

3pRealizarea corectă a mersului razelor pentru construcţia imaginii prin lentilă 2p

bMărirea liniară transversală este 4β = minus 1p

3pDin 2

2 1 21

4 20cmy y y yy

β = rArr = minus rArr = minus 2p

c

Din reprezentarea grafică observăm1 2d x x= minus + 1p

4pCum 2

2 1 11

4 4 5x x x d xx

β = = minus rArr = minus rArr = minus 2p

1

2 1

16cm54 64cm

dx

x x

= minus = minus

= minus =1p

d

Utilizăm formula lentilelor subţiri 1 2

2 1 1 2

1 1 1 ( 16) 64 128 cm16 64

x xfx x f x x

minus sdotminus = rArr = = =

minus minus minus2p

5pDistanţa focală pentru lentila biconvexă este 1

1 1 2( 1)( 1) ( )

Rfnn

R R

= =minus minus minus minus

2p

2( 1) 128 cmR n f= minus = 1p



54


a

Reprezentarea mersului razei de lumină

3p 3p

b

302

i π= minus α = deg 1p

3pLegea refracţiei icircn punctul de incidenţă aer( 1)n = sin sini n r= 1p

sinsin irn

rArr =

3sin8

r = 1p

c

Mărimea umbrei este x OI AB= + 1p

5p

tg hOI

α = rArr 1038mtg 3

h hOI = = =α 1p

Icircn triunghiul IAB tg ABrH

= rArr2

sintg 0809m1 sin

rAB H r Hr

= sdot = =minus

2p

Rezultă 1847mx = 1p

d

Reprezentarea razelor SI = raza incidentăIR = raza refractatăIRprime = raza reflectată 1p

4p

Deoarece raza IRprime este perpendiculară pe IR rArr 90 darr r i rprime primerArr + = deg = rArr 90r i= deg minus 1p

Aplicacircnd legea refracţiei aer( 1)n = sin sin sin sin (90 )4sin (90 ) cos tg 1333

53

i n r i n i

i i i n

i

= rArr = minus

deg minus = rArr = = =

rArr =

2p



55

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj

1 a 32 d 33 a 34 b 35 a 3



aG1 = m1g 1p

3pG2 = m2g 1prezultat final G1 = 2 N G2 = 3 N 1p

b

Aplicacircnd principiile mecanicii clasice pentru cele două corpuri se obţin relaţiile m2g minus T = m2a 1p

4pT minus m1g = m1a 1p

Adunacircnd relaţiile anterioare se obţine ( )2 1

1 2

g m ma

m mminus

=+

1p

rezultat final a = 2 ms2 1p

c

Din relaţiile de la punctul anterior se poate scrie T = m1(a + g) 1p

4p1 2

1 2

2m m gTm m

=+

2p

rezultat finalT = 24 N 1p

d

eF k l= sdot ∆ 2p

4p2Tlk

∆ = 1p

rezultat final 48 cml∆ = 1p



a

LFf = minus Ff ∙ l 1p

3pFf = μN = μmg cos α cos xl

α = 2 2x l h= minus 1p

rezultat final 2 2fFL mg l h= minusmicro minus 2700J

fFL = minus 1p

b

totalcE L∆ = 1p

4p

2

2cmvE∆ =

total fG FL L L= + GL mgh= 1p

( )2 cosv g h l= minus micro α 1p

rezultat final 6 5 134 msv m s=

1p


56

c

util

consumat

LL

η = utilL mgh= consumatL F l= sdot 1p

4pcos sinf tF F G mg mg= + = micro α + α 1p

11 ctg

η =+ micro α

ctg xh

α = 1p

rezultat final 714η = 1p

d

c totalE L∆ = 1p

4p

2

2cmvE∆ = minus

total fFL L= 1p

2

2mv mgd= micro

2

2vd

g=

micro 1p

rezultat final 30md = 1p



57


Subiectul I Punctaj

1 d 32 d 33 d 34 a 35 c 3



a

1 1 1p V RT= ν 1p

3p

11

1

RTVp

ν=

1p

3 31 831 10 mV minussdot 1p

b

Legea procesului izobar 1 2

1 2

V VT T

= 2p

4p1 22

1

TVTV

=

1p

rezultat final 2 360KT = 1p

c

2 1∆ρ = ρ minus ρ 1p

4p11

mV

ρ =

22

mV

ρ = 2p

rezultat final 3064 kgm∆ρ = minus 1p

d

( )12 1 2 1L p V R T T= ∆ = υ minus 1p

4p( )23 12 1 2VL U C T T= minus∆ = minusυ minus

52VC R= 1p

( )12 23 2 172totalL L L R T T= + = υ minus 1p

rezultat final total 1750 JL = 1p



58


a

11

NnV

= 1p

3p1 1 1p V RT= ν A

NN

υ = 1p

rezultat final 11

1

Ap NnRT

= 24 31 24 10n mminussdot 1p

b

123 3 1U U U∆ = minus 1p

4p1 1VU C T= ν

3 3VU C T= ν 1p

( )123 3 1VU C T T∆ = ν minus 1p

rezultat final 7123 125 10 JU∆ sdot 1p

c

43-4 3

3

ln VL RTV

= ν 1p

4pLegea procesului izoterm 3 3 4 4p V p V= 1p

33-4 3

4

ln pL RTp

= ν 34 3 ln 2L RT= ν 1p

rezultat final 734 046 10 JL sdot

1p

d

cedat 4-1Q Q= 1p

4p4-1 1 4( )pQ C T T= ν minus

1p

4p VC C R R= + = 1p

rezultat final cedat 1 44 ( )Q R T T= ν minus 7cedat 17 10 JQ minus sdot 1p



59


Subiectul I Punctaj

1 d 32 a 33 d 34 a 35 d 3



a

0

0p

R RRR R

=+

2p

4pe pR R R= + 1p

rezultat final 9eR = Ω 1p

b

Intensitatea curentului electric prin acumulator este e

EIR r

=+

1p

4pAplicacircnd teorema a doua a lui Kirchhoff pe un ochi de reţea se obţine ( ) 0AE I r R I R= + + 1p

( )0

A

E I r RI

Rminus +

= 1p

rezultat final 04 AAI = 1p

c

V eU R I= 1p

4pu r I= sdot 1p

rezultat final 18 VVU = 2Vu = 2p

d

A

EIr

= 1p

3p 0VU = 1p




a

21 1 1P R I= sdot

22 2 2P R I= sdot 1p

4p1

1

EIR r

=+

22

EIR r

=+

1p

Egalacircnd cele două puteri se obţine 21 2R R rsdot = 1p

rezultat final 6r = Ω 1p


60

b scEIr

=

2p3p

rezultat final E = 120 V 1p

c

ext s

totala s

P RP R r

η = =+

2p

4p1 2sR R R= + 1p

rezultat final 13 68419

η = = 1p

d rEP4

2

max =

2p

4pR = r 1prezultat final Pmax = 600 W R = 6 Ω 1p



61


1 a 32 d 33 c 34 b 35 b 3



a1

1

1Cf

=

11

1fC

=

2p3p

rezultat final f1 = 25 cm 1p

breprezentarea grafică a imaginii 2p

4prezultat final imaginea este reală răsturnată şi mai mare decacirct obiectul 2p

c

112

111fxx

=minus

1p

4p2 2

1 1

y xy x

β = =

1p

rezultat final x2 = 150 cm y2 = minus10 cm 2p

d

1 2

1 1 1

sf f f= + 1p

4p2 1

1 1 1

sx x fminus = 1p

rezultat final x2prime = minus75 cm 2p



a reprezentarea graficului 3p 3p

b c exth E Lν = + 2p4p

rezultat final 19332 10 JextL minus= sdot 2p

c0extL h= ν 2p

4prezultat final 14

0 5 10 Hzν = sdot 2p

dc sE e U= sdot 2p

4p

rezultat final Us = 04 V 2p



62

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj

1 a 0

0

FllE s

∆ =sdot

3

2 c 2

m4 ms 22 s s

va at

∆= rArr = =

∆2 NF m a F= sdot rArr =

3

3 b

[ ] 1 N m 1 JL F dL= sdot

= sdot =

3

4d

355 10 2750 N20

PP F v Fv

F

= sdot rArr =

sdot= =

3

5b Condiţia de deplasare cu viteză constantă impune

cos ( sin )cos sin

mgF mg F F microα = micro minus α rArr =

α + micro α 3



a

x24 = Aria subgraficului [24] 1p

3px24 = 3 ∙ (4 minus 2) 1p

x24 = 6 m 1p

b

2

m(3 1) s(2 1) s

m2s

va at

a

∆ minus= rArr = rArr

∆ minus

=

2p

4p

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

= 2p

c

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms

(3 1)1Aria trapez 2 m2

3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +

= sdot ∆ rArr = sdot =

+= rArr = =

= sdot ∆ rArr = sdot ==

1p

5p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1ptotal 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =

= sdot ∆ rArr = sdot == 2p

d

total

total

9 m2254 s

m

m

xvA

v

=

= =

2p

3ptotal

total

9 m2254 s

m

m

xvA

v

=

= = 1p



63


a

Consideracircnd nivelul 0 de energie potenţială gravitaţională la baza planului icircnclinat EpA = 0 1p

3p2

0 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

=

1p20 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

= 1p

b

Aplicacircnd teorema de variaţie a energiei mecanice rezultă 2

0

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L

minus = minusmicro sdot αα

rArr =micro α + α

=

minus = 1p

4p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

c 056B

B

p

p

E mgh

E J

=

=

2p4p

056B

B

p

p

E mgh

E J

=

= 2p

d

Potrivit teoremei de variaţie a energiei mecanice L = ∆Ec rezultă 1p

4p024 J

f AB

f AB

pF B A

pF

L E E

L

= minus

= minus

2p

024 Jf AB

f AB

pF B A

pF

L E E

L

= minus

= minus 1p



64


Subiectul I Punctaj

1 c Icircn procese izoterme U = const 3

2 d [ ]SI

J1mol K

C =sdot

3

3c 2

1

ln

1929 J

VL RTV

L

= ν

= minus

3

4 b Ordonata p Abscisa V 3

5 a A

NN

ν = 3



a

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

2p

4p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

1p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot 1p

b0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

c

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

2p

4p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

1p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr = 1p

d

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν

ν = = rArr ν =micro

νrArr =

=

1p

4p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

= 1p



65


a

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

4p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

= 1p

b

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

2p

4p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

1p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr = 1p

c

L12341 = Aria12341 1p

4p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot 1p

d( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot

2p3p( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot 1p



66


Subiectul I Punctaj

1 c Rezistenţa electrică a unui conductor este direct proporţională cu lungimea sa 3

2 a [ρ]SI = 1 Ω m 3

3 c q = I Δt 3

4 b 0 0 00

112 3 6eR R RR R= + + = 3

5 d 6 2003

UR RI

= rArr = = Ω 3



a2 3

12 3

11

e

e

R RR RR R

R

= ++

= Ω

2p3p

2 31

2 3

11

e

e

R RR RR R

R

= ++

= Ω 1p

b

Potrivit legilor Kirchhoff I = I2 + I3 1p

4p

I2R2 = I3R 3rArr I3 = I2

2

3

RR

1p

rArr 22

3

1 R

I RI =

+

1p

rArr 32

2 3

IRIR R

=+ rArr I = 04 A 1p

c

e

EIr R

= rArr+ 2p

4pe

Er RI

= minus 1p

rArr 15 11 4r = minus = Ω 1p

d

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

2p

4p

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

1p2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =1p



67


a

Din legea Joule

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

3p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr = 1p

b

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1p

4p

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I

= +rArr = minus rArr =

1 2

2 1 2

Din 3 A

I I II I I I

= +rArr = minus rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I


c

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

2p

4p

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

1p2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω 1p

d

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

2p

4p

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr = 2p



68


1 c 1

SI

m ms s

cminus

= = ν sdot 3

2 d Nr fotoelectronilor leneşi ~ frecvenţa 3

3 c Negativă 3

4 d Obiectul trebuie aşezat icircnaintea focarului obiect pentru a obţine imaginea reală pe ecran dmin = 4 m

3

5 a sin 2 141sin

in nr

= rArr = = 3



a 3p 3p

b

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

2p

4p

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p

c

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =

rArr = sdot =sdot minus

2p

4p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

d

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

4p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1psistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr = 1p



69


a

00

814

0 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot

2p

3p0

08

140 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot1p

b 8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

2p

3p8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

1p

c

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ

sdot sdot sdotrArr =

sdot= sdot = sdot

1p

4p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot 1p

d

Din ecuaţia Einstein

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =

sdot sdot sdot minus sdot=

sdotrArr =

2p

5p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr = 1p



70

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj

1 a 32 a 33 b 34 c 35 d 3



a

Pentru reprezentarea corectă a forţelor ce acţionează asupra lui m1

2p4p

Pentru reprezentarea corectă a forţelor ce acţionează asupra lui m2 2p

b

Aplicacircnd principiul al doilea al dinamicii( ) 2 2 2 2

1 1 1 1

1

(2) ( ) ( )

G T m a G T m a

G G T m m a G G T m m a

+ = rArr minus + =

+ + = + rArr + minus = +

1p

4p

Corpul de masă ( )1 m m+ coboară accelerat iar corpul de masă 2m urcă Din ecuaţiile (1) şi (2) se obţine acceleraţia

1 2 2 1( ) ( )m m m a m m m g+ + = minus + rArr 2 1

1 2 1

( )2

m m m g mgam m m m m

minus += =

+ + +

1p

respectiv tensiunea icircn fir2 1

2 1 1 21 2

1 ( ) ( ) ( )2

m m m gT m m m m m m gm m m

minus += minus minus + + +

+ +rArr 1 1

1

2 ( )2

m m m gTm m

+=

+ 1p

rezultat final 2

m02s

a = 25NT = 1p

cForţa de apăsare asupra axului scripetelui se observă din reprezentarea grafică 1 1

1

4 ( )22

m m m gF Tm m

+= =

+ 2p

3p

rezultat final 5NF = 1p

d

Scriind principiul al doilea al dinamicii pentru corpul de masă m

( )G N ma G N ma mg N maN m g a

+ = rArr minus = rArr minus == minus

2p

4p1

1

22

m mgNm m

=+

1p

rezultat final 298 10 NN minus= sdot

1p


71



a

Se aplică principiul al doilea al dinamicii pentru a se determina acceleraţia corpului după care din ecuaţia de mişcare se determină icircnălţimea la care ajunge corpul după 6 s

FG F ma G F ma mg F ma a gm

+ = rArr minus + = rArr minus + = rArr = minus

2p

4p2 2

90 m2 2

at F th gm

= = minus =

1p

pE mgh= rezultat final 4500 JpE =

1p

b

Se calculează spaţiul parcurs după icircncetarea forţei apoi icircnălţimea maximă atinsă de corp aplicacircnd ecuaţiile de mişcare

22

2

45 m2

2

FF g tv g t mm h hg

v gh

minus = minus rArr = rArr = =

2p

4p

max 135 mh h h= + = max maxpE mgh= 1p

rezultat final max 6750 JpE = 1p

c

Aplicacircnd conservarea energiei se obţine 2

max max 2p cmvE E mgh= rArr = 1p

3pmax2v gh= 1p

rezultat final 52 msv = 1p

d

Din momentul icircncetării acţiunii forţei şi pacircnă se atinge icircnălţimea maximă greutatea face lucru mecanic rezis-

tent apoi pacircnă la sol face lucrul mecanic motor 1p

4p2 maxL mgh= 1p

maxL mgh mgh= minus + 1p

rezultat final 4500 JL = 1p



72


Subiectul I Punctaj

1 c 32 a 33 d 34 c 35 b 3



a

Din ecuaţia de stare se obţine

11

1

AA

A

pV RTNpV RTm NN

NN RTVpN

= ν rArr =ν = = micro

=

3p4p

rezultat final 3 37479 10 mV minus= sdot 1p

b

Din ecuaţia de stare se obţine1 1

11 A A

m N NmN N

microν = = rArr =

micro

1p2p

rezultat final 84 gm = 1p

c

Pentru fiecare gaz se scrie numărul de moli ca fiind 1

1A

NN

ν = 2 12 1

2 2A A

N NN N

ν = = = ν respectiv 3 13 1

3 3A A

N NN N

ν = = = ν 3p

5pMasa molară a amestecului se calculează cu relaţia 1 1 2 2 3 3 1 2 3

1 2 3

2 36

amestecamestec

total

m micro ν + micro ν + micro ν micro + micro + micromicro = = =

ν ν + ν + ν 1p

rezultat final 3533 gmolamestecmicro = 1p

d

Numărul de moli este 1 2 3 16totalν = ν + ν + ν = ν 1p

4piar volumul recipientului este 1RTV

pν

= 1p

Din ecuaţia de stare 1 ( )totalp V R T T= ν + ∆ se obţine prin icircnlocuire 1( ) 6 ( )total R T T p T TpV T

ν + ∆ + ∆= = 1p

rezultat final 5 21 2 10 Nmp = sdot 1p



73


a

Pentru reprezentare corectă

3p 3p

b

Căldura primită pe procesul izocor 1 2 2 1( )V VQ C T C T Tminus = ν ∆ = ν minus 1p

4p

Se determină din ecuaţia de stare numărul de moli 1 11 1 1

1

p Vp V RTRT

= ν rArr ν = 1p

Din ecuaţia procesului izocor unde se cunoaşte 2 13p p= se determină temperatura T21 2

2 11 2

3p p T TT T

= rArr = 1p

rezultat final 1 2 1 13 12000JQ p Vminus = = 1p

c

Pentru a determina lucrul mecanic pe icircntreg ciclul se aplică primul principiu al termodinamicii U Q L∆ = minus şi se ţine cont de faptul că icircntr-un proces ciclic 0U∆ = Aşadar 1 2 2 3 3 4 4 1L Q Q Q Qminus minus minus minus= + + + 1p

5p

Pentru procesul izoterm din ecuaţia procesului se determină V3

2 22 2 3 3 3 2

3

32

p Vp V p V V Vp

= rArr = =

1p

după care se calculează căldura schimbată de gaz cu mediul ca fiind 3

2 3 2 1 12

3ln 3 ln 4800 J2

VQ RT p VVminus = ν = = 1p

Temperatura icircn starea 4 este 4 34 1

3

32

p TT Tp

= = iar căldura schimbată de gaz cu mediul icircn procesul izo-

cor este 3 4 4 3( ) 9000 JVQ C T Tminus = ν minus = minus

1p

Icircn procesul 4rarr1 4 1 1 4( ) 5000 JpQ C T Tminus = ν minus = minus rezultat final 2800 JL =

1p

dDin relaţia de definiţie a randamentului

1 2 2 3primit

L LQ Q Qminus minus

η = =+

1p3p




74


Subiectul I Punctaj

1 a 32 b 33 b 34 c 35 d 3



aDin formula de definiţie a dacă intensităţii curentului de scurtcircuit sc

sc

E EI rr I

= rArr = 2p3p


b

Rezistoarele R3 şi R4 sunt legate icircn serie 34 3 4R R R= + 1p

4p

Gruparea este icircn paralel cu R2 avacircnd rezistenţa echivalentă 2 3 4234

2 3 4

( )R R RRR R R

+=

+ + 1p

Cele trei rezistoare sunt icircnseriate cu R1 şi R5

Rezistenţa echivalentă a grupării de rezistoare este 2 3 41 5

2 3 4

( )e

R R RR R RR R R

+= + +

+ +1p


c

Din legea lui Ohm se obţine curentul prin circuit e

EIR r

=+

şi tensiunea la bornele grupării de rezistoare

eU IR= 1p

4p

Rezistoarele R1 şi R5 sunt străbătute de acelaşi curent I dar au tensiuni diferite la borne U1 respectiv U5 Ten-siunea la bornele rezistorului R2 este egală cu tensiunea la bornele grupării de rezistoare R3 şi R4 curenţii pe ramuri fiind icircnsă diferiţi suma lor fiind I I I+ =

1p

2 2 3 4 ( )U I R I R R= = +

1 2 5 1 5 2( ) U U U U I R R I R= + + = + + 1p

1 5

2

( ) eI R R RIR

minus minus= rezultat final 075 AI = 1p

d

Curentul pe ramura ce conţine rezistorul R3 este I I I= minus 1p

4piar tensiunea la bornele lui va fi 3 3U I R= 2p

rezultat final 3 1 VU = 1p



75


a

Căldurile disipate pe cei doi rezistori sunt 1 1W UI t= respectiv 2 2W UI t= Cum energia totală reprezintă suma dintre energiile disipate pe cele două rezistoare şi se cunoaşte raportul

acestor energii se poate scrie 1 2 1 133

WW W W W W= + = rArr =

2p

4p

Prin icircnlocuire se obţine 11 3

W WIUt Ut

= = şi 22

23

W WIUt Ut

= =

rezultat final 1 067AI = şi 2 133 AI =

2p

b

Din legea lui Ohm 1 1 2 2U R I R I= = se determină valorile rezistenţelor şi 11

18URI

= Ω 22

9URI

= Ω 2p

4pCele două rezistoare fiind legate icircn paralel rezistenţa echivalentă se calculează cu relaţia 1 2

1 2p

R RRR R

=+

1p

rezultat final 6pR = Ω 1p

cPuterile consumate icircn circuitul exterior vor fi 2 1

1 1 1WP R It

= = 2 21 2 2

WP R It

= = 2p4p

rezultat final 1 8 WP = şi 1 16 WP = 2p

dRandamentul circuitului se poate calcula fie cu relaţia

p

p

RR r

η =+ fie cu putil

consumat

R EPP EI

η = = 2p3p




76


1 d 32 c 33 a 34 d 35 a 3



a

Imaginea fiind micşorată şi reală 2 2 22 1

1 1 1

0 4 4y x x x xy x x

β = = lt rArr = minus rArr = minus

Distanţa obiect ndashecran este 1 2d x x= minus +

2p4p

rezultat final 1 25 cmx = minus şi 2 100 cmx = 2p

b

Convergenţa sistemului se determină ca suma convergenţelor celor două lentile

Pentru determinarea convergenţei lentilei L1 se scrie legea lentilelor 11 2 1

1 1 1 5Cf x x

= = minus = δ 1p

3p

Convergenţa sistemului format de cele două lentile este 1 2C C C= + 1p

rezultat final 2C = δ 1p

c

Dacă lentilele sunt alipite lentila echivalentă are distanţa focală 1 50 cmFC

= = 1p

4pUtilizacircnd formula lentilelor

2 1

1 1 1F x x

= minus 1p

se obţine poziţia imaginii finale date de sistem 12

1

FxxF x

=+

1p

rezultat final Imaginea finală este reală 2 166 cmx = 1p

d

Dacă a doua lentilă va avea convergenţa 2 3C = δ atunci distanţa focală a sistemului devine 1 125 cm

FC

= = 1p

4piar imaginea se va forma la distanţa 12

1

25 cm

F xXF x

= =+

1p

Deci imaginea finală se depărtează de sistemul de lentile cu 2 2D X x= minus 1p

rezultat final D = 84 cm 1p



a

Pentru hchε = ν =λ 1p

3prezultat final 1483 10 Hzhε

ν = = sdot 1p

9360 10 mhc minusλ = = sdotε

1p


77

b

Legea conservării energiei ext ch L Eν = + 1p

4pse scrie icircn condiţiile date sub forma 2

2extmvLε = + 1p

de unde se determină viteza fotoelectronilor emişi ( )2 extLv

mε minus

= Rezultat final 571 10 m sv = sdot 2p

c

Legea conservării energiei ext sL eUε = + 1p

3pde unde se obţine exts

LUe

ε minus= 1p

rezultat final 144 VsU = 1p

d

Din legea conservării energiei ext cL Eε = + se obţine energie cinetică a fotoelectronilor emişi icircn condiţiile

date iniţial c extE L= ε minus 1p

5p

Dacă lungimea de undă a radiaţiei incidente devine un sfert din lungimea de undă de prag se poate scrie 0

1 4λ

λ = (1) 1p

Din lucru mecanic de extracţie 00

exthcL h= ν =λ

se determină lungimea de undă de prag ca fiind 0ext

hcL

λ = (2)

Din (1) şi (2) se obţine 1 4 ext

hcL

λ =

1p

1 1 11

4 3ext c ext ext c c exthc L E L L E E L= + rArr = + rArr =λ

Variaţia de energie cinetică a fotoelectronilor emişi va fi 1 4c c c extE E E L∆ = minus = minus ε 1p

rezultat final 19729 10 JcE minus∆ = sdot 1p



78

TESTUL 11

A MECANICAtilde

Subiectul I Punctaj

1 b 2 2

2 22

m1kg kg m s2 sSI

m v minussdot= sdot = sdot sdot 3

2 b 3

3 c 1 2tL L L= + 120 JdreptA L l= sdot = ( ) 60 J

2trapezB b hA + sdot

= = minus 60 JtL = 3

4 b 0c c tE E G lminus = sdot

m2 sin 10s

v g h= sdot sdot α sdot = 3

5 b 0a = tF G= 10 KWL m g hPt t

sdot sdot= = =

∆ ∆ 3



a

Pentru reprezentarea corectă şi completă a forţelor

1p

4p

Pentru fiecare corp icircn parte putem scriem1 T minus microm1g = m1am2 m2g minus T = m2a

Adunacircnd relaţiile obţinem 2 1 2 1

1 2 1 2

( )m g m g g m mam m m m

sdot minus micro sdot sdot sdot minus micro sdot= =

+ +

2p

Rezultă a = 6 ms2 1p

bDin expresia scrisă pentru m2 2 ( )T m g a= sdot minus 2p

3pRezultă 16 NT = 1p

c

Sistemul se deplaseaza cu viteză constantă deciv ct= 0a = 11 0fT Fminus =

1 2 0T Gminus + = 12 fG F= 2p

4pSe obţine expresia masei suplimentare 2 1( )m g m m gsdot = micro + sdot 2 1m mm minus micro sdot

=micro

1p

Rezultă 18 kg m = 1p

d

Noua tensiune icircn fir va fi 0T Fminus = 2 4 NT m g= sdot =

1p

4p

Forţa din scripete se calculează cu relaţia 2 2 2 2 cos90 2F T T T T= + + sdot sdot deg = 2p

Rezultă 564 NF = 1p



79


aEnergia totală a schiorului

A A At c pE E E m g h= + = sdot sdot 2p3p

Rezultă 42500 JAt

E = 1p

b

Aplicacircnd teorema de variaţie a energiei cinetice c tE L∆ = 1p

4p

2 2

2 2 f

B AG F

m v m v L Lsdot sdotminus = +

1p

Obţinem 2

2f

BF

m vL m g hsdot= minus sdot sdot 1p

Rezultă 159375 JfFL = minus 1p

c

Din expresia lucrului mecanic al forţei de frecare cosfF fL F d m g d= minus sdot = minusmicro sdot sdot sdot α sdot 1p

4pConform desenului distanţa parcursă se exprimă

sinhd =

α1p

Obţinem fFLm g h ctg

micro =minus sdot sdot sdot α

1p

Rezultă 0216micro = 1p

d

Aplicăm teorema de conservare a energiei totale

A C Bt t tE E E= = A C Ct P CE E E= +

C CP CE E= 2A Ct PE E= sdot

2p

4pObţinem 1 2

AtE

h sm g

=sdot sdot

1p

Rezultă 1 25 mh = 1p



80


Subiectul I Punctaj

1d 1 1J kg K

SI

Q Jc cm T kg K

minus minus= rArr = = sdot sdotsdot ∆ sdot

3

2 c mp V R Tssdot = sdot sdotmicro

R Tp ρsdot sdot

=micro 3

kg8m

pR T

sdotmicroρ = =

sdot3

3 b V T pentru p ct= 3

4 c 2 1 2 2 1 1 1 13 3( ) ( ) 152 2v vQ C T R T T p v p v p v= ν sdot sdot ∆ = sdotν sdot sdot minus = sdot sdot minus sdot = sdot 3

5 b 3



a

Pentru o moleculă de oxigen 1N = 0

A

m NN

=micro

2p

4pRezultă 0A

mNmicro

= 1p

Obţinem 260 533 10 kgm minus= sdot 1p

bPentru starea iniţială putem scrie 1 1 1p V R Tsdot = ν sdot sdot rezultă 1 1

1p VT

Rsdot

=ν sdot

2p3p

Obţinem 1 200 KT = 1p

c

Temperatura icircn starea 2 va fi 2 2 2p V R Tsdot = ν sdot sdot 2 1 2 12 2p p T T= sdot rArr = sdot 1p

4p

Densităţile corespunzătoare stărilor 2 şi 3 sunt 22

2

pR T

sdotmicroρ =

sdot

33

3

pR T

sdotmicroρ =

sdot 1p

Raportul densităţilor va fi 2 2 1

3 1 1

2T TT T

ρ sdot= =

ρ1p

Rezultă 2

3

2ρ=

ρ 1p

d

Pentru starea finală putem scrie 3 3 3A

Np V R TN

sdot = sdot sdot 1p

4p

Obţinem 1 12 4A

Np V R TN

sdot = sdot sdot sdot 1p

Icircn final obţinem 1

1

24

Ap NnR T

sdot sdot=

sdot sdot1p

Rezultă 263

mol036 10m

n = sdot 1p



81


a

Pentru procesul izoterm 2rarr3 3 3 2 2p V p Vsdot = sdot 1p

4pRezultă 3 11

22 R TV Vp

sdot ν sdot sdot= sdot = 2p

Obţinem 3 33 50 10 mV minusasymp sdot 1p

b

Lucrul mecanic total efectuat de sistem asupra mediului exterior este

12 23tL L L= + 12 0L = 3

23 22

ln VL R TV

= ν sdot sdot sdot 2p

4pObţinem 23 12 ln 2L R T= sdotν sdot sdot sdot 1p

Rezultă 69804 JtL = 1p

cCăldura schimbată de sistem icircn procesul 1rarr2 12 2 1 1( ) 3vQ C T T R T= ν sdot sdot minus = sdotν sdot sdot 2p

3pRezultă 12 14958 JQ = 1p

d

Pentru starea 2 energia internă corespunzătoare este 2 232

U R T= sdotν sdot sdot 1p

4pDin procesul izocor 1rarr2 obţinem T2 = 2T1 2 13U R T= sdotν sdot sdot 2p

Rezultă 2 14958 JU = 1p



82


Subiectul I Punctaj

1 d 2W R I t= sdot sdot ∆

2 1J A sSI

R minus minus= sdot sdot 3

2 c 100U U U tR QI Qt

sdot ∆= = = = Ω

∆

3

3 b 2 2

3 3eR R RR

Rsdot sdot sdot

= =sdot

3

2 3e

E EIR r R r

sdot= =

+ sdot + sdot3

4 d 3

5 c 2W R I t= sdot sdot ∆ 120

WIR t

= =sdot ∆

50 mAI = 3



a

Rezistenţa internă a grupării celor două surse o determinăm din relaţia 1 2Sr r r= + 2 Sr = Ω

Tensiunea electromotoare echivalentă a grupării celor două surse este 2 40eE E V= sdot =1p

4pPentru icircntreg circuitul 2 5 33eRR R R= + = 1p

Rezultă 1 1 1U I R= sdot unde 1e

e e

EIR r

=+

= 5A 1p

Obţinem 1= 15 VU 1p

b

Scriem legile lui Kirchhoff

2 32RI R I= sdot 1 2 3I I I= +2p

4p

Obţinem 1 33I I= sdot 1p

Rezultă 3 166 AI = 1p

cDin expresia rezistenţei obţinem 1

1lR

Sρsdot

= 2

4dS πsdot

= 2

1 11 4

R S R dl sdot sdot π sdot= =

ρ sdotρ2p

3pRezultă 1 15 ml = 1p

dDacă icircntre punctele A şi B se leagă un fir de rezistenţă neglijabilă 1e eR R R scade I creste= rArr rArr 2p

4pJustificare

1

2 666Ate

EI I cresteR r

sdot= = rArr

+2p



83


a

Icircn condiţii normale de funcţionare 2b bP R I= sdot 2p

4pRezultă 2bb

PRI

= 1p

Obţinem 15bR = Ω 1p

b

Din desen b RU U U= +

1p

4pRezultă b

Rb

PU U sI

= minus 1p

RU R I= sdot

R

b

URI

= 1p

Obţinem 35R = Ω 1p

c

b b bW U I t= sdot sdot ∆ 1p

4pDeci b b bU I Psdot = b bW P t= sdot ∆ 2p

Rezultă 108 kJbW = 1p

d

2R bP R I= sdot 2p

3p

Rezultă 140 WRP = 1p



84


1 c 2 h hcE mc mc pcν

= rArr = rArr =λ

3

2 d 3

3 c 22 1 1 1

1

3sin sin 2 31sin sin

2

i n in n n nr n r

= rArr = = sdot = 3

4 c 1 1 1 02 m = 20 cm5

C ff C

= rArr = = = 3

5 b 3



aConvergenţa lentilei este 2

1 125 10

Cf mminus= =

sdot2p

3p

Rezultă 4C = δ 1p

b

Din prima formulă fundamentală a lentilelor 12

2 1 2 1 1

1 1 1 1 1 1 fxxx x f x f x f x

minus = rArr = + rArr =+

1 2 1 21y y x x= rArr β = minus rArr = minus2p

4pobţinem 2

2 22

2fxx x ff xminus

= rArr =minus

1p

Rezultă 2 50 cmx = 1p

c Desen corect şi complet 4p 4p

d

Convergenţa sistemului de lentile 1 2C C C= + 1p

4pRezultă 1 2

1 1 100 10025 20

Cf f

= + = minus 2p

Obţinem C = minus1 dioptrie 1p



a

Energia fotonului incident este ch hε = ν =λ

2p

4pRezultă 25 8

199

66 10 3 10 199 10450 10 4

minusminus

minus

sdot sdot sdotε = = sdot

sdot1p

Obţinem 195 10 Jminusε sdot

1p


85

bLucrul mecanic de extracţie 0 0

0

hcL h= ν =λ

2p3p

Rezultă 190 44 10 JL minus= sdot 1p

c

0eU L= ε minus 1p

4pTensiunea de stopare va fi 0LU

eε minus

= 1p

Obţinem 19

19

(49 44) 1016 10

Uminus

minus

minus sdot=

sdot1P

Rezultă 031 VU = 1p

d

Energia cinetică a fotoelectronilor emişi de către catodul celulei fotoelectrice este2

2cmE eU ν

= =

1p

4pRezultă 02 2( )eU L

m mε minus

ν = = 2p

Obţinem v = 033 106 ms 1p



86

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj

1 b 32 c 33 b 34 a 35 a 3



aPentru punctul de pe fir icircn care acţionează forţa 1 0F Tminus = 2p

3pRezultă 1 10 NT F= = 1p

bConsideracircnd rezultatul de mai sus se poate scrie 1

12

m3 s

F mg maFa gm

minus micro =

= minus micro =

2p

4p1

12

m3 s

F mg maFa gm

minus micro =

= minus micro = 2p

c

Pentru un sistem de axe cu axa Ox orizontală se poate scrie 2

2 2

2

cos 0

sin 0f

f

F FN F mgF N

α minus =

+ α minus == micro

2p4p

Rezultă 33 Ncos sin

mgF micro= =

α + micro α 2p

dPentru ansamblul celor două corpuri aflate icircn mişcare uniformă scriem ( ) 0F m m gminus micro + ∆ = 2p

4pRezultă 3 kgFm

g∆ = =

micro2p



aEnergia se conservă

2

2MvMgh = 2p

4p

Rezultă 2 10 msv gh= = 2p

b 50 Nsp Mv= = 3P 3P

cSe aplică teorema de variaţie a energiei cinetice

2

0 cos1802c

mvE L mgd∆ = minus = micro deg 2p

4p

Rezultă 25 mhd = =micro

2p

dUtilizacircnd teorema de variaţie a energiei cinetice se scrie

2 2

( )cos1802 2 2m v mv mg d x minus = micro minus deg

2p

4p

Rezultă 625 m4hx = =micro

2p



87


Subiectul I Punctaj

1 c 32 c 33 a 34 d 35 b 3



Din ecuaţia de stare mpV RT=micro

1p3p

rezultă 0128 kgpVmRT

micro= = 2p

b

Din pV RT= ν 2p

3p

rezultă 4molpVRT

ν = = 4 moli 1p

c

Se schimbă energia internă şi presiunea 1 1VU C T= ν 2 2VU C T= ν 2 1 1U U fU= minus 2p

5p

1 2 2

1 1

2871 1 004(3) 4(3)300

U U TfU Tminus

= = minus = minus = = scade cu f 1p

1 1 1 1p V RT p V RT= ν = ν 2 1 1 1(1 )p p kp p k= minus = minus 1p

2

1 2 2 2

11 1 1

1 1 1 4(3)

RTp p p TVk RTp p T

V

νminus

= = minus = minus = minus =ν

scade cu k 1p

d

Icircn butelie rămacircne oxigen la presiunea atmosferei de afară icircntr-o primă instanţă 2p

4p1 0

0 1 134 gm p Vp V RT mRT

micro= = =

micro2p



88


a

Procesul 1 rarr 2 este izocor 2 2 22 1 2

1 1 1

8atm 300K 1200 1200 273 927 degC2atm

p T pT T K tp T p

= = = = = minus = 1p

4p

Procesul 2rarr3 este izoterm deci 3 2 927 Ct t= = deg 1p

Graficul

2p

b

1 1 1p V RT= ν 1p

3p1 1

1

p VRT

ν =

1P

v cong 012 moli 1p

c

323 2

2

ln VL RTV

= ν

1p

4pPentru starea 2 2 2 1RT p Vν = 1P

Pentru procesul 2rarr3 3 22 2 3 3 3 1 2 2 1 3

2 1

V pp V p V p p p V p VV p

= = = = 1p

223 2 1

1

V ln 11088 JpL pp

= = 1P

d

1abs ced ced

abs abs abs

Q Q QLQ Q Q

minusη = = = minus 1p

4p2

12 23 2 1 21

( ) lnabs VpQ Q Q C T T RTp

= + = ν minus + ν 1p

3 1 3 2( )ced pQ C T T T T= ν minus = 1P

025 25η cong = 1p



89


Subiectul I Punctaj

1 b 3

2

a Pentru cele trei situaţii curenţii au intensităţile 1 2 32 3

2 3E E EI I I

R r R r R r= = =

+ + +

Rezolvacircnd se obţin pentru rezistenţe valorile 2 1 1 2

2 2 2E E E Er RI I I I

= minus = minus iar pentru intensitatea curentului cerut

1 23

1 2

3 3 2727A3 4

E I IIR r I I

= = =+ minus

3

3 d 3

4

d Pentru circuitul cu linia de alimentare randamentul este 1l

RR R r

η =+ +

(unde lR este rezistenţa liniei r este rezis-

tenţa internă a sursei) iar cu fire de alimentare scurte (cu rezistenţa de linie neglijabilă) randamentul este 2R

R rη =

+

Rezolvacircnd rezultă 1 1

4lR RR = minus = Ωη η

3

5

c Pentru o rezistenţă externă oarecare impunem condiţia din problemă 2 2

2

8 8( ) 9 9 4m

RE EPR r r

= = sdot+

Din această relaţie rezultă expresia rezistenţei R icircn funcţie de rezistenţa internă a sursei 2

5 34

2

rr rR r

plusmn = =

Se vede că aceeaşi putere se obţine şi pentru rezistenţa externă pentru 2R r= dar şi pentru 2rR =

3



a

12 3

12

1 AEI R Rr RR R

= =+ +

+1p

4p2 3

12

2 VBAR RU I

R R= =

+ 1p

22

2 A3

BAUIR

= = 1p

33

1 A3

BAUIR

= = 1p

b

22 2 2qI q I t

t= = sdot ∆

∆2p

4p2

2 A 30 s 20 C3

q = sdot = 2p

c

Icircntre A şi B se creează un scurtcircuit echivalent cu scoaterea celor două rezistenţe din montajul iniţial 1p

3pNoul curent prin sursă va fi 1

scEI

R r=

+ 1p

Numeric 4 1(3) A3scI = = 1p

d

p

EIR r

=+ 1p

4p1 2 3

1 1 1 1 142pp

RR R R R

= + + rArr = Ω 1p

U = IRp 1pU cong 47 V 1p



90


a

Icircn cele două cazuri intensitatea curentului electric se scrie 2

2LL

U UI RR R R=

+ + 2p

4pDe aici rezultă 2LR R= iar pentru un singur fir conductor 1 200R R= = Ω 1p

Randamentul este 2

2

1 05 50( ) I 2L

RIR R

η = = = =+

1p

b

Energia consumată este 2

2 290400J3U tW I Rt

R= = =

Deoarece 51 kWh 36 10 J= sdot 2p

4penergia se poate scrie sub forma 5

290400 003 kWh36 10

W = congsdot

1p

Costul este lei003 kWh 25 0075 lei 75 bani

kWhC = sdot = = 1p

c

Rezistenţa electrică a unei porţiuni de linie de lungime x este

xxrS

= ρ iar a unei linii simple icircntregi este

lRS

= ρ

Din cele două relaţii se obţine xRxrl

=

Intensitatea curentului prin dispozitiv este 2(1 )

UI xRl

=+

1p

4p

Puterea debitată de dispozitiv va fi 2

2(1 )

UP xRl

=+

1p

Pentru limitele extreme ale liniei avem

- La capătul unde se află sursa 2

max0 605Ux P WR

= = =

- La capătul opus 2

min 20163Ux l P W

R= = cong

1p

Graficul este un segment de hiperbolă

1p

dRezistenţa electrică a unui bec este r = ui rarr r = 20025 = 80 ΩLa conectarea a n becuri icircn serie intensitatea curentului prin fiecare este i = Unr rarr n = Uri = Uu = 55 2p

3pPuterea consumată de becuri este P = nri2 = 55 ∙ 80 ∙ 625 ∙ 10-6 =275W 1p



91


1 c 32 d 33 c 34 d 35 c 3



a

Lama cu feţe plane şi paralele constituie un ansamblu de doi dioptri plani pentru care este valabilă relaţia 2 1

2 1

n nx x

=

Dacă minusx1 este coordonata unui obiect faţă de primul dioptru aplicacircnd succesiv relaţia se obţine coordonata x2

a imaginii obiectului faţă de al doilea dioptru icircn condiţiile problemei 2 1ex xn

= minus

2p

3p

Distanţa cu care se apropie imaginea de observator este 1 125 cmne

nminus

δ = = 1p

b

Folosim prima formulă fundamentală a lentilelor 2 1

1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+ Aici avem -x1 = 875 cm 1p

Rezultă 2 2592 cm 026 mx = cong 1p

c

Mărirea liniară transversală este 2 2

1 1

x yx y

β = = 1p

5p

Rezultă 22 1

1

xy yx

= 1p

Rezultă 2 11

fy yx f

=+

1p

Rezultă 2 177 cmy = minus 1p

Imagine răsturnată 1p

dDin relaţia care arată cu cacirct se apropie imaginea de observator se vede că nu intervine coordonata obiectului 15p

3pPoziţia imaginii icircn lentilă nu se modifică 15p



a Coordonatele pentru maxime de interferenţă sunt max 2kkDx

lλ

= 2p3p

Rezultă 5max 688 mmx cong 1p

b Coordonatele pentru minime de interferenţă sunt min(2 1)

4kk Dx

l+ λ

= 2p4p

Rezultă 3min 481 mmx cong 2p


92

c

Interfranja se calculează ca diferenţă a coordonatelor a două franje succesive de acelaşi tip maxime sau mini-

me 1k ki x x+= minus 1p

4pSe obţine

2Di

lλ

= 2p

Rezultă 1375 mmi = 1p

d

Variaţia relativă a grosimii interfrajei este i i i

i i∆ minus

ε = = 1p

4pRezultă 1D DDminus δ

ε = minus 1p

Rezultă DD

δε = minus 1p

Se obţine 40ε = minus 1p


TESTE DE NIVEL MEDIU


94

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj1 d 32 c 33 b 34 c 35 a 3



aDeoarece N = Gn 1p

3unde Gn = m middot g ∙ cos α 1prezultă N = 865 N 1p

b

Acceleraţia cutiei o exprimăm din relaţia 0

0

v vat t

minus=

minus 2p

4La momentul t0 = 0 cutia are viteza v0 = 0 iar la momentul t cutia are viteza v = 2t 1p


c

Pe direcţia mişcării cutiei avem Gt minus Ff = m ∙ a 1p

4unde Gt = m ∙ g ∙ sin α 1pDeci Ff = m ∙ (g ∙ sin α minus a) 1pRezultă Ff = 3N 1p

d

La urcarea uniformă a cutiei pe planul icircnclinat pe direcţia mişcării cutiei avem F minus Ff minus Gt = 0 2p

4Deci F = m ∙ g ∙ sin α + Ff 1p

Rezultă F = 8 N 1pTOTAL pentru Subiectul al II-lea 15p


a

Puterea sistemului autoturismndashremorcă este maximă dacă a = 0 şi viteza lui este vmax deci pmax = Fr ∙ vmax 2p

5unde Fr = f ∙ m∙ g 1p

Impulsul maxim al sistemului autoturismndashremorcă este max maxPp m v

f g= sdot =

sdot1p

Rezultă pmax = 55200 kg ∙ ms 1p

b Energia cinetică maximă a sistemului autoturismndashremorcă este 2 2max max

max 2 2cm v pE

msdot

= = 2p3p

Rezultă Ec max = 1 015 680 J 1p

c

Aplicăm teorema de variaţie a energiei cineticercf ci FE E Lminus = 1p

3punde Ecf = 0 Eci = Ec max 1p

Rezultă 1015680 JrFL = minus 1p

d

Deoarece rF rL F d= minus sdot 2p

4obţinem rF

r

Ld

F= minus 1p

Rezultă 67712 md = 1p


Bareme teste nivel mediu

95


Subiectul I Punctaj1 a 32 c 33 b 34 d 35 a 3



a

Din ecuaţia termică de stare 0 0 p V R Tsdot = ν sdot sdot 1p

4unde V L S= sdot 1p

obţinem 0

0

p L SR Tsdot sdot

ν =sdot

1p

Rezultă 22 10 moliminusν cong sdot 1p

b

Deoarece 1 2 ν = ν obţinem 1 2

1 2

m m=

micro micro1p

4Dar 1 2 ν = ν + ν sau 1 2 1 2

1 2

m m m m+= +

micro micro micro 1p

Deci 1 2

2micro + micro

micro = 1p

Rezultă 36 gmolmicro = 1p

c

Densitatea amestecului este 0

0

m mm R TV

p

ρ = =sdot

sdotmicro

1p

3pDeci 0

0

pR T

sdot microρ =

sdot1p

Rezultă 3159 kgmρ cong 1p

d

Icircn acest caz maxp V R Tsdot = ν sdot sdot 2p

4Deci 0max

0

p TTpsdot

= 1p

Rezultă max 1365 KT = 1p



96


a


31

1 11

2 10 Pa

= 4 dm

9627 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

4


32

2 22

4 10 Pa

= 4 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p


33

3 33

4 10 Pa

= 8 dm

38508 K

p

Vp VT

R

= sdot

sdot= cong

ν sdot

1p


34

4 44

2 10 Pa

= 8 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

b

Căldura cedată de gaz mediului exterior icircn decursul unui ciclu este cedat 34 41Q Q Q= + 1p

4

unde 34 V 4 3 4 35( ) ( )2

Q C T T R T T= ν sdot sdot minus = ν sdot sdot minus 1p

şi 41 P 1 4 V 1 4 1 47( ) ( ) ( ) ( )2

Q C T T C R T T R T T= ν sdot sdot minus = ν sdot + sdot minus = ν sdot sdot minus 1p

Rezultă cedat 6800 JQ = minus 1p

c

Lucrul mecanic schimbat de gaz cu mediul exterior icircn decursul unui ciclu este

12341 2 1 3 2( ) ( )L p p V V= minus sdot minus 2p3p

Rezultă 12341 800 JL = 1p

d

Randamentul ciclului termodinamic este 12341

primit

LQ

η = 1p

4unde 12341 primit cedatL Q Q= minus 1p

Deci 12341

12341 cedat

LL Q

η =+ 1p

Rezultă 1053 η cong 1p



97


Subiectul I Punctaj1 b 32 a 33 d 34 c 35 a 3



a

Rezistenţa internă a grupării celor două surse o determinăm din relaţia 1 2

1 1 1

Pr r r= + 1p

3pObţinem 1 2

1 2P

r rrr r

sdot=

+1p

Rezultă 05 Pr = Ω 1p

b

Tensiunea electromotoare echivalentă a grupării celor două surse este 1 2 2 1

1 2P

E r E rEr r

sdot + sdot=

+1p

4Intensitatea curentului prin rezistorul 1R cacircnd icircntrerupătorul k se află icircn poziţia deschis este

P

S P

EIR r

=+

1p

Unde 1 2SR R R= + 1p

Rezultă 05 AI = 1p

c

Intensitatea curentului prin rezistorul 1R cacircnd icircntrerupătorul k se află icircn poziţia IcircNCHIS este

2

P

P

EIR r

prime =+

3p4

Rezultă 064 AI prime cong 1p

d

Aplicăm legea a doua a lui Kirchhoff 2 2 2 2E I r I Rprime= sdot + sdot 2p

4Deci 2 22

2

E I RIr

primeminus sdot= 1p

Rezultă 2 064 AI cong 1p



a

Tensiunea electromotoare a sursei echivalente cu bateria dată este SE N E= sdot 1p

4Rezultă 15 SE V= 1p

Rezistenţa internă a sursei echivalente cu bateria dată Sr N r= sdot 1p



98

b

Puterea dezvoltată de rezistorul de rezistenţă R poate fi scrisă sub forma 2 P R I= sdot 1p

4

unde S

S

EIR r

=+

1p

După efectuarea calculelor obţinem 2 22 4

2S S S SE P r E E P r

RP

minus sdot plusmn minus sdot= 1p

Rezultă 25 R = Ω 1p

cRandamentul circuitului electric este

S

RR r

η =+

2p3p

Rezultă 50 η = 1p

d

Rezistenţa echivalentă a grupării de rezistori este 2PRR = 1p

4Icircn acest caz intensitatea curentului prin circuit este S

P

EIR r

prime =+ 1p

Puterea totală furnizată de baterie pentru gruparea paralel a rezistorilor este Ptotală SE I prime= sdot 1p

Rezultă Ptotală 60 W= 1p



99

DOPTICAtilde

Subiectul I Punctaj1 b 32 a 33 d 34 c 35 c 3



a

Din prima formulă fundamentală a lentilelor subţiri 2 1 1

1 1 1x x f

minus = 2p

4Obţinem 1 21

1 2

x xfx x

sdot=

minus1p

Rezultă 1 8 cm 008 mf = = 1p

bConvergenţa primei lentile este 1

1

1Cf

= 2p3p

Rezultă 11 125 mC minus= 1p

c

Din prima formulă fundamentală a lentilelor subţiri pentru sistemul de lentile 2 1

1 1 1 x x f

minus =prime 1p

4

cu 1 2

1 1 1 f f f

+ = 1p

obţinem 12

1

f fff f

sdot=

minus sau 2 2

22 2

x xfx x

primesdot=

primeminus 1p

Rezultă 2 109 cm 0109 mf cong minus = minus 1p

d

Pentru sistemul de lentile mărirea transversală este 2 2

1 1

y xy xprime prime

β = = 1p

4

Pentru prima lentilă mărirea transversală este 2 21

1 1

y xy x

β = = 1p

Deci 2 2

2 2

y xy xprime prime

= 1p

Rezultă 2

2

12yyprime

= 1p



100


a

Deoarece x N i= sdot 1p

3Obţinem xiN

= 1p

Rezultă 500 mi = micro 1p

b

Din expresia interfranjei 2

Dil

λ sdot= 2p

4Obţinem

2l iD

sdotλ = 1p

Rezultă 500 nmλ = 1p

c

Maximul de ordinul al II-lea se află faţă de franja centrală la distanţa 22

2Dx

lλ sdot

=1p

4Maximul de ordinul al IV-lea se află faţă de franja centrală la distanţa 4

42

Dxl

λ sdot= 1p

Distanţa dintre maximul de ordinul al IV-lea şi maximul de ordinul al II-lea aflate de aceeaşi parte a figurii de

interferenţă este 4 22

2Dx x x

lλ sdot

∆ = minus =1p

Rezultă 1 mmx∆ = 1p

d

Icircn acest caz interfranja este dată de relaţia 11

2Di

lλ sdot

= 1p

4unde 1

1 cn n

λλ = sdot =

ν1p

Deci 1 12

D inl i i

λ sdot= =

sdot1p

Rezultă 133n cong 1p



101

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1 c Spaţiul parcurs de corp este 15 mh∆ = 2 2 3 s

2gt hh t

g∆

∆ = rArr = = 3p

2 b pF F t pt

∆= rArr ∆ = ∆

∆

3p

3

c Se ridică graficul F = F(x) şi se calculează lucrul mecanic ca arie a suprafeţei evidenţiate

L = 22 J 3p

4 a Din legea spaţiului x = 2t2 + 6t + 8 (m) obţinem a = 4 ms2F = ma = 4000 N 3p

5 a N = 0 Fy = G sin 60 NsinmgF mg Fα = rArr = =

α 3p


Subiectul al II-lea Parţial Punctaja Ridicarea graficului a = a (Δm) 3p 3p

b

Studiem mişcarea fiecărui corp separat Reprezentăm forţele care acţionează asupra fiecărui corp şi aplicăm principiul al doilea al dinamicii

Considerăm cazul icircn care 2 1gtm m

2p

3p

2 2

1 1

G T m aT G m a

minus =minus = 1 2

g mam m

∆rArr =

+ 2p

c

Pentru prima pereche de valori a = 125 ms2 şi Δm = 01 kg Din expresia acceleraţiei deducem masa totală a

sistemului 1 2 08 kgg mm ma∆

+ = = 1p

4pCu ajutorul ecuaţiilor 1 2

2 1

08 kg01 kg

m mm m

+ =minus =

2p

Rezultă 1 2035 kg 045 kgm m= = 1p

dSe calculează viteza sistemului după h = 1 m 2 25 msv ah= =

2p

4p

1 2( ) 126 kg ms= + = sdotp m m v 2p



102


aReprezentăm forţele care acţionează asupra corpului şi aplicăm principiul al doilea al dinamicii

t fG F m aminus = sdot

2p3p

2(sin 03) 2 msa grArr = α minus = 1p

b

GL mg h= ∆

1p

4psinh x∆ = ∆ α unde Δx = spaţiul parcurs icircn secunda a doua

2

2atx = 1p

pentru t1 = 1 s 1 1 mx = pentru t2 = 2 s 2 4 mx = 3 m 15 mx h∆ = rArr ∆ = 1p

45 JGL = 1p

c

Dacă mişcarea corpului pe planul icircnclinat durează 2 s lungimea planului este x2 = 4 m

1 11 1c t c G fE L E L L∆ = rArr ∆ = + 1p

4p

21 2 2c f

h xE mg F= minus

1p

21 (sin 03)

2cmgxE = α minus

1p

1 12 JcE = 1p

d

22 2

1 12c

cc

p p EEm p E

= rArr =

1p

4p


2 2 (sin 03)cE mgx= α minus

2 24 JcE =

1p

2

1

2pp

=

1p



103

B ELEMENTE DE TERMODINAMICAtilde

Subiectul I Punctaj1 Ecuaţia calorică de stare b 3p2

1 2

1 2

m m+micro =

ν + ν

1 1 2 21 2

1 2

A A

N Nm mN N

N N

micro micro= =

=

1 2

2micro + micro

micro = b

3p

3( )V f iL C T T= minusν minus

1

1 1 ii i f f f i

f

VTV T V T TV

γminus

γminus γminus

= rArr =

1

1iV i

f

VL C TV

γminus = minusν minus

2 1 2 1 2 1γ gt γ gt rArr gtv vC C L L

c

3p

41 1 2 2

1 32 2 5

1 1

3 3

p V p V

V VV V

γ γ

γ

=

= rArr = b

3p

52 2

1 1

22

1

1 1

065 260 K

Q TQ T

T TT

η = minus = minus

rArr = rArr =

b3p



a

1 2

1 1 2 2( )

m m m

m p V p VRT

= +micro

= +3p

4p

3464 10 kgm minus= sdot 1p

b

Din ecuaţia 1 2 fν + ν = ν 1p

4pRezultă 1 1 2 2

1 2

p V p VpV V

+=

+2p

5 28 10 Nm3

p = sdot

1p

c

mpV RT=micro 2p

4p1 21 2pV pVm m

RT RTmicro micro

= = 1p

3 31 2154 10 kg 31 10 kgm mminus minus= sdot = sdot 1p


104

d

2 22

2

m RTp V

=micro 2p

3p5 2

2 241 10 Nmp = sdot 1p



a

reprezentarea grafică

3p 3p

b

3 31

1

2 10 m minus= = sdotρmV 5 21

11

186 10 NmmRTpV

= = sdotmicro 1p

4p

3 312 2 12 4 10 m

2V V minusρ

ρ = rArr = = sdot

1p

5 21 12

2

93 10 Nmp VpV

= = sdot 1p

2 1

212 1

1

ln 25915 J

T Tm VL RT

V

=

= =micro

1p

c

13 450 K

2TT = = 1p

5p

11

3 323 2

3

224 10 mTV VT

γminusminus

= = sdot

1p

5 233

3

083 10 NmmRTpV

= = sdotmicro 1p

3 3 5 24 1 4 32 10 m 083 10 Nmminus= = sdot = = sdotV V p p 1p

4 44 40 Kp VT

mRmicro

= = 1p

d

min

max

1CTT

η = minus 1p

3p4

1

1CTT

η = minus 1p

95Cη = 1p



105


Subiectul I Punctaj

1 Formula corectă b 3

21

2

1 A

AA

A

EI Rr R

EIR R r

= rArr = Ω+

= =+ + a

3

3 124pRR R= rArr = Ω

c3

4 22

225 W4

ER r Ir

EPr

= rArr =

= = c

3

5( 1) 10 ka V a

V

UR R n n RU

= minus = rArr = Ω

Ra = 10 kΩ icircn serie cu voltmetrul c

3



a

e

UIR

= 3 4 15sR R R= + = Ω 1p

4p1

1

5sp

s

R RRR R

= = Ω+ 1p

2 10e pR R R= + = Ω 1p

1I A= 1p

be

EIR r

=+ 2p

4p

2eEr RI

= minus = Ω 1p

c

2 3

2 3

25pR RR

R R= = Ω

+ 1p

4p

1 10s pR R R= + = Ω 1p

4

4

5se

s

R RR R R

= = Ω+ 1p

12 A7e

EIR r

= =+

857 VeU IR = = 1p


106

d

Aplicăm legile lui Kirchhoff

1p

4p1 2

2 1 4

1 26 A7

s

I I I I R I R

I I

= +=

rArr = =

2 3 4

3 2 4 3

3 46 A

14

I I I I R I R

I I

= +=

rArr = =1p

1 4AI I I = + 1p

9 A7AI = 1p



a

max 1 2P r R RrArr = + 1p

4p1 3 1 1 2 3( )P P r R R R R= rArr = + + 1p

( )21 2 1 1 2 3( )R R R R R R+ = + + 1p

rezolvarea ecuaţiei rezultă R2 = 6 Ω 1p

b

2

max 84EP r

r= = Ω 1p

3pPmax = 450 W 1p

1 2

1 2

05R RR R r

+η = =

+ + 1p

c

2 Ann

n

PIU

= = 1p

4p1( ) 6 AnE I R r U I= + + rArr = 1p

4 AR n RI I I I= + rArr = 1p

4 15n

R

URI

= = Ω 1p

d

P UI= 1p

4p72 VU E Ir= minus = 2p

432 WP = 1p



107

DOPTICAtilde

Subiectul I Punctaj

1

2 22 1

1 1

12

1

2

53

23 3

y x y yy x

fxx ff x

fy

β = = rArr = β

= =+

rArr β = minus rArr = b

3

22 5π

∆ϕ = δ = πλ c 3

3 375 nmaan

λλ = = b 3

4 206 nmexex

hc hcL eUL eU

= + rArr λ = =λ +

a 3

5 00

295 nmexhc L= rArr λ =λ

c 3



a

Din grafic se observă că C = 4δ şi R = 025 m 1p

4

2( 1)C nR

= minus pentru lentila biconvexă 1p

( 1)2

CRnrArr minus = 1p

15n = 1p

b

2 1 2 1

1 1 1 1 1 Cx x f x x

minus = hArr minus = 1p

3p12

1

15 m1

xxCx

= =+

cu 1 03 mx = minus 1p

2

1

5xx

β = = minus 1p

c

Convergenţa lentilei divergente 12

6 3CC = minus = minus δ 1p

4

12( 1)

l

nCn R

= minus 1p

1 12

lnn C R=

+ 1p

163ln = 1p


108

d

Construcţia corectă a imaginii 1p

4p

13 1 1 3 1

1 1 1 1 1 Cx x f x x


13

1 1

025 m1

xxC x

= = minus+ 1p

3

1

083xx

β = = 1p



a

max minx x x∆ = minus 1p

4pDistanţa de la o franjă luminoasă de ordin k la franja luminoasă centrală este max max 8kx ki x i= rArr = 1p

Distanţa de la o franjă icircntunecoasă de ordin k la franja luminoasă centrală este min min(2 1) 35

2kk ix x iminus

= rArr = 1p

45 2 mmx i i∆ = rArr = 1p

b

2Dil

λ= 1p

3p2liD =λ

1p

142 mD = 1p

c

Prin aşezarea unei lame de sticlă de grosime e şi indice de refracţie n icircn dreptul fantei superioare se introduce icircn calea razei care provine de la această fantă un drum optic suplimentar ( 1)e nδ = minus 1p

4p

2 ( 1)klx e n kD

δ = minus minus = λ

Pentru maximul central 2 ( 1) 0klx e nD

δ = minus minus = 1p

Pentru 33

1kx i en

λ= rArr =

minus 1p

42 me = micro 1p

d

1 2r vk kδ = λ = λ 1p

4p2 12k k= 1p

1 21 2k k= = 1p



109

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj1 d 32 c 33 b 34 c 35 d 3



a

1

2

AI dtgIB d

α = = 1p

4p1 0 =d v t

2

2 2atd = 2p

Rezultat final 2 tg

=sdot α

vta

15s=t 1p

b

2 0v v at at= + = 1p

3p1

21 1 135 kJ2

= =cm vE 1p

2

22 2 90 kJ2

= =cm vE 1p

c1 1= ∆ = minus

fF c cL E E 2p3p

Rezultat final 135 kJ= minusfFL 1p

d

minus minus =f tF G ma 1p

5p= microfF N sin = αtG mg cosnN G mg= = α 2p

(sin cos )= minus α + micro αa g 1p

Rezultat final267 mscong minusa 1p



a

0T F G+ + = 1p

4pcossin

T mgT F

α =α =

1p

Rezultat final sin cos

α=

αF mg

3 N 09 N2

= congF 2p

b cosmgT =

α2p

3p

Rezultat final 1 N=T 1p


110

c

Conservarea energiei mecanice A BE E= 1p

3p=AE mgh

2

2BmvE = 1p

(1 cos )= minus αh l 1p

Rezultat final 2 (1 cos )= minus αv gl 2 ms 14 ms= =v 1p

d

Conservarea impulsului pentru sistemul de bile ce se ciocnesc plastic initial finalp p= 1p

4

p mv= 1p

final2mv mv= 1p

Rezultat final final 07 ms2

= =vv 1p



111


Subiectul I Punctaj1 c 32 c 33 d 34 b 35 a 3



a

pV RT= ν 1p

4pν =

microm m

Vρ =

1p

p RTmicro = ρ 1p

Rezultat final 33 024 kgmρ = 1p

b

1 2 3 1 2 3( ) ( )p V V V RT+ + = ν + ν + ν 1p

4p

1 1 2 2 3 3

1 2 3

p V p V p VpV V V

+ +=

+ + 1p

173pp = 1p

Rezultat final 523 10p Pa= sdot 1p

c

1 1 1 ∆ = minusU U U 1 1 1 15 5 2 2

= ν =U RT p V 1 15 2

U RT= ν 1p

4p1 1 1 1

5 2

= ν =pV RT U pV

1 1 1 15 2

= ν =pV RT U pV 1p

1 1 1 1 15 10( )2 3

U V p p p V∆ = minus = 1p

Rezultat final 1 3333U J∆ = 1p

d

1 1 2 2 3 3

1 2 3

micro ν + micro ν + micro νmicro =

ν + ν + ν p

3p1 2 34 9

14micro + micro + micro

micro = 1p

Rezultat final 116 gmolmicro = 1p



112


a

2 1 1 1( )(3 )= minus minusL p p V V 1p

4p

2 1 1 1 2 13 3= =p V p V p p 2 1 1 1 2 13 3= =p V p V p p 1p

1 1 1 1 12 2 4 4L p V p V RT= = = ν

14LRT

ν =

1p

Rezultat final v = 12 moli 1p

b


4p

12 1 13 2 32

U R T RT∆ = ν sdot = ν 1p

12 34LU∆ = 1p

Rezultat final 12 75 kJ∆ =U

c

23 3 2( )PQ C T T= ν minus 1p

3p23 1 15 6 152

Q R T RT= ν sdot = ν 1p

Rezultat final 2315 375 kJ4

= sdot =Q L 1p

d

P

LQ

η = 1p

4p1 2 2 3minus minus= +pQ Q Q 1p

1 2 2 1 1 23( ) 4minus minus= ν minus = ∆ =vLQ C T T U

92PQ L= 1p


η = = 1p



113

C PRODUCEREA ordfI UTILIZAREA CURENTULUI CONTINUUSubiectul I Punctaj1 c 32 b 33 c 34 c 35 d 3



aAplicacircnd teorema a doua a lui Kirchhoff pentru un ochi de reţea se obţine relaţia 1 2E U U= + 2p

3pRezultat final 200 V=E 1p

bAplicacircnd teorema a doua a lui Kirchhoff pentru alt ochi de reţea se obţine relaţia 1 1 2( )= +E I R R 2p

3pRezultat final 1 004 A=I 1p

c

K deschis 1 1 = sdotU r I 2 2 = sdotU r I 1 1

2 2

23

U rU r

= = 1p

5pK icircnchis 1 11

1 1

=+p

r RRr R

2 22

2 2

=+p

r RRr R

11 2p p

U R RI

= = 2p

Rezultat final 1 2 k = Ωr 2 3 k= Ωr 2p

d

1

1G

p

U IR

= 2p

4p1 11

1 1

12 k= = Ω+p

r RRr R 1p

Rezultat final 833 mA=GI 1p



a

2

b

UPR

= 2p3p

Rezultat final Rb = 6 Ω 1p

b

nE = Ib(nr + Rb + Ra) 1p

5p

2 A= =bPIU

1p

minn = dacă Ra = 0 1p

rezultă 342= =minus b

UnE I r

1p

Rezultat final min 4n = 1p

c4 (4 )= + +b b aE I r R R 2p

3pRezultat final Ra = 1 Ω 1p

d

2

max 4

=EP

r cacircnd R r= 2p

4p

Rezultat final max 83 W=P şi 075= ΩR 2p



114

DOPTICAtilde

Subiectul I Punctaj1 b 32 a 33 c 34 d 35 b 3

TOTAL pentru Subiectul I 15p Subiectul al II-lea Parţial Punctaja Construcţia imaginii 4p 4p

b 2 1 1

1 1 1x x f

minus = 2p3p

Rezultat final 2 30 cm=x 1p

c

1 2 1 10 cm= minus =x x f 2 1 2 80 cmx x f= minus = minus 2p

4p2 1 2

1 1 1x x f

minus = 1p

Rezultat final 2 89 cm= minusf 1p

d

Imaginea formată de lentila convergentă se formează pe oglindă Astfel oglinda va reflecta lumina ce va trece din nou prin lentila convergentă 1p

4pNoul obiect pentru lentilă se află pe oglindă la distanţa 1 2ndashx x= 1p

1 12

1 1

x fx x f

=+ 1p

Rezultat final 2 60 cmx = 1p



aDistanţa dintre două maxime sau două minime succesive este

2Dil

λ= 2p

3p

Rezultat final 2 m=D 1p

b

Distanţa de la maximul central la al doilea maxim de interferenţă este 2i 1p

4p2 2 4d i i i= + = 2p

Rezultat final 24 mm=d 1p

c

Intensitatea luminoasă icircntr-un punct P aflat pe ecran este 2 2 cosPI I I= + ∆ϕ 1p

4p

2 π∆∆ϕ =

λr 2

3sdot λ

∆ = =y lr

D

Rezultă 2 3π

∆ϕ = astfel se obţine PI I=1p

Intensitatea este maximă dacă cos 1∆ϕ = deci max 4I I=

Rezultat final max 4P

II

= 1p


115

d

Lentila formează imagini reale ale fantelor dispozitivului Young aflate la o distanţă x2 faţă de lentilă 1

21

15 cm= =+

x fxx f

unde 1 30 cm= minusx

Figura de interferenţă aflată pe ecran se datorează imaginilor formate de lentilă Astfel distanţa de la noile

surse la ecran este 2 = minus minusD D d x iar distanţa dintre aceste surse este 2

1

22 sdot=

minusl xl

x

2p

4p

Noua interfranjă este 2Dil

λ= 1p

Rezultat final 093 mmi = 1p



116

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1

b8

3

m 3 10 s

1 al 631 10 ua

1 an

c

dc dt

t

= sdot = rArr = = sdot

∆ ∆ =

3

2 d 3

3

a

m m

vF m a m

t∆

= sdot = sdot∆

m5s

v v∆ = =

= 25 NF

3

4

c

fR N F= +

2 2 2 2 1 1 fR N F N m g= + = sdot + micro = sdot sdot + micro

Notacircnd cu θ unghiul pe care R

icircl face cu verticala tg fFN

θ = = micro

3

5

cAplicacircnd teorema de variaţie a energiei cinetice icircn cele două situaţii obţinem ecuaţiile

2

22 2

2 42

m v F dv v

m v F d

sdot= sdot primerArr = sdotprimesdot sdot = sdot sdot

3



a

1

1

0const2m

d va vt

+= rArr = =

∆ 2p

4p11

1

2 100 m sdv vt

minussdot= rArr = sdot

∆ 2p

b

22

2

50 sdv tt

= rArr ∆ =∆ 2p

4p

1 2 110 st t t t∆ = ∆ + ∆ rArr ∆ = 2p

c2 sin = 500 mh d h= sdot α rArr 3p 3p

d11 2 727 m sm m

d dv vt

minus+= rArr = sdot

∆ 4p 4p



117


a 20 JG GL m g h L= minus sdot sdot rArr = minus 3p 3p

b

Aplicacircnd teorema variaţiei energiei cinetice icircntre stările iniţială şi cea corespunzătoare icircnălţimii h2 2

0

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus

sdot sdot∆ = rArr minus = minus sdot sdot

= sdot sdot

sdot sdot = sdot

1p

2p

4p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

1p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p 1p

c

Notăm cu 1d distanţa parcursă de corp pe planul icircnclinat pacircnă la icircnălţimea h şi cu 2d distanţa parcursă icircn continuare pe porţiunea cu frecare pacircnă la oprirea pe planul icircnclinat

1 1 231 msin

hd d= rArr =α

1p

4pAplicăm teorema variaţiei energiei cinetice icircntre punctele A ndash aflat la icircnălţimea h şi B ndash icircn care corpul se opreşte

( )

2

2 2

2

2 2

0 sin - cos 2

237 m2 sin cos

ABAB R

m vEc L m g d m g d

vd dg

sdot∆ = rArr minus = minus sdot sdot sdot α micro sdot sdot sdot sdot α rArr

rArr = rArr =sdot sdot α + micro sdot α

2p

1 2 868 mD L d d D= + + rArr = 1p

d

2 sin 405 m 41 mH h d H= + sdot α rArr = cong 1p

4p

Energia potenţială gravitaţională a corpului aflat la icircnălţimea H este 405 Jp pE m g H E= sdot sdot rArr = 1p

2 2 1p

2 = 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr = 1p2 2

1p2

= 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr =

1p



118


Subiectul I Punctaj1 a Scriind expresia cantităţii de substanţă pentru cele două gaze

11

1 1 2

2 2 12

2

1 8

A

A

m NN N

m N NN

ν = = micro micro rArr = =microν = =

micro

3

2

d

0 V

Q U LL U C T

Q= ∆ +

rArr = minus ∆ = minus ν sdot sdot ∆=

1 mol 15 100 KVC R Tν = = sdot ∆ = minus

= 12465 JL

3

3

c

12 12

12 12

1213 13 13

13

12 13

31 2 32

5 61 3 80 1202 5

V

V

RC C RQ R T

Q C TQQ C T R T KQ

T T

= + = sdot rarr rArr = sdot ν sdot sdot ∆ = ν sdot sdot ∆ rarr = ν sdot sdot ∆ = sdot sdot ∆ = minus rArr = =∆ = ∆

3

4 d Exprimăm variaţia energiei interne a gazului icircn fiecare situaţie vU C T∆ = ν sdot sdot ∆ şi comparăm valorile acesteia ţinacircnd cont de relaţia dintre temperaturi determinată din grafic

3

5

c Scriem expresiile randamentului unui motor termic respectiv randamentul ciclului Carnot pentru situaţiile descrise icircn problemă

2

1

1C

p

LQ

TT

η =

η = minus

şi rezolvăm sistemul de ecuaţii obţinut

3



119


a 32 gm m mν = rArr = ν sdotmicro rArr =micro

3p 3p

b51

1 1 1 1 15 10 PaR Tp V R T p pV

ν sdot sdotsdot = ν sdot sdot rArr = rArr = sdot 3p 3p

c

1 N f N= sdot molecule disociază 2 f NrArr sdot sdot particule 1 2 frArr ν = sdot sdot ν - cantitatea de substanţă monoato-mică ( )2 1 N f N= minus sdot molecule rămacircn nedisociate ( )1 f NrArr minus sdot particule ( )2 1 frArr ν = minus sdot ν - cantitatea de substanţă biatomicăIcircn urma disocierii noua cantitate de substanţă din sistem devine ( )1 2 = 1 1f pprimeν = ν + ν ν sdot +

1p

5p

0const 0 0 1i f

QV L U U U pQ U L

= = rArr = rArr ∆ = rArr == ∆ +

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T

= ν sdot sdot = sdot primeprime= ν sdot sdot = sdot sdot ν sdot + ν sdot minus sdot sdot rArr= sdot = sdot

rArr ν sdot sdot sdot = sdot sdot ν sdot sdot + ν sdot minus sdot sdot sdot rArr

sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p 1p

d ( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa

p V R T p T pp V R T p T

Tp f p pT

p

primesdot = ν sdot sdot prime νrArr rArr sdotsdot = ν sdot sdot ν

= + sdot sdot

= sdot 1p

2p

4p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

1p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p 1p



a

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa

p V R T p T Tp pp V R T p T T

p

sdot = ν sdot sdot rArr = rArr = sdotsdot = ν sdot sdot

= sdot

1p

4p

Pentru procesul 2rarr3 2 1 1 3 p V p Vγ γsdot = sdot 1p

1

2 3 3 23

53 51 1 1 1

1

2 = 8 = 15

5 3

p

v

p V V pVp V V pVC

C

γγ

= rArr = rArr =

γ = =

1p

13 1 3

1

15 15 374 lR TV V Vp

ν sdot sdot= sdot = sdot rArr = 1p


120

b

2p

4p

( )12 2 1

12

37395 J

VQ C T TQ

= ν sdot sdot minus

=2p

c

1231 12 23 31 1pL L L L= + +

12 0 L =1p

4p

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

( )

( )

23 23 3 2

23 1 1 13 315 22 4

vL U C T T

L R T T R T

= minus∆ = minusν sdot sdot minus

= minus sdot ν sdot sdot sdot minus sdot = sdot ν sdot sdot

1p

23 186975 J 1pL =

( )31 1 3 105 L R T T R T= ν sdot sdot minus = minus sdot ν sdot sdot1p

31 12465 J 1pL = minus

1231 62325 JL =1p

d

2rarrArarr3 - proces izoterm urmat de unul izocor2rarrBrarr3 - proces izoterm urmat de unul izobar

2 3 2 3

2 3 2 3

A A A

B B B

Q Q QQ Q Q

= += +

32 2

2

lnAVQ R TV

= ν sdot sdot sdot

( )3 3 2A vQ C T T= ν sdot sdot minus

32 2 2

2 3

ln ln BB

V VQ R T R TV V

= ν sdot sdot sdot + ν sdot sdot sdot

( )3 3 2B pQ C T T= ν sdot sdot minus

2 3 2 3 B AQ Qgt ( )2 3 2 3 0 1 B AQ Qminus gt

Icircnlocuind expresiile de mai sus icircn relaţia (1) şi efectuacircnd calculele folosind datele din problemă obţinem

( )2 3 23

ln 0BVR T R T TV

ν sdot sdot sdot minus ν sdot sdot minus gt 2 2 1 2 12 2B Bp V p V V V V= rArr = =

( )2 3 2 2 3 23

1

2 3 2 3

4ln ln 3

42 ln 15 2 0 3

B

B A

VR T R T T R T T TV

R T

Q Q

ν sdot sdot sdot minus ν sdot sdot minus = ν sdot sdot sdot minus + =

= ν sdot sdot sdot sdot minus + gt rArr

rArr gt

3p 3p



121


Subiectul I Punctaj

1

b

1

2

21

2

156

lRl LLR

l lR lR L

= ρ sdotsdot

= ρ sdotsdot

= =

3

2

a U R I= sdot

2

lRmS R

m m d Sd ll S d S

= ρ sdot rArr = ρ sdotsdot= rArr =

sdot sdot

2 mU Id S

= ρ sdot sdotsdot

= 3 VU

3

3

d2 W R I t= sdot sdot ∆

2 WRI t

=sdot ∆

1 R = Ω

3

4

bAlegacircnd un ochi convenabil şi aplicacircnd teorema a II-a a lui Kirchhoff

1 2 3 4E E E E I R+ minus minus = sdot

02 AI =

3

5

b2

max 4

p

p

EP

r=

sdot

1 2

1 2

6 12 5p pr rr r

r rsdot

= rArr = Ω = Ω+ 1 2

1 2

26 Vpp

p

E E E Er r r

= + rArr =

max 14083 WP =

3



a

1

3 1 1

075 A

EI THORN IR R r

= =+ + 1p

3p1

1

1

45 C 1p

R

R

qI p

tq

=∆

=

1p1

1

1

45 C 1p

R

R

qI p

tq

=∆

= 1p

b

2

3 2 2

06 A

EI IR R r

prime prime= rArr =+ + 2p

4p2 2 1

84 1U E I r pU V p

prime= minus sdot=

1p2 2 1

84 1U E I r pU V p

prime= minus sdot= 1p


122

c

Ambele icircntrerupătoare fiind icircnchise sistemul celor două surse poate fi redus la o singură sursă de tensiune avacircnd polaritatea sursei 1 şi caracteristicile

( ) ( )( )

1 1 2 2

1 1 2 2

90 19p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1p

4p1 2

1 1 2 2

18 V 19

pp

p

E E E Er R r R r

= minus rArr =+ + 1p

3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= =

1p3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= = 1p

d

Consideracircnd că pe ramurile pe care se află cele două surse curentul electric circulă icircn sensul impus de către

acestea iar prin 3R curentul circulă icircn sensul impus de sursa echivalentă cu cele două surse şi aplicacircnd teoremele lui Kirchhoff obţinem

( )3

3

2 1

1 1 1 1

1

2

115 106

R

R

I I I

E I R r I

I AI A

+ =

= sdot + +

==

2p

4p

1 1 2 2 1 1883 1

AB

AB

U I R I R pU V p

= minus sdot minus sdot= minus

1p1 1 2 2 1

1883 1AB

AB

U I R I R pU V p

= minus sdot minus sdot= minus 1p



a

K icircn poziţia 11 2

1 2 1 2

085 A

E EI IR R r r

minus= rArr =

+ + +3p 3p

b


3 2 2

2 A 1p

EI IR R r

prime prime= rArr =+ +

1p

3p115 A 1I I I I pprime∆ = minus rArr ∆ = 1p

iar curentul icircşi schimbă sensul 1p


123

c

[ ] 015 mint isin ( ) 1 1 1

685 V

AB

AB

U E I R rU

= minus sdot +

=1p

6p

[ ] 1545 mint isin1 10 VAB ABU E U= rArr =

1p

[ ] 4560 mint isinNotăm cu I primeprime intensitatea curentului electric prin rezistorul cu rezistenţa R3Cele două surse sunt legate icircn paralel şi sunt echivalente cu o singură sursă avacircnd caracteristicile

( ) ( )( )

1 1 2 2

1 1 2 2

079

p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1 2

1 1 2 2

7 V

pp

p

E E E Er R r R r

= + rArr =+ +

3

25 A 1p

p

p

EI I

R rprimeprime primeprime= rArr =

+

1p

3 5 V 1pAB ABU I R Uprimeprime= sdot rArr = 1p

2p

d 3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W

prime primeprime= sdot sdot ∆ + sdot sdot ∆

=

2p3p3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


= 1p



124

DOPTICAtilde

Subiectul I Punctaj

1

b90 37i = deg minus α = deg

0 1 1 2 2sin sin sin n i n r n rsdot = sdot = sdot

02

2

sinsin n irnsdot

=

2sin 04r =

Folosind datele din table 2 237r = deg

3

2

cnotacircnd cu g factorul care ţine de geometria lentilei şi care nu se modifică prin introducerea lentilei icircn apă

1 0

1 1ln gf n

= minus sdot

2

1 1l

a

n gf n

= minus sdot

Raportacircnd cele două relaţii vom obţine( )

( )2

1

14a l

l a

n nff n n

sdot minus= =

minus

2 1 4f f= sdot

3

3 c 3

4

d

Scriind ecuaţia lui Einstein pentru cele două situaţii 21

1

22

2

2

2

extr

extr

m vL

m vL

sdotε = +

sdotε = +

şi prelucracircnd ecuaţiile obţinem 1 2 2v v= sdot

3

5

c

2

11 2

2k

Dx kl

λ sdot= sdot sdot

sdot 2

2 2k

Dx kl

λ sdot= sdot

sdot

22 1 k k

d x x= minus

( )1 2 2 - 02k Dd

lsdot

= sdot sdot λ λ =sdot

3



a n0 sdot sin i = n sdot sin r Icircnlocuind obţinem r = 30deg 3p 3p

b

n0 sdot sin 90deg = n sdot sin l 1p

4p

Icircnlocuind obţinem l = 45deg 1pIcircn condiţiile punctului a raza de lumină cade sub un unghi iprime = 60deg pe suprafaţa cilindrică laterală a fibrei optice Deoarece acest unghi este mai mare decacirct unghiul limită la trecerea luminii din fibra optică icircn aer lumina se reflectă total rămacircnacircnd icircn interiorul fibrei optice

Raza fibrei optice 0 1 cm2dr = =

0 01 1

1

sin 2 cmsin

r rr d dd r

= rArr = rArr =

1p

( )0 1 02 1 82 cmd N d d= sdot + sdot rArr = 1p


125

c

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv

= ∆ δ = rArr δ = sdot∆

=

δ = sdot

3p4p

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot 1p

d

60 30iα = deg rArr = deg 1p

4p

0 sin sin n i n rsdot = sdot (1) 1p

90 90r l l r+ = deg rArr = deg minus

0 0 sin 90 sin cos n n l n n rsdot deg = sdot rArr = sdot (2) 1p

Ridicacircnd la pătrat relaţiile (1) şi (2) şi adunacircndu-le obţinem 5 112

2n = = 1p



aPoziţia maximului de ordinul 3 pentru radiaţia 1 este dată de

1

13 1 3 3

2Dx ilλ

λ sdot= sdot = sdot

sdot 2p

3px3λ1

= 36 mm 1p

b

x4λ2

= 36 mm 1p

4p

x4λ2

=i2 1p

1 2 1 23 4 3 4i isdot = sdot rArr sdot λ = sdot λ 1p

12

2

3 4

450 nm 1p

sdot λλ =

λ =1p

c

3

1

3 1

34

13 3

4 3

4 2

3 360 nm2

DxlDxl

x x

λ

λ

λ λ

λ sdot = sdot sdot λ sdot = sdot rArr λ =

sdot =

3p4p

λ3 nu aparţine domeniului vizibil 1p

d1 1

14 4 4 6 mm

08 2Dx x

lλ λ

λ sdotprime prime= sdot rArr =sdot sdot

4p 4p



126

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj1 d 32 d 33 c 34 d 35 c 3



a

Corpul coboară uniform pe plan ndash aplicăm principiul II al mecanicii newtoniene

0N Gn Gt Ff+ + + =

1p

4pFăcacircnd proiecţia pe axe obţinem cosN mg= α sin cos 0 tgmg mgα minus micro α = rArr α = micro

2p

1p

b

Aplicăm principiul II al mecanicii newtoniene

Pentru corpul A A At nA AG N T Ff G m a+ + + + = sdot

Pentru corpul B B BG T m a+ = sdot

1p

4pFăcacircnd proiecţia pe axe obţinem B BT G m aminus = A AGt Ff T m aminus minus = 2p

Rezolvacircnd sistemul obţinem ( ) 2m2 sin cos 1 053 s3

ga = α minus micro α minus = 1p

c

Forţa de apăsare pe axul scripetelui este egală ca mărime şi de sens opus cu rezultanta celor două tensiuni

N T T= + 1p

4p( ) 306BT m a g N= + = 1p

( )( )2 22 1 cos 90N T= + minus α 1p

5911N N= 1p

d

Aplicăm teorema de variaţie a energiei cinetice pentru corpul B ∆Ec= LUnde L reprezintă lucru mecanic efectuat de rezultanta forţelor ce acţionează asupra punctului material 1p

3p( )2

2B B

B Bm v m gh m a g h= minus + + 1p

072m sBv 1p



127


a

LFf = minusFf ∙ d 1p

3pConform legilor a frecării Ff = μN 1p

LFf = minusμmgd = minus4 J 1p

b

Aplicăm legea conservării energiei mecanice pentru sistemul dat A BE E= 1p

4pEnergia mecanică icircn punctul A A A AE Ep Ec mgh= + = 1p

Energia mecanică icircn punctul B 2

2B

B B BmvE Ep Ec= + = 1p

Rezultă 632msBv = 1p

c

Conform definiţiei 2

2

cc

mvEc p mv= = 2p

4pPentru a calcula viteza icircn punctul C aplicăm teorema variaţiei energiei cinetice 2 2

2 2C B

Ffmv mv Lminus = 1p

Rezultă

6 m sCv = 12 m sCp = 36 JCEc = 1p

d

Aplicacircnd teorema energiei cinetice obţinem 2 2

max

2 2Cmv kx

=

Variaţia energiei cinetice este egală cu lucru mecanic efectuat de forţa elastică2p

4p

max cmx vk

= 1p

Rezultă xmax= 06 m 1p



128


Subiectul I Punctaj1 c 32 c 33 b 34 a 35 a 3



aConform definiţiei He

HemNamicro

= 2p3p

Rezultă mHe = 066 middot 10-26 kg 1p

b

Ecuaţia de stare pentru incinta 1 p1V1 = ν1RT1 1p

4pEcuaţia de stare pentru incinta 2 p2V2 = ν2RT2 1pRezultă ν1 = 04 mol 1pRezultă ν2 = 12 mol 1p

c

Gazul va trece din incinta (2) icircn incinta (1) (p2 gt p1) pacircnă cacircnd se va ajunge la aceeaşi presiune prsquo icircn ambele incinte şi aplicacircnd conservarea numărului de moliν 1 + ν2 = ν1prime + ν2prime

1p

4p

Ecuaţia de stare pentru incinta 1 finală pprimeV1 = ν1primeRT1 1pEcuaţia de stare pentru incinta 2 finalăpprimeV2 = ν2primeRT2

1p

Obţinem

1 1 1

2 2 2 1 2

1 2 1 2

047 mol 113 molV TV T

ν= ν =gt ν = ν =

ν + ν = ν + ν

rArr ν1prime = 047 moli ν2prime = 113 moli 1p

d

Variaţia energiei interne ΔU = ΔU1 + ΔU2 1p

4pΔU = T1Cv(ν1rsquo ndash ν 1) + Cv (ν2rsquo T2 ndash ν2T1) 2p

ΔU = minus705 J 1pTOTAL pentru Subiectul al II-lea 15p


a

Pentru procesul izocor 1rarr2 1 2

1 2

p pT T

= 1p

4pPentru procesul izocor 3rarr4 2 1

3 4

p pT T

= 1p

Din cele două relaţii şi ţinacircnd cont de faptul că T2 = T4 1p

T2 = T4 = 1 3 360KTT = 1p

b

LTOT = L12 + L23 + L34 + L41 1p

3pLTOT = νR(T1 + T3 ndash 2T2) 2pLTOT = 997 J 1p


129

c

TOT

p

LQ

η = 1p

4pQp = Q12+Q23 1pQp=νCv(T2 minus T1 ) + νCp(T3 minus T2) 1p

Rezultă Qp = 2214 Jη = 45 1p

d

min

max

1 TT

η = minus 1p

3p1

3

1 TT

η = minus 1p

30η = 1p



130


Subiectul I Punctaj1 a 32 a 33 b 34 b 35 b 3



a

Cele 3 surse legate icircn paralel pot fi icircnlocuite cu o sursă echivalentă de tensiunea electromotoare Ep şi rezis-tenţa internă rp

1 2 3

1 2 3

1 2 3

36V1 1 1

E E Er r rEp

r r r

+ += =

+ +

1p

3p

1 2 3

1 1 1 1 1pp

rr r r r

= + + =gt = Ω 1p

Aplicacircnd legea lui Ohm pentru un circuit simplu 189p

p

EI

R r= =

+A 1p

b

Aplicam legile lui Kirchhoff

1 2

1 1 1

2 2 2

I I IE I r I RE I r I R

= +

= sdot + sdot = sdot + sdot

2p

5p

Rezolvacircnd sistemul obţinem ( )1 2 1 2

11 2 1 2

E r E E RI

r R r R r rsdot + minus sdot

=sdot + sdot + sdot

2 1 2 12

1 2 1 2

( )E r E E RIr R r R r r

+ minus=

+ + 2p

Icircnlocuind valorile numerice rezultă I1 = 4 A I2 = minus2A (I2 este negativă deci sensul curentului prin latură este invers decacirct cel considerat) 1p

c

Deoarece

ABU I R= sdot

1 2I I I= + 1p

4p1 2

1 2

1 2 3

1 1 1AB

E Er rU

r r r

+=

+ +2p

UAB = 40 V 1p

d

Icircn relaţia

1 2

1 2

1 2

1 1 1AB

E Er rU

r r R

+=

+ +1p

3p

Impunem condiţia vR rarr infin sau 1 0

vRrarr 1p

Rezultă 40VABU = 1p



131


a

Conform legii I a lui Kirchhoff I = I1 + I2 1p

4p

Din datele problemei

Q1 = f sdot Q = 1U I tsdot sdot rArr 2f QIU t

sdot=

sdot

Q2 = (1 ndash f ) Q = U middot I2 middot t rArr 2(1 )f QI

U tminus sdot

=sdot

2p

5AQIU t

= =sdot

1p

b

Q = I2 Rp middot t 2p

4pRezultă 2pQRI t

= 1p

Rp = 22 Ω 1p

cE = U + I middot r 2p

3pE = 120 V 1p

d

Pentru ca puterea debitată de sursă icircn circuitul exterior să fie maximă trebuie ca rezistenţa circuitului exte-rior să fie egală cu rezistenţa internă a surseiRex= rAcest lucru este posibil dacă legăm rezistorul icircn paralel cu cele două rezistenţe

x p

x p

R Rr

R R=

+

2p

4p

px

p

R rR

R r=

minus 1p

R s = 22 Ω 1pTOTAL pentru Subiectul al III-lea 15p


132

DOPTICAtilde

Subiectul I Punctaj1 a 32 b 33 a 34 d 35 b 3



a

Aplicăm ecuaţia lui Einstein şi ţinem cont de relaţia dintre energia cinetică maximă a fotoelectronilor emişi

şi tensiunea de stopare 2

2 smv e U= sdot 1p

4p11

22

S ex

S ex

ch eU L

ch eU L

= + λ = + λ

2p

Rezolvacircnd sistemul obţinem h = 64 sdot1034 Js 1p

b

Aplicăm ecuaţia lui Einstein

11

S exch eU L= +λ

2p

4p

11

ex ShcL eU= minusλ 1p

Rezultă Lex = 289 middot 10-19 J 1p

c

Conform legii a treia a efectului fotoelectric extern h 0 exLυ = 2p

4p0exLh

υ = 1p

15 10 045 10 sminusυ = sdot 1p

d2

2

cW Nh Nh= υ =λ 2p

3p

Rezultă N = 26 middot 1015 cuante 1p



a1 2

11 11e

m

fnn R R

=

minus minus

2p3p

Rezultă f = 40 cm 1p

b

Aplicăm formula punctelor conjugate ţinacircnd cont de expresia distanţei dintre obiect şi imagine

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2p

4pDistanţa d este minimă pentru acea valoare a lui x1 pentru care derivata lui d icircn raport cu x1 este egală cu 0dprime(x) = 0 rArr x1 = 2f 1p

minus x1 = 80 cm 1p


133

c


1 22 1

1 1 1d x xx x f

= minus + minus =

1 22 1

1 1 1d x xx x f

= minus + minus = 1p

4pExpresia măririi liniar transversale 2 2

11

1y xy x

β = = = minusminus

2p

Rezolvacircnd sistemul obţinem 1 280cm 80cmx x= minus = 1p

d

Sistem optic centrat cu distanţa focală F

1 2

2

1

2 1

1 1 1

2

1 1 1

F f fxx

x x F

= +

minusβ = =

minus

minus =

2p4p

Rezultă f2= minus533 cm 2pTOTAL pentru Subiectul al III-lea 15p


134

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj1 a 32 a 33 c 34 b 35 c 3



atg 1

3tg ϕ = micro = 2p

3p

30ϕ = 1p

b

t fG Fa

mminus

= 1p

4psin cosmg mga

mα minus micro α

= 1p

a = g(sin α minus μcos α) 1p

2310 =577ms3

a = 1p

c

Notăm forța suplimentară cu F

Sensul ei este opus lui tG

1p

4pt fG F F= + 1p

t fF G F ma= minus = 1p

310 3 10 100 N3

F = sdot = 1p

d



1p

4p

t fG F F+ = 1p

sintG mg= α 1p

3 1 1 100 3 100 3 200 N2 23

F = sdot + sdot sdot = 1p



135


a

sin 5

1000m54 =15ms3600s

hpl

v

= = α =

= sdot1p

5pLa coboracircre sintg

cosUcirc α

α = micro = microα

rArr sintg cos

Ucirc αα = micro = micro

α1p

La urcare P = Ft vFt = Gt + Ff pentru că v este constant 1p

(sin cos ) (sin sin ) 2 sintF mg mg mg= α + micro α = α + α = α 1p

4 352 sin 2 10 15 15 10 W=15kW100

P mgv= α = sdot sdot sdot = sdot 1p

b

tP F v= sdot t fF F mg= = micro 2p

4psincos

P mgv mgvα= micro =

α1p

7500W=75kWP 1p

c cos(90 ) singravitP G v G v G v= sdot = sdot sdot minus α = sdot sdot α

2p3p

7500W =75kWgravitP = 1p

dsinpE E mgh mgl∆ = ∆ = = α 2p

3p25kJpE E∆ = ∆ = 1p



136


Subiectul I Punctaj1 c 32 b 33 d 34 d 35 a 3



a

1

1 1 2 1 21

22 1 2 2 1

2

2 540 2 54 4 083 450 3 45 5

pVV T V TRT

pV T V V TRT

ν= = sdot = sdot = sdot = sdot = =

ν2p

3p

1

2

4 085

ν= =

ν 1p

b

1

1 1

22 2

p VVRT

p V VRT

ν= =

ν 2p

3p

1

2

08VV

= 1p

c

1 1 11 2 1 1 2 2

2 2 2

( ) ( )pV RT

p V V R T TpV RT

= ν rArr + = ν + ν= ν

1p

5p

1 1

1 2 1 22 2

( ) ( )

p V RTp V V RT

p V RT

= ν rArr + = ν + ν= ν

1p

1 1 2 2 2 2

1 2 2

5 08 450 540 ( ) 3 18

p T Tp T T

ν + ν ν sdot + ν sdot= hArr =

ν + ν sdot ν 2p

o300K 27 CT t= rArr = 1p

d

1 36lV =

12

2

2 3 36 3 18 54l3 2

V VV

sdot= rArr = = sdot =

1p

4p 1 2 1 2

1

1 22

9

08 08

V V V V lV V VV

+ = + =

= rArr =2p

2 2 2 2 108 9 18 9 5l 4lV V V V V+ = hArr = hArr = rArr = 1p



137


a 4p 4p

b

212 1

1

41 1 4 1 4

ln 0

5( - ) ( ) 02V

VQ RTV

RQ C T T T T

= ν gt

= ν = ν minus gt2p

4p

12 41 229356JprimitQ Q Q= + = 2p

c

23 4 1 4 1

134 4

2

5( ) ( ) 02

ln 0

VRQ C T T T T

VQ RTV

= ν minus = ν minus = lt

= ν = lt2p

4p

23 34 159552JcedatQ Q Q= + = minus 2p

d1 cedat

primit

QQ

η = minus 2p3p

3043η = 1p



138


Subiectul I Punctaj1 a 32 b 33 a 34 b 35 d 3



a

2AP R I= sdot 2p

3p25 WP = 1p

b

0

0 0

0

A

A

A

I I IxI R I Rl

l xE I R I Rl

= + = minus

= +

3p5p

05 mx = 2p

c

ee

E EI RR I

= rArr = 1p

4p0

00

05A

A

A

I I IR lI Ix R

= +sdot

= =sdot

2p

151 151eI A R= = = Ω 1p

dW U I t E I t= sdot sdot ∆ = sdot sdot ∆ 2p

3p2250Wh 225 kWhW = = 1p



139


a

2 40 2 60 200WtotalP = sdot + sdot = 2p

3p200 4A50

totaltotal

PP U I IU

= sdot rArr = = = 1p

b

Legacircnd un voltmetru ideal ( )VR rarr infin icircntre A şi B putem aplica legea a II-a Kirchhoff pentru ochiul CABC

1 1 2 20 ABI R U I R= + minus 1p

5p

Dar 1 2 2II I= = datorită simetriei celor două ramuri CAD şi CBD

2 1( )2ABIU R R= minus

1p

21

1 1 1 2

22

2 2 2 2

40 102 4

2

60 152 4

2

I PP R RI

I PP R RI

= rArr = = = Ω

= rArr = = = Ω

2p

10VABU = 1p

ctotal totalW P t= sdot ∆ 2p

3p472 10 720 kJtotalW J= sdot = 1p

d

1

1 01 1 3

0

191 1800 C

5 10

RR Rt tR minus

minus= + α rArr = = =

α sdot

2p

4p

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot= t2 =

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot

2p



140

DOPTICAtilde

Subiectul I Punctaj1 c 32 b 33 a 34 c 35 d 3



a

1 2

1 1 1( 1) 8 cmaeraer

n ff R R

= minus minus rArr = minus

2p

4p

1 2

1 1 11 32cmapaapa apa

n ff n R R


2p

b

2 1

1 1 1

aerx x f= + 1p

4p2

24 cm 48cm5

x = minus = minus 1p

2

1

xx

β = 1p

04β = 1p

c

2 1

1 1 1

apax x f= + 1p

4p

2

96 cm 192cm5


21

xx

β = 1p

04β = 1p

d2p

3p

Imaginea este virtuală dreaptă și micșorată 1p



141


a

412 10 m 12mm2Dil

minusλ= = sdot = 2p

3p

0 0 0mmx i= sdot = 1p

b

λapăapatilde nλ

λ = 1p

3piapă

12 3 09mm4apatilde

iin

= = sdot = 1p

xapă= 0 ∙ iapă = 0 mm 1p

c

Introducerea unui strat subţire (lamă film peliculă) icircn calea unuia din fasciculele luminoase care interferă conduce la deplasarea figurii de interferenţă spre acel fasciculDe exemplu dacă se introduce o lamă transparentă cu grosimea e şi indicele de refracţie n icircn calea fascicu-lului 1 atunci noua diferenţă de drum optic icircntre razele care interferă este

2 1 2 1) ( 1)( e ne e nr r r r rminus + minus minus∆ = minus = minus

2p

5pDeci icircn centrul ecranului unde diferenţa r2-r1 este nulă apare o diferenţă de drum suplimentară e(n-1) Punacircnd condiţia de maxim găsim noul ordin al maximului plasat icircn centrul ecranului

( 1)( 1) nou noue ne n k k minus

minus = sdot λ rArr =λ

2p

10nouk = 1p

d

Figura de interferenţă are aceeaşi interfranjă ca şi icircn cazul a) deci i =12 mm 2p

4pFigura de interferenţă este deplasată icircn sensul fasciculului acoperit deci noul maxim central va avea abscisa

0 nou 10 12 12mmnoux x k i= ∆ = sdot = sdot = 2p



142

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj1 c 32 a 3

3

Viteza medie este 1 2

md dvt t t

= =+

Cum 112 6

d dtv v

= = şi 222 2

d dtv v

= =

6 154

6 2

md vv vd d

v v

rArr = = =+

a

3

4

Calculăm acceleraţia din reprezentarea grafică 28 msvat

∆= = minus

∆

Calculăm acceleraţia la urcare pe planul icircnclinat fG F N ma+ + =

OX t fG F maminus minus =

OY 0 cosnN G N mgminus = rArr = α

Cum cosf fF N F mg= micro rArr = micro α

sin cos (sin cos )mg mg ma a grArr minus α minus micro α = rArr = minus α + micro α

Icircnlocuind 35

rArr micro =c

3

5

Aplicăm legea conservării energiei mecanice 2

11 2

2mvmgh v gh= rArr =

Cum 2 1 2 2v kv v k gh= rArr = Variaţia impulsului este

2 1 2 1( ) ( 1) 2 16 Ns 800 Nmpp p p mv mv m k gh Ft

∆∆ = minus = minus minus = + = = =

∆ b

3



a

Reprezentarea forţelor care acţionează icircn sistem

3p 3p

b

Lucrul mecanic util Lu = mgh 1p

4p

Lucrul mecanic consumat Lc = T l

T = 2F rArr Lc = 2Fl 1p

Randamentul este 2

u

c

L mghL Fl

η = = 1p

Rezultă forţa de tracţiune 500 N2mghF

l= =

η1p


143

c

Aplicăm principiul fundamental al dinamicii ţinacircnd cont că 0v ct a= rArr =

0fG T N F+ + + =

pe Ox sin 0 sinf fT F mg F T mgminus minus α = rArr = minus α

2p

5p12 2T T F= = iar sin h

lα =

Rezultă 2fhF F mgl

= minus

2p

600 NfF = 1p

dcos

cosf

f f

FF N F mg

mg= micro rArr = micro α rArr micro =

α 2p3p

065rArr micro = 1p



a

Reprezentarea forţele care acţionează asupra corpului pe porţiunea orizontală şi aplicarea teorema variaţiei energiei cinetice

cE L∆ = 1p

4p

Lucrul mecanic total L = LG + LN + LFf

dar LG = 0 LN = 0 iar LFf = minusFf d

1p

Ff = μN = μmg

Rezultă 2 2

200 2 21 ms 458ms

2 2mv mv mgd v v gdminus = minusmicro rArr = minus micro = =

2p

b

Pe porţiunea MN mişcarea avacircnd loc fără frecare aplicăm conservarea energiei mecanice

M N CM PM CN PNE E E E E E= rArr + = + Considerăm M nivel de referinţă 0PME =

1p

4pNotăm cu 1v viteza corpului icircn punctul N rezultă 2 2

1

2 2mv mv mgR= + 1p

Viteza icircn N este 21 2v v gR= minus 1p

1 11 ms 331msv = = 1p

c

Pentru icircnălţimea maximă la care ar putea să ajungă corpul faţă de punctul M aplicăm legea conservării

energiei mecanice 2 2

max max2 2mv vmgh h

g= rArr =

2p3p

max 105msh = 1p

d

Relaţia dintre cele două icircnălţimi este 2

max 081 m2vh fh fg

= = = 1p

4pNotăm cu 2v viteza corpului la icircnălţimea h şi aplicăm legea conservării energiei mecanice faţă de punctul M

( nivel de referinţă) Ţinacircnd cont că 0PME = rezultă 2 2

222 2

2 2mv mv mgh v v gh= + rArr = minus

2p

2 204 v m s= 1p



144


Subiectul I Punctaj1 b 32 b 33 c 3

4Calculăm lucrul mecanic pe cele două transformări prin metoda grafică

1 2 2 1 1 21 2 2 3 2 3 2

2 3

( )( ) 150 J ( ) 200J 0752

p p V V LL L p V VL

rarrrarr rarr

rarr

+ minus= = = minus = =

c

3

5

Randamentul ciclului Carnot min

max

1CTT

η = minus

Utilizacircnd formula randamentului termic min

1 max 1

1L T LQ T Q

η = rArr minus =

Din 1 06 16Q L L L= + = rezultă maxmin

3 150 K8

TT = =d

3



a

Scriem legea transformării izobare pentru fiecare gaz

1 1

0 1

V VT T

= şi

2 2

0 2

V VT T

= 1p

4p

dar 1 2V V V+ = şi 2 1 12

3VV V V= rArr = şi 2

23VV = iar

1 2 2VV V= = 1p

Icircnlocuind rezultă 01

32TT = şi 0

234TT = 1p

Rezultă 1 4095KT = şi 2 20475KT = 1p

b

Scriem legea transformării izoterme pentru fiecare gaz 0 1 1 1 0 2 2 2p V p V p V p V= = 1p

4p

1 1 ( )

2 2l lV S V x S= = +

1 ( )2lV x S= + şi respectiv

2 2 ( )2 2l lV S V x S= = minus 1p

Obţinem 50 00 1 1 2

500 2 2 0 2

2 N( ) 066 102 2 2 3 m

N( ) 2 2 102 2 2 m

l l p l pp S p x S pl x

l l p lp S p x S p pl x

= + rArr = = = sdot+

= minus rArr = = = sdotminus

1p

51 2

52 2

N066 10m

N2 10m

p

p

= sdot

= sdot1p

c 2 1( )F p p S= minus 2p3p

536 NF = 1p

d

Scriem ecuaţiile de stare sub forma 10 1 1

mp V RT=micro

şi 2 2 2 20 2 2

1 1 1

m V m Tp V RTV m T

= rArr =micro 2p

5p

Dar 0 01 2

3 32 4T TT T= = şi 1 2

23 3V VV V= =

1p

2 2

1 1

2m Vm V

rArr = 1p

2

1

2mm

rArr = 1p



145


a

Identificăm transformările1 2rarr şi 3 4rarr sunt transformări izobare2 3rarr şi 4 1rarr sunt trandformări izocore

1p

3p

Reprezentările ciclului icircn coordonate (p V) şi (V T)

2p

b

Aplicăm legea transformării izobare pe 1 2rarr 1 2 1 12 1

1 2 1 2

3 3V V V V T TT T T T

= rArr = rArr = 1p

4pAplicăm legea transformării izocore pe 2 3rarr 2 3

2 3

p pT T

=

Cum 3 1T T= (din reprezentarea grafică) 1 3 1

31 13 3

p p ppT T

rArr = rArr =

2p

5 23 10 Nmp = 1p

c

Energia internă 1 2 2 1 1( ) 2v vU C T T C Trarr∆ = υ minus = υ 1p

4p1 1

32 32

U RT RTrArr ∆ = υ = υ 1p

1 1 1 1 2 1 13p V RT U p Vrarr= υ rArr ∆ = 1p

1 2 2700JU rarr∆ = 1p

d

1

LQ

η =

Lucrul mecanic total schimbat de sistem este 1 4 2 1( )( )L p p V V= minus minus dar 14 3 3

pp p= = şi 2 13V V= 1 143p VLrArr =

1p

4pCăldura primită este 1 1 2 4 1 2 1 1 4( ) ( )p vQ Q Q C T T C T Trarr rarr= + = υ minus + υ minus

52p vC C R R= + = şi aplicacircnd legea transformării izocore pe 4 1rarr 4 1 3 1 1

44 1 4 1 3

p p p p TTT T T T

= hArr = rArr =

Rezultă 1 1 1 1 1 15 6 6Q RT RT RT p V= υ + υ = υ =

2p

Randamentul este 1 1

1 1

4 2 02218 9

p Vp V

η = = = 22rArr η = 1p



146


Subiectul I Punctaj1 d 32 b 33 b 3

41 2sR R= 1 1 1 2

2 3pp

RRR R R

= + rArr = 22 53 3sR RR R= + =

1 1 23AB

RR R

= +

5 1258ABRRrArr = = Ω

b

3

5 a 3TOTAL pentru Subiectul I 15p


a

1 2R R R R= + = 1p

4p

22 1

1 1 1 3 22 3

RRR R R R

= + = rArr = 1p

3 22 53 3R RR R R R= + = + = 1p

3

1 1 1 3 1 85 5

5 6258

AB

AB

R R R R R RRR

= + = + =

= = Ω1p

b

Tensiunea la bornele sursei 32AABAB

UU IR IR

= rArr = = 2p

4pAplicăm legea lui Ohm pe icircntregul circuit

( ) 264ABAB

EI E I R r VR r

= rArr = + =+

2p

c

ABRprime rezistenţa echivalentă a circuitului icircn condiţiile icircn care 2R R=

10 1258ABRRprimerArr = = Ω 1p

4pCalculăm intensitatea I aplicacircnd legea lui Ohm pe icircntregul circuit 18A

AB

EIR r

= =+

1p

Tensiunea la bornele sursei devine 225 VABU I R= = 1p

Relaţia dintre cele două tensiuni este

0125 125U UU U fU f fUminus

= + rArr = = rArr = creşterea de tensiune 1p

d

2

max 4EP

r= 2p

3p

max 8712WP = 1p



147


a

Reprezentăm intensităţile curenţilor pe ramuri şi ale-gem sensurile de parcursAplicăm legea I a lui Kirchhoff pentru nodurile N şi B

2 4 1

3 4 5

00

I I II I Iminus + minus =

minus minus =

2p

6pAplicăm legea a II-a a lui Kirchhoff pentru ochiurile de reţea

1 3 1 1 3 1 2 2

2 3 2 4 4 2 2

2 3 2 5 5

( )E E I R R r I RE I r I R I RE I r I R

minus = + + minus= + += +

3p

Rezultă 1 2 3 4 515 05 5 2 3I A I A I A I A I A= = = = = 1p

b1 1 3 3( )MNU I R R E= minus + minus 3p

4p19MNU V= minus 1p

c

24 4 4P I R= 2p

3p4 8P W= 1p

d2

1 1 1W I R t= 2p3p

1 10800 J 108 kJW = = 1p



148

DOPTICAtilde

Subiectul I Punctaj1

680nma a aa

nnλ

λ = rArr λ = λ =

d 3

2 c 3

3

Notăm cu P poziţia ochiului Din construcţia imaginii aplicăm legile reflexiei se observă că PO OO= imaginea şi obiectul sunt simetrice 2PO POrArr =

Din asemănarea triunghiurilor PMO şi PAprimeOprime rezultă

2MO PO AOMOAO PO

= rArr =

2ON PO O BONO B PO

= rArr =

Icircnălţimea oglinzii

=09 m2 2

A B Hh MO ON h= + = = rArr

b

3

4

Maximul luminos de ordinul k se formează la distanţa 2k

k Dxl

λ=

Pentru primele două franje luminoase avem 1

2

1 1 mm222 2 mm2

Dk xl

Dk xl

λ= rArr = =

λ= rArr = =

c3

5

Aplicăm formula lentilelor subţiri 1 2

2 1 1 2

1 1 1 x xfx x f x x

minus = rArr =minus

Din formula mărimii transversale 22 1

1

x x xx

β = rArr = β 1 1

11 1

x xf βrArr = =

minus β minusβ

Din grafic 11 1 20 cmx= minus = minusβ

20 10 cm1 1

f minusrArr = =

minus minusb

3



a

Avem o lentilă plan convexă R1 rarr infin R2 = minusR 1p

4pFormula convergenţei

1 2

1 1 1 1( 1)( ) n nC n RR R R C

minus minus= minus minus = rArr = 2p

R = 10 cm 1p

b

Din 1 1 20C f cmf C

= rArr = = cm 1p

5pAplicăm formula lentilelor subţiri 12

2 1 1

1 1 1 fxxx x f f x

minus = rArr =+

2p

Aplicăm formula lentilelor subţiri 220( 40) 40 cm20 40

x minus= =

minus2p


149

c1 2d x x= minus +

2p3p

( 40) 40 80cmd = minus minus + = 1p

d


1 1

y xy x

β = = 1p

3p22 1

1

xy yx

= 1p

2 2 cmy = minus

Imaginea este reală răsturnată egală cu obiectul 1p



a

Scriem ecuaţia lui Einstein pentru cele două radiaţii

1 11 1

2 12 2

( )(1 )

c ch L eU h L eU

c ch L eU h L eU f

= + minus = λ λ = = + minus = minus λ λ

2p

5p

11

1

2

08

ch LeU

c eUh L

minusλ

=minus

λ

1 2

1 2

0802

L hc λ minus λrArr =

λ λ2p

2166 10 041L E eVminus= sdot = 2166 10 041L E eVminus= sdot = 1p


LL hh

= sdot ν rArr ν = 2p3p

1410 HzL = 1p

cEnergia unui foton corespunzător radiaţiei λ1 1 1

1

ch hε = ν =λ

2p3p

191 011 10 J 006eVminusε = sdot = 1p

d

Relaţia dintre energia cinetică maximă şi tensiunea de stopare2

11

22

2

2

2

e

e

m v eU

m v eU

=

=

2p

4p1 1 1

2 2 108v U Uv U U

= = 1p

1

2

54

vv

rArr = 1p



150

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj1 b 32 b 33 d 34 a 35 b 3



a

Condiţia de icircntacirclnire 1 2x x= 1p

4p2

1 2atx = 2 0x v t= 2p

rezultat final 02 8svt ta

= = 1p

b0v v at= + 2p

3p16msv = 1p

c

0 = + fracircnare fracircnarev a t 2p

3prezultat final

-=fracircnare

fracircnare

va

t 2-4 fracircnarea m s=minus4ms2 1p

d

1 2= +d d d 2p

5p1 0=d v t 1 1 2= = = 64d x x m 64 m 2p

2 = oprired d 2 2= + 2oprire franare oprirev v a d 2

2 2 fracircnare

vdaminus

= 2 = 32d m

rezultat final d = 96m1p



aE mgh= 2p

3prezultat final E = 4500 J 1p

b

c totalE L∆ = 1p

4p

2

2cmvE∆ = = +

ftotal G FL L L 1p

=GL mgh ∙ cos ∙sinfF f

hL F l mg= minus = minusmicro αα

1p

rezultat final 2 (1 middotctg )v gh= minus micro α v cong 13 ms 1p


151

c

c totalE L∆ =

1p

4p 0cE∆ = GL mgh=

f

totaltotal G FL L L= + 2p

rezultat final = -f

totalFL mgh = -4500

f

totalFL J 1p

d

= +f f f

total orizontalatildeF F FL L L 1p

4p2f

A Bf forizontalatilde

F oprire

F FL d

+= minus 1p

01 0Af A AF mg x mg= micro = = 01∙ 01∙B

f B B oprireF mg x mg d mg= micro = = 1p

rezultat final ( )2 1- middot01oprire

hd ctg= micro α 13oprired m 1p



152


Subiectul I Punctaj1 b 32 a 33 b 34 d 35 d 3



a

pV RT= ν 1p

4p1 1 1 2

2 2 2 1

p mp m

ν micro= = sdot

ν micro2p

rezultat final 1

2

74

pp

= 1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p

127 amestec g molmicro = 1p

c

0

2p V RT= ν 0 f finalp V RT= ν 1p

4pVf = 2 V 1p

rezultat final 4fT T= 1200fT K= 2p

d

Icircn decursul transformării gazul parcurge o icircncălzire izocoră pacircnă cacircnd presiunea devine p0 apoi se dilată izobar pacircnă la volumul 2V

0 0middot middot 2L p V p V RT= ∆ = = ν

rezultat final 196L kJ=

4p 4p



a1 1

1 1 11

p Vp V RTRT

= ν rArr ν = 2p3p

rezultat final v = 12 moli 1p

b

ΔU12 = vCVΔT12 = vCV(T2 minus T1) 1p

4p

1 2

1 2 12 1

2 1

34

V

p pV ctT T U C T

T T

= rArr =rArr ∆ = ν

=2p

rezultat final 12 1 1 129 18kJ2

U p V U∆ = rArr ∆ = 1p


153

c

min

max

1 TT

η = minus 1p

4pmin 1=T T 1p

max 2 1= = 4T T T 1p


η = = 1p

d

P

LQ

η = 1p

4p

1-2 2-3pQ Q Q= + 1-2 2 1( )vQ C T T= ν minus 32 3 2

1

ln VQ RTVminus = ν 1p

1-2 2-3 3-1L L L L= + + 1-2 0L = 32-3 2

1

ln VL RTV

= ν 3-1 1 3( )L R T T= ν minus 1p

rezultat final 8ln 2 38ln 2 45

minusη =

+ 25η 1p



154


Subiectul I Punctaj1 d 32 d 33 c 34 a 35 b 3



a

Aplicacircnd teorema a doua a lui Kirchhoff pentru un ochi de reţea se obţine relaţia

( )1 1 1 3E I R R= + de unde 1 3AI =2p

4pTensiunea electrică UAM = R1I1 1prezultat final UAM = 3 V 1p

b

Aplicacircnd teorema a doua a lui Kirchhoff pentru alt ochi de reţea se obţine relaţiaE1 = I2(R2 + R4) de unde I2 = 4 A 2p

4pEnergia electrică consumată de rezistori tIRRtIRRW ∆sdot++∆sdot+= 2

2422

131 )()( 1p

rezultat final W = 3024 103 J 1p

c

Aplicacircnd teorema a doua a lui Kirchhoff pentru al III-a ochi de reţea se obţineE2 = I1R3 minus I2R4

2p4p

rezultat final E2 = 5 V 2p

d

Aplicacircnd teorema a doua a lui Kirchhoff pentru ochiul format de sursa a doua şi voltmetrul ideal se obţine E2 = Uvoltmetru

2p3p

Tensiunea indicată de voltmetrul ideal este egală cu tensiunea electromotoare a sursei = 5U V 1p



a

Puterea electrică consumată de cele două becuri se poate scrie P1 + P2 = U I unde U este tensiunea la bornele becurilor iar I este intensitatea curentului electric prin sursă 1p

4pPentru circuit se aplică teorema a doua a lui Kirchhoff E = I (R + r) + U 1pSe obţine ecuaţia I 2 minus 10I + 25 = 0 care are soluţia I = 5 A 2p

bP1 + P2 = Rp I

2 2p3p


c

echivalent

echivalent

RR r

η =+ 2p

4p= +echivalent pR R R 1p

rezultat final η cong 89 1p

d

= middotbaterieP E I 1p

4p

Prin conectarea ampermetrului ideal la bornele becurilor curentul electric nu va mai trece prin becuri(scurtcircuit) astfel 1p

EIR r

=+

1p

rezultat final = 240baterieP W 1p



155

DOPTICAtilde

Subiectul I Punctaj1 c 32 b 33 b 34 a 35 d 3


Subiectul al II-lea Parţial Punctaja Construcţia imaginii 3p 3p

b

2 1

1 1 1x x f

minus = 1p

4p2

1

xx

β = 1p

rezultate finale x2 = 60 cm 2β = minus 2p

c

21 CCCsistem += 1p

4pdioptriiCsistem 52= 1p

2 1

1 1sistemC

x xminus = 1p

rezultat final cmx 1202 minus= cm 1p

d

1 2= -x d x 2p

4p

1 22

1 2

x fxx f

=+

1p

rezultat final 2 = -20x cm cm 1p



a

Distanţa dintre două maxime sau două minime succesive este i1 1p

3p11 10

yi = 1p

rezultat final i1 = 06 mm 1p

b

Distanţa de la maximul central la primul minim de interferenţă este 1

2i 1p

4p1

12 22id i= + rArr d = 3i1

2p

rezultat final d = 18 mm 1p


156

c

22 29

2iy i= + 2

22 072 mm19yi = = 1p

4pPentru acelaşi dispozitiv Young 1

1 2Dil

λ= 2

2 2Dil

λ=

1 22

1

ii

λλ = 1p

rezultat final 2 720 nmλ = 1p

d

Pentru dispozitivul Young introdus icircntr-un mediu optic interfranja devine 11

iin

= 2p

4pDar interfranja scade cu 25 deci

1 1

1

-025 i ii

= 1p

rezultat final4 1333

n = 1p



157

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj1 b 32 b 33 a 34 b 3

5

21 1 2 2

12 1 2

2

2 2m 1156s

m v m v

mv v vm

=

= =3



a 3p 3p

b

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s

m N G T Ff m aOy N G N m gOx Ff m a

m G T m am mOy G T m a m g T m a a gm m

a

+ + + = sdotminus = rArr = sdot

minus =

+ = sdotminus micro

minus = rArr minus = rArr =+

=

1p

4p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

c2 2

08 NT m g m aT

= minus=

2p3p2 2

08 NT m g m aT

= minus= 1p


158

d

1 0 1 1 0

1 1 0

1 1 0

( ) 0

( )g 0 ( )

fm m N m m g T FOy N m mN m m g

+ + + + + =

minus + == +

1p

5p

Ox T prime minus Ff prime = 0T prime = μ(m1 + m0) g

1p

2 2

2

2

0 0

m G TOy G Tm g T

+ =minus =

=

1p

2 1 0

2 10

( )m g m m gm mm

= micro +minus micro

=micro

1p

m0 = 03 kg 1p



aEA = m1gh 2p

3pEA = 1 J 1p

b

ndashf AB

f AB

B A F

B A F

E E L

E E L

=

= +2p

4p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE

= minus micro sdot αα

= minus micro α=

1p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p

c

Din legea conservării impulsului rezultă ( )1 1 1 2 1m v m m v= + 1p

4p

( )( )

1 1 1 2 1

1 1 1 2

1 1

1 2

m v m m v

m v m m vm vv

m m

= +

= +

=+

1p

unde 1m25s

v = 1p

01 255 m08503 s

v sdot= = 1p

d

Lungimea resortului fiind foarte mică icircn comparaţie cu lungimea planului icircnclinat se poate neglija valoarea lucrului mecanic efectuat de către forţa de frecare icircn timpul comprimării resortului precum şi variaţia ener-giei potenţiale gravitaţionale

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

2p

4p( ) 22

1 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

1p

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

= 1p



159



VU VC T∆ = ∆ b 3

2 0U∆ = icircn procese ciclicice d 3

3pRT

microρ =

[ ] 3

kgmSi

ρ = c

3

41 1

1 1

33 22

L p VLU p VU

=

∆ = rArr =∆

a

3

5 36 gA A

m N Nm mN N

= rArr = sdotmicro rArr =micro

a 3



a15 mol

mν =

microν =

2p3p

15 mol

mν =

microν = 1p

b

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

2p

4p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

1p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ = 1p

c

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

3p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot 1p

d

2 2

3 3

mp V RT

m mp V RT

=micro

minus ∆=

micro

1p

5p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=

minus ∆ minusminus=

minus ∆ minusrArr ∆ =

∆ = minus sdot

1p( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot 1p



160


a

L1231=Aria1231 1p

3p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot

1p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot 1p

b

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

5p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot05p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot 05p

c

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p

3p

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot 1p

d

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= =

1p

4p

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= = 1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η =

1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η = 1p



161



0 ER I U E= rArr = rArr =ν

d 3

2 2W RI t= ∆ b 3

3 [ ] 2 1SI

V1 11 N m A sA

R minus minus= = sdot sdot sdot a 3

4

( )

( )

Aria trapez2

200 300 2500 500 mC

2

b B hQ

Q Q

+ sdot= =

+ sdot= = rArr = a

3

5

( )0

2

0

0

1

288

2110 C

R R t

UR RP

R Rt tR

= + α

= rArr = Ω

minus= rArr = deg

α

(Valoarea de 2110 degC a fost obţinută consideracircnd că R0 = 28 Ω icircn loc de R0 = 152 Ω)

3



a

Ampermetrul conectat icircntre A şi B are indicaţia de curent nul dacă R1R4 = R2R3 1p

3p

1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω

1p1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω 1p

b

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R

= + rArr = ΩrArr = + rArr = Ω

sdotrArr = rArr

+= Ω

1p

4p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω 1p

c

Potrivit legii Ohm

22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

2p

4p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

1p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

= 1p


162

d

1p

4pPotrivit legilor Kirchhoff

( ) ( )2 4

2 1 2 4 3 4

2 4

I I II R R I R R

I I

= +

+ = +

rArr =

1p( ) ( )

2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr =

1p

( ) ( )2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr = 1p



a

1 2

1 2

1 2

1 1 1

07 A

E Er rI

RR r r

I

+=

+ +

=

3p

4p

1

2

2E EE E

==

1

2

2r rr r

==

1p

bU = R I 3p

4pU = 7 V 1p

cP = RI2 3p

4pP = 49 W 1p

d88

e

RR r

η =+

η =

2p3p

88e

RR r

η =+

η = 1p



163

DOPTICAtilde

Subiectul I Punctaj

12

2 smv eU= rArr

2

2SI

2 ms

S

e

eUm

=

d 3

2 2

1

sinsin

i nr n

=

a 3

3 Lentila divergentă aflată icircn aer are focare virtuale c 3

4 0 00

275 nmextext

hc hcLL

= rArr λ = rArr λ =λ

b 3

5Interferenţa localizată la infinit a luminii se poate obţine pe lame transparente cu feţe plane şi paralele prin reflexie sau prin transmisie c 3



a

1

1 01 m10

fC

f

=

= =

2p

3p

1

1 01 m10

fC

f

=

= = 1p

b

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R

= minus minus rarr infin

= minus

rArr = minus sdot

= sdot = rArr =

1p

4p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr = 1p

c

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+

rArr β = = = minusminus minus

1p

4p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+

rArr β = = = minusminus minus 1p

d

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =

+ minusprime primeminus = rArr = minus

minusprimeprime prime= rArr = =

prime+ minus

2p

4p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

1p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus1p



164


a

Din grafic15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot

1p2p15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot 1p

b

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

5p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot1p

c

Din ecuaţia Einstein hν = L + Ec unde Ec = eUs 1p

4p

rArr

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot1p

d

( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p

4p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot= 1p



165

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj1 d 32 a 33 c 34 c 35 b 3



a

Icircnălţimea la care urcă ascensorul se determină ca suma spaţiilor parcurse icircn fiecare etapă d1- spaţiul parcurs uniform accelerat timp de 3s d2=36m spaţiul parcurs icircn mişcare uniformă şi d3 spaţiul parcurs icircn ultimele 4s

Icircn prima etapă ascensorul porneşte din repaus atingacircnd viteza 21

2

7 msdvt

= = cu o acceleraţie conform

legii vitezei 211 1 1 1

1

2 msvv a t at

= rArr = =

1p

4p

Spaţiul parcurs se calculează aplicacircnd legea lui Galilei2

2 11 1 1 1

1

2 9m2vv a d da

= rArr = =

Icircn a treia etapă din legea vitezei se determină acceleraţia iar din legea lui Galilei se determină spaţiul parcurs2 2

1 3 3 21 3 3

1 3 31 3 3

22

0

final

final

final

v v a dv a d

v v a tv a t

v

= + = minus= + rArr

= minus= 21

33

21

33

15ms

12m2

vat

vda

= minus = minus = minus =

1p

Icircnălţimea H la care urcă ascensorul 1 2 3H d d d= + + 1p

rezultat final 57 mH = p

b

Viteza cu o secundă icircnainte de oprire se determină din legea vitezei acceleraţia fiind a3 mişcarea uni-form icircncetinită avacircnd loc de 3 s

4 1 3 41 3 4

4 1 4 113 3 3

3

v v a tv t tv v t vva t t

t

= + minus rArr = minus == minus

3p4p

rezultat final 4 15msv = 1p

c

Se scrie principiul al doilea al dinamicii pentru fiecare din cele trei etape

1 1 1 1 1 1( )G T ma T G ma T m a g+ = rArr minus = rArr = +

1p

4p2 2 20 0G T T G T mg+ = rArr minus = rArr =

1p

3 3 3 3 3 3( )G T ma T G ma T m g a+ = rArr minus + = rArr = minus

1p

rezultat final 1 3000 NT = 2 2500 NT = 1 2125 NT = 1p


166

d

Se scrie principiul al doilea al dinamicii pentru corpul de masă m pentru fiecare etapă

2 20G N N G+ = rArr = 1p

4p1 1 1 1 1 1( )G N ma G N ma N m a g+ = rArr minus + = rArr = +

2p

rezultat final 1 960 NN = 2 800NN = 1 680NN =

1p



a

Din legea de mişcare se determină spaţiul parcurs de corp icircn primele 3s iar din legea lui Galilei se determină viteza după primele 3s punacircnd condiţia ca viteza iniţială să fie zero

2

2 20

45 m2

2 2

gth

v v g h v g h

∆ = =

= + ∆ rArr = ∆

2p

4p2

2cmvE mg h= = ∆

1p

rezultat final 450 JcE = 1p

b

Energia cinetică a corpului după cele trei secunde se calculează cu formula de la punctul aDupă primele 3s corpul se va afla faţă de sol la o icircnălţime egală cu h h h= minus ∆

( )c

p

E mg h hE mg h h h h

∆ ∆= =

minus ∆ minus ∆

2p

3p

rezultat final 016c

p

EE

= 1p

c

Pacircnă la adacircncimea d = 02 m corpul pierde icircntreaga sa energie datorită forţei de rezistenţă din partea solului Aplicacircnd teorema de variaţie a energiei cinetice

2

22

c

c

L E Fd

mvE

v gh

= ∆ =

∆ =

=

2p

4p

2

2 2mv mghF

d= = 1p

rezultat final 16 kNF = 1p

d

Forţa de rezistenţă din parte aerului făcacircnd un lucru mecanic rezistent se aplică teorema de variaţie a ener-giei

2

( 1)

2

f i

i f i

f

E E E LE mgh E E fmgh mgh mgh f

mvE fmgh

∆ = minus == rArr minus = minus = minus= =

2p

4p

( 1)( 1) ( 1)

L mgh fL Fh mgh f F mg f

= minus= = minus rArr = minus

1p

rezultat final 9NF = minus 1p



167


Subiectul I Punctaj1 c 32 d 33 a 34 b 35 b 3



a

Din ecuaţia de stare şi formula de definiţie a densităţii se obţine pV RT

mpV RTm= ν

rArr =ν = micromicro pVmRT

micro=

2p3p

rezultat final 2096 10m kgminus= sdot 1p

bDin ecuaţia de stare se obţine A

A

N N pVpV RT NN RT

= rArr = 2p3p

rezultat final 21144 10N molecule= sdot 1p

c

Din ecuaţia de stare se exprimă masa de heliu pentru cele două stări 11

1

p VmRT

micro= respectiv 2

22

p VmRT

micro= 2p

4pMasa ce trebuie adăugată se calculează ca diferenţă 2 1

2 12 1

V p pm m mR T T

micro∆ = minus = minus

1p

rezultat final 39 10 kgm minus∆ = sdot 1p

d

După adăugarea masei suplimentare parametrii corespunzători stării (2) sunt 2 2( )p V T iar icircn starea finală (3) volumul creşte la 3 1(1 ) 16V V f V= + =

2p

5ppresiunea scade la 3 2 2 2(1 ) 06p p f p= minus = 1p

Din ecuaţia Clapeyron Mendeleev 2 3 3 3 3 23 2 1 2

2 3 2

(1 )(1 )p V p V p V TT f f TT T p V

= rArr = = minus + 1p

rezultat final 3 2976 KT = 1p



a

Din formula de definiţie a randamentului ciclului 1 cedatciclu

primit primit

QLQ Q

η = = minus 2p

4pse obţine (1 ) (1 )cedat primit

LQ Q= minus η = minus ηη

1p

rezultat final 3600 JcedatQ = 1p

b

Scriind randamentul ciclului funcţie de temperaturile extreme 2

1 1

1 T TT T

∆η = minus = 1

TT ∆=

η1p

4p2 (1 )TT ∆= minus η

η1p


2 300 KT = p


168

cDin ecuaţia de stare 2

2 22

m m RTp V RT Vp

= rArr =micro micro

2p3p

rezultat final 3 3623 10V mminus= sdot 1p

d

Cantitatea de căldură degajată la se răcirea izocoră este 2 3( )V V VQ C T C T T= ν ∆ = ν minus 1p

4p

Din ecuaţia transformării izocore 2 3 33 2

2 3 2

p p pT TT T p

= rArr = 1p

Prin icircnlocuire se obţine 32

2

1V Vm pQ C T

p

= minus micro

1p

rezultat final 233718 JVQ = 1p



169


Subiectul I Punctaj1 a 32 b 33 b 34 d 35 a 3



a

Se calculează rezistenţa R3 aplicacircnd formula 3 4lRSρ

= = Ω 1p

4pRezistorii R2 şi R3 sunt icircn paralel rezistenţa echivalentă fiind 2 3

232 3

R RRR R

=+ 1p

Rezistenţa echivalentă a grupării de rezistoare va fi 2 31 23 1

2 3

R RR R R RR R

= + = ++

1p

rezultat final 84R = Ω 1p

b

Tensiunea electromotoare echivalentă a grupării de acumulatoare icircn serie este 9sE nE V= =

Rezistenţa internă a grupării serie de acumulatoare este 06sr nr= = Ω 1p

4pCurentul pe ramura principală este dat de legea lui Ohm 1 As

s

EIR r

= =+

1p

Tensiunea icircntre bornele a şi b icircn circuit icircnchis va fi abU IR= 1p

rezultat final 84VabU = 1p

c

Legile lui Kirchhoff pentru nodul de reţea respectiv pentru ochiul format de cele două rezistoare se vor scrie2 3

2 2 3 3

I I IR I R I

= +=

2p3p

Din rezolvarea sistemului se obţine intensitatea curentului prin rezistorul R3 rezultat final 3 06 AI =

1p

d

Prin icircnlocuirea rezistorului R2 cu un voltmetru ideal se va modifica rezistenţa echivalentă a grupării de re-zistoare

1 3R R R= +1p

4pşi implicit valoarea curentului

1 2

093As

e

EIR R r

= =+ +

1p

Tensiunea indicată de un voltmetru ideal montat icircn locul rezistorului R2 va reprezenta tensiunea la bornele rezistorului R3

3 3U R I= 1p




170


a

Din parametrii nominali se obţin valorile curenţilor pentru funcţionare normală1

1 1 1 11

33 3 3 3

3

125 A

05 A

PP U I IU

PP U I IU

= rArr = =

= rArr = =

2p

5pDeci prin rezistenţa R trebuie să circule un curent de 05A la borne avacircnd aplicată o tensiune 1 2 3( ) 4 VRU U U U= + minus =

1p

Din legea lui Ohm R

R

URI

= 1p


b

Intensitatea curentului prin baterie este 1 3 175I I I A= + = 1p

3pTensiunea electromotoare a bateriei este E UE U u U Ir r

Iminus

= + = + rArr = 1p


cEnergia disipată pe R icircn 15mint∆ = este 2W RI t= ∆ 2p

3prezultat final 1800 JW = 1p

d

Puterea consumată reprezintă puterea consumată de becuri şi de rezistor 2

consumat becuri 1 2 3( ) 42 WRP P P P P P RI= + = + + + = 2p

4pPuterea utilă este reprezentată de puterea consumată de becuri Deci randamentul va fi becuri

consumat

095PP

η = = 1p




171

DOPTICAtilde

Subiectul I Punctaj1 b 32 d 33 c 34 b 35 d 3



a

Distanţa focală a lentilei L1 este 11

1 20 cmfC

= =

Lentila L1 formează o imagine reală situată la 1 12

1 1

30 cmf xxf x

= =+

faţă de lentilă1p

4pCum distanţa dintre lentile este de 40cm faţă de lentila L2 imaginea este situată la distanţa

1 2 10 cmx D x= minus = Distanţa focală a lentilei L2 este 22

1 25 cmfC

= = Aşadar lentila L2 se comportă ca

o lupă formacircnd o imagine finală virtuală situată la distanţa 2 12

2 1

f xxf x

=+

faţă de lentila L2

1p

rezultat final 2 166 cmx = minusImaginea se formează icircntre cele două lentile şi este o imagine virtuală

2p

b

Convergenţa sistemului format prin alipirea celor două lentile este 1 2 9C C C= + = δ 1p

3pLentila echivalentă are distanţa focală 1FC

= 1p

rezultat final 011 mF = 1p

c

Prin alipirea celor două poziţia imaginii finale va fi 12

1

1346 cmFxxF x

= =+

1p

4pMărirea liniară transversală a sistemului este 2

1

y xy x

β = = 1p

Icircnălţimea imaginii finale va fi 2

1

xy yx

= 1p

rezultat final yprime= 157 cm 1p

d

Dacă imaginea reală formată este de patru ori mai mare decacirct obiectul 2

1

4xx

β = = minus 1p


2 1

1 1 1F x x

= minus 1p

unde icircnlocuind 2 14x x= minus se obţine poziţia imaginii finale prin sistem 154Fx = minus 1p

rezultat final 1 1388mx = 1p



172


a

Legea conservării energiei ext ch L Eν = + se scrie icircn condiţiile date sub forma 1

1ext s

hc L eU= +λ

(1) 1p

4prespectiv 2

2ext s

hc L eU= +λ

(2) 1p

Din (1) şi (2) se obţine scăzacircnd cele două relaţii ( ) ( )( )1 2 1 2

1 21 2 1 2

1 1 s ss s

e U Uhc e U U h

cminus λ λ

minus = minus rArr = λ λ λ minus λ

1p

rezultat final 34662 10h Jsminus= sdot 1p

b

Din legea conservării energiei max1

ext chc L E= +λ consideracircnd maxc sE eU= 1p

4pşi relaţia pentru lucru mecanic de extracţie 0extL h= ν 1p

se obţine 10

1

sc eUh

ν = minusλ

1p

rezultat final 140 458 10 Hzν = sdot 1p

c

Pentru a determina raportul vitezelor maxime ale fotoelectronilor emişi se consideră 21

1 max 12c smvE eU= = 1p

4p

Respectiv 22


Se obţine 1 1

2 2

s

s

v Uv U

= 1p

rezultat final 1

2

073vv

= 1p

d

Legea conservării energiei se scrie pentru cele două radiaţii

11

ext shc L eU= +λ

(1) respectiv 33

ext shc L eU= +λ

(2)

Variaţia tensiunii de stopare a fotoelectronilor emişi 3 3s s sU U U∆ = minus

1p

3p

3 1

1 1s

hcUe

∆ = minus

λ λ 1p

rezultat final 086 VsU∆ = 1p



173

TESTUL 11

A MECANICAtilde


c 2

2 2mkg sN s s=

kg m kg m

sdot sdotsdotsdot sdot

= adimensional 3

2 c 3

3

b Din grafic rezultă 0h = 25 JcE =

5h = 0cE =2

0

2m v m g hsdot

= sdot sdot 0m2 10s

v g h= sdot sdot =

2

2 05 kgc

o

Emvsdot

= =

3

4 c 3

5 d ( ) 60 kW

2m trapezB b hP A + sdot

= = = 3



a

Reprezentarea corectă a forţelor care acţionează asupra sistemului

3p 3p

bPentru expresia acceleraţiei 1 2 ( )t f f t

F F F F g m MaM m M mminus minus minus micro sdot sdot +

= =+ +

3p

4p

Rezultă 2

m75s

a = 1p

c

Din legea deformărilor elastice se obţine 0F lSE l

sdot=

sdot ∆

Expresia secţiunii firului 2

4dS πsdot

=

3p4p

Rezultă 3346 10 md minus= sdot 1p

d

Pentru viteza constantă tP F v= sdot 1 2

0t f fF F Fminus minus = 2p

4pRezultă 1 2

( ) ( )f fP F F v g M m v= + sdot = micro sdot sdot + sdot 1p

Obţinem 375 10 WP = sdot 1p



174


a

Din conservarea energiei se obţine energia potenţială maximă a corpului

ti tfE E= 2

0

2 cf pfm v E Esdot

= +

20

max 2pf pm vE E sdot

= =

2p

3p

Rezultă max 36 JpE = 1p

b

Energia totală a bilei icircntre cele două stari1 1ci pi c pE E E E+ = + 1p

4pRezultă 1 1

14ci p pE E E= sdot + din condiţiile problemei 2p

Obţinem 144 mh = 1p

c

Din condiţiile problemei cA A pE E= deci energia totală a bilei este

max maxpE m g h= sdot sdot max 18 mh =

Icircnălţimea va fi max1 045 m4

h h= sdot =

1p

4pImpulsul la icircnălţimea h se obţine din

2 20

2 2m v m v m g hsdot sdot

= + sdot sdot

20 2v v g h= minus sdot sdot p = mv

2p

Rezultă m1038 kgs

p = sdot 1p

d

Lucrul mecanic efectuat de greutatea corpului din momentul lansării pacircnă la atingerea solului esteu cG G GL L L= + 1p

4pDeci max maxGL m g h m g h= minus sdot sdot + sdot sdot 2p

Rezultă 0GL = 1p



175



d 2 3

2 3

N kgN 1m mol mJ m N m mkg k

mol k

minussdot

= = =sdot sdotsdot sdot

sdot

3

2 c p vRc c= +micro

p vRc cminus =micro

3

3c 0Q = 1T V ctγminussdot = 1

2 8VV = 12

1

8TT

γminus= 53

p

v

cc

γ = =

2 232 3 3

2 11

8 (2 ) 4 4T T TT

= = = rArr = sdot 3

4 b 3

5 d 12 23 310 U U U= ∆ + ∆ + ∆ 1252vU C T p V∆ = ν sdot sdot ∆ = sdot sdot 31

72pU C T p V∆ = ν sdot sdot ∆ = minus sdot sdot 23 12 310U U U p V∆ = minus ∆ minus ∆ = sdot 3



a

Din ecuaţia termică de stare 1 1 1

mp V R Tsdot = sdot sdotmicro

3 311

1

025 10 mm R TVp

minussdot sdot= = sdot

sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro 1p

3pSe obţine 3 31

11

025 10 mm R TVp


sdotmicro1p

Rezultă 3 31 025 10 mV minus= sdot 1p

b

Numărul de moli poate fi exprimat atacirct din relaţia A

NN

ν = 1p

4pcacirct şi din relaţia

mν =

micro1p

Din relaţiile precedente obţinem pentru o moleculă de aer 1N = 0A

mNmicro

= 1p

Rezultă 260 48 10 kgm minus= sdot 1p

c

Pentru starea iniţială 1 1 1A

Np V R TN

sdot = sdot sdot 1p

4pConcentraţia se defineşte Nn

V= 1p

Se obţine 1

1

Ap NnR T

sdot=

sdot1p

Rezultă 263024 10 molecn m= sdot 1p


176

d

Pistonul rămacircne icircn poziţia iniţială dacă 1 2p p= 1 2V V= 1 1 2

1 2

( )m R T m m R TV V

sdot sdot minus sdot sdot=

micro sdot micro sdot1 019 kgm = 1 2V V=

1 1 2

1 2

( )m R T m m R TV V


micro sdot micro sdot 1 019 kgm =

2p

4punde 1

11

m R TpV

sdot sdot=

micro sdot pentru starea 1

1 22

2

( )m m R TpV

minus sdot sdot=


1p

Icircn final obţinem 1 1 2

1 2

( )m R T m m R TV V


micro sdot micro sdot

Rezultă 1 019 kgm = 1p



a

Pentru icircntregul proces ciclic 12 23 34 41tL L L L L= + + +

2 4 ln 2 3 28 18tL pV pV pV pV pV pV= + minus = minus = 1p

5p

Pentru fiecare stare se obţine 1 2rarr 12 0V ct L= rArr =

23 3 3 2( ) 2 (2 ) 2L p V V p V V pV= minus = minus =

4 1

34 3 3 33 1

4ln ln 2 2 ln 2 4 ln 22

V VL vRT p V p V pVV V

= = = =

41 1 1 4( ) ( 4 ) 3L p V V p V V pV= minus = minus = minus

2p

Obţinem 18tL pV= 1p

b

Căldura cedată de gaz icircntr-un ciclu de funcţionare este 41 1 4 1 47( ) ( )2c pQ Q vC T T vRT vRT= = minus = minus 1p

4pDin ecuaţia termică de stare 41 1 1 4 4

7 ( )2cQ Q p V p V= = minus 1p

Obţinem 417 ( 4 )2cQ Q pV pV= = minus 1p

Rezultă 41 105cQ Q pV= = minus 1p

c

Randamentul unui Ciclu Carnot este min

max

1 TT

η = minus 1p

4p1 1

min 1p VT TvR

= =

3 3max 3

p VT TvR

= =

1p

Obţinem 1 1

3 3

1 12 2

p V pVp V p V

η = minus = minus 1p

Rezultă 75η = 1p

d

Trasarea corectă a graficului

3p 3p



177



c SI

V RA

= 2W R I t= sdot sdot ∆ 2 12 J A s

SI

WR RI t

minus minus= rArr = sdot sdotsdot ∆

3

2 d 11

lRS

ρsdot= 2

2

lRSρsdot

= 1

2

14

RR

= 3

3 b 34 d 3

5 c 572 10 JW P t= sdot = sdot 2 kWhW = 3



a

Stabilirea corectă a sensului curenţilor

1p

4p

Legile lui Kirchhoff

1 1 1 3 3 1 ( )I I R r I R Esdot + + sdot = 2 2 2 3 3 2 ( )II I R r I R Esdot + minus sdot = 1 2 3A I I I= + 3p

bDin ochiul I se obţine 1 1 1 1( ) 78 VABU E I R r= minus sdot + = 2p

3pRezultă 78 VABU = 1p

c

Pentru nodul A 2 1 3I I I= minus 3

3

13 AABUIR

= = 2 24 AI = 2p

4pIndicaţia voltmetrului se obţine din 2 2 2MNE U I r= + sdot 1p

Rezultă 152 VMNU = 1p

dDaca se icircnlocuieşte R3 cu un voltmetru ideal pe ramura respectivă nu mai trece curent deci putem scrie

1 2 1 2 1 2( )I R R r r E Esdot + + + = + 3p4p

Rezultă 312 AI = 1p



a

Pe circuitul exterior max

4extPP =

22

16ER I

rsdot =

sdot1p

4pSe icircnlocuieşte 2

2

P

EIR r

= +

1p

Pentru R = r obţinem relaţia 2 28 4 0R Rminus sdot + = 768∆ = 1p

Rezultă 12 2784 016R = Ω Ω 1p


178

b

Pentru puterea maxima 2R r= = Ω 1p

4pIntensitatea va fi 5 A2EI

r= =

sdot

Energia disipata de rezistorul R este 2W R I t= sdot sdot ∆ 1p

Rezultă 15 kJW =

c

Icircn acest cazvaloarea intensitatii va fi 1 2 A5SCII = = 1p

3pDeci 1P U I= sdot unde 1 16 VU E r I= minus sdot = 1p

Rezultă 32 WP = 1p

d

Intensitatea curentului de scurtcircuit al sursei este1

1 1 1 11

8E r IE R I r I RI

minus sdot= sdot + sdot rArr = = Ω

1

1

RR r

η =+

2p3p

Rezultă 80η = 1p



179

DOPTICAtilde


a 22

m 1 m1 1 m ss s ssic minuslt sdotν gt = sdot = = sdot 3

2 c 3

3 d sin 1 3sin 3sin4sin 4 8

3

i rir

= rArr = = 3

4 c 15

2 1214

1 2

1

5 10 252 10

c

cλ ν sdotν

= = = =λ ν sdot

ν

3

5 c 9

43

600 10 2 6 10 06 mm2 2 10Dil

minusminus

minus

λ sdot sdot= = = sdot =

sdot3



a


1 1 1x x f

minus = 1p

4Rezultă 1

21

20 ( 30)20 30

fxxf x

sdot minus= =

+ minus2p

Obţinem 2 60 cmx = 1p

b

Distanţa focală 2

1 1 2

1 1 11nf n R R

= minus minus

unde 1 1n =

( )1 1 11nf R R

= minus minus minus

2p

4p

obţinem ( )2 11 1

2n Rn

f R fminus

= rArr minus = 1p

Rezultă 20 1 152 20

n = + =sdot

1p

c

Realizarea corectă a desenului

3p 3p

d

Convergenţa pentru cele două medii

1 2 1

2( 1)

nCf n R

C nR

= = minus

= minus

2p

4pPrin icircmpărţire obţinem

1 15 1151 112

C nnCn

minus minus= =

minus minus1p

Rezultă 2CC

= 1p



180


aFrecvenţa radiaţiei monocromatice de frecvenţă 1ν este

8

1 91

3 10600 10

cminus

sdotν = =

λ sdot2p

3p


b

Pentru radiaţia de frecvenţă 1ν avem 1 1ex Sh L e Usdot ν = + sdot 1 0 1h h eUν = ν + 8

16 152 9

2

3 10 1 10 055 10 Hz54 10 18

cminus

sdotν = = = sdot = sdot

λ sdot 1p

4pPentru radiaţia de frecvenţă 2ν avem 2 0 1( )h h e U Uν = ν + + ∆ 1p

Din relaţiile precedente obţinem prin adunarea lor 2 1

2 1

( )( )

h e UhU

e

ν minus ν = ∆ν minus ν

∆ = unde 152

2


λ1p

Rezultă 02VU∆ = 1p

c

Lucrul mecanic de extracţie este 0 00

cL h h= ν =λ

1p

4pObţinem 17

0 0044 10L Jminus= sdot 1p

Dar 17

190 19

0044 101 16 1016 10

eV J Lminus

minusminus

sdot= sdot rArr =

sdot1p

Rezultă 0 275L eV= 1p

d

Pentru radiaţia de frecvenţă 1ν energia cinetică a fotoelectronilor emişi de către catodul celulei fotoelectrice este

11 0 ch h Eν = ν +1p

4pPentru radiaţia de frecvenţă 2ν energia cinetică a fotoelectronilor emişi de către catodul celulei fotoelectrice este 2 0 1( )c chv hv E E= + + ∆

1p

Variaţia energiei cinetice a fotoelectronilor emişi de către catodul celulei fotoelectrice se obţine scăzacircnd relaţiile 2 1( ) ch v v Eminus = ∆ 1p

Rezultă 19033 10 JcE minus∆ = sdot 1p



181

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj1 a 32 d 33 b 34 d 35 c 3



a

Pentru corpul cu masa m1 se scrie 1

1 1

1 1

0

f

F T m aN m gF N

minus =minus == micro

2p

5pPentru corpul cu masa m2 2 2

2 2

2 2

0f

f

T F m aN m gF N

minus =

minus == micro

2p

Din sistemul de mai sus se deduce acceleraţia sistemului 1 1 2 22

1 2

( ) m125s

F g m mam m

minus micro + micro= =

+1p

b Tensiunea va fi 2 2( ) 675NT m a g= + micro = 3p 3p

c

Putem presupune că icircn urma acţiunii forţei F sistemul a ajuns la o anumită viteză Dacă forţa icircncetează corpu-rile se vor mişca uniform icircncetinit acceleraţiile lor fiind

1 1 2

m2s

a g= minusmicro = minus respectiv 2 2 2

m1s

a g= minusmicro = minus 2p

4p

Deoarece acceleraţia de fracircnare a primului corp este mai mare rezultă că firul de legătura se detensionează deci tensiunea va fi nulă 2p

d Putem raţiona astfel deoarece forţa exterioară sistemului nu se schimbă acceleraţia sistemului nu se schimbă Vom obţine deci 1 1 ( ) 225NT m a g= + micro = 3p 3p



aAplicăm teorema de variaţie a energiei cinetice

2 21 cos180

2 2mv mv mgdminus = micro deg 3p

5p

Rezultă 2 2

1 0482

v vgdminus

micro = = 2p

bLa sfacircrşitul perioadei de fracircnare impulsul vagonului este 1p mv= 2p

3pp = 8000 Ns 1p

c

Icircn timpul opririi vagonului icircn urma ciocnirii celor patru tampoane energia cinetică a vagonului se transformă icircn energie potenţială elastică a tampoanelor Un singur tampon va icircnmagazina energia

21

112 2

MvW =

2p3p

1 8000JW = 1p

d

Energia mecanică se conservă pentru un resort scriem2 211

4 2 2Mv kx

= 2p

4p

Rezultă 212

kN3204 mMvk

x= = 2p



182


Subiectul I Punctaj1 b 32 b 33 c 34 c 35 a 3



a

Scriem ecuaţie de stare pentru cele două baloane 1 21 2

1 2

m mp V RT p V RT= =micro micro

1p

3p1 1 2 21 2p V p Vm m

RT RTmicro micro

= = 1p

1 1 1

2 2 2

35m pm p

micro= =

micro1p

b

Prin deschiderea robinetului gazele se amestecă iar cantităţile de substanţă se adună1 2

1 2

(2 ) ( )m mp V RTsdot = +micro micro

1p

4p

1 2

21 2

2 2

( )

2

m m RTm RTp V

V p

+micro micro

= =micro

1p

Forţăm factor comun 2m

12

2 2 1 12 1 21 2

2 1 2 2 2

2 2

1 1( )1 1 1( ) (p p )

2 22

mmp pmp m RT p

p

sdot +micro micromicro micro

= = sdot + = +micro micro micro

micro

1p

1 (1atm 2atm) 15atm2

p = + =

Observaţie dacă se scrie rezultatul final direct nu se ia icircn consideraţie

1p

c

Masa molară medie (aparentă) a amestecului este 1 2

1 2

1 2

masa amestecului m mm msuma molilor

+micro= =

+micro micro

2p

4pSe forţează factor comun m2

12

2

12

2 1 2

( 1)

1 1( )

mmm

mmm

+micro =

sdot +micro micro

1p

şi se icircnlocuieşte raportul maselor de la punctul a) Se obţine 1 1 2 2

1 2

kg12kmol

p pp p

micro + micromicro = =

+1p

d

0 2 rmp V RT=micro

1p

4p02r

p VmRTmicro

= 1p

0 0

1 2 1 1 2 2 1 2

2 2 2 0(6)3

rm p pm m p p p p

micro= = = =

micro + micro +2p



183


a

1 (273 27)K 300KT = + =

1p

3p

Pentru transformarea 1rarr2 1 12 1

2 1

2 2 600Kp p T TT T

= = = 1p

Pentru transformarea 2rarr3 3 1

3 2

V VT T

= pentru transformarea 3rarr1 3 1

3 1

p pV V

=

Rezultă 3 14 1200 KT T= =

1p

b

Pentru transformarea 1 rarr 3 aplicăm primul principiu al termodinamicii

13 13 13 3 1 3 11 sau ( )( )2VQ U L C T C T p p V V= ∆ + ν ∆ =ν ∆ + + minus

1p

4p

Ţinacircnd cont de ecuaţia de stare pentru 1 şi 3 şi de ecuaţia dreptei 1 rarr 3 lucrul mecanic se scrie

13 3 3 3 1 1 3 1 1 3 11 1( ) ( )2 2

L p V p V p V p V RT RT= minus + minus = ν minus ν 1p

Rezultă 12VC T C T R Tν ∆ =ν ∆ + ∆ 1p

Rezultat final J2 16620

2 kmol KVRC C R= + = =

sdot1p

c

123 12 23L L L= +

12 23 2 3 1 2 3 2 10 ( )L L p V V p V p V= = minus = minus2p

4p2 3 3 1 2 1 1p 4 V RT RT p V RT= ν = ν = ν 1p

123 1 1 13 3 300JL RT p V= ν = = 1p

d

1 ced

abs

QQ

η = minus 1p

4p31 3 1( )( )

2ced VRQ Q C T T= = ν + minus 1p

12 23 2 1 3 2( ) ( )abs V pQ Q Q C T T C T T= + = ν minus + ν minus 1p

1 1

1 1 1 1

2 (4 ) 11 773 5 13(2 ) (4 2 )2 2

R T TR RT T T T

ν sdot minusη = minus =

ν sdot minus + ν sdot minus

1p



184


Subiectul I Punctaj1 d 3

2

b

Intensitatea curentului prin circuit este EI RxrR x

=+

+

Puterea debitată este P = Rechivalentă exterioară I2

2

2

( )( )

E Rx R xRx rR rx

+=

+ +

Funcţia ( )P R are un extrem (icircn acest caz un maxim) dacă derivata de ordin I se anulează 0dPdx

=

Rezultă RrxR r

=minus

3

3

c

Prin definiţie intensitatea curentului electric este dQIdt

=

Considerăm o porţiune de conductor electric cu aria secţiunii transversale S de lungime dx icircn care se află purtători de sarcină cu concentraţia n care se deplasează relativ uniform cu viteza v şi avacircnd sarcina q Este valabilă următoarea succesiune de relaţii

dQ dN q n dV q n q S dxI nqSvdt dt dt dt

dQ dN dxq n dV Sdx vdN dV dt

sdot sdot sdot sdot sdot sdot = = = = = = = = =

3

4

c

Pentru circuitul serie avem 12

2EI

R r=

+

Pentru circuitul cu sursele dispuse icircn paralel se scrie 2

22

EI rR=

+

Raportul curenţilor este 1

2

22 2122 7

rRI EI R r E

+= sdot =

+

3

5

dCa să lumineze normal trebuie ca becurile să aibă aceeaşi rezistenţă nominală cacircnd sunt parcurse de acelaşi curent nomi-

nal 1

2nn n

E EIR r R r

= =+ +

Rezultă 174

E E=3



aIndiferent de poziţia extremă a cursorului acesta scurtcircuitează punctele A şi B 2p

4pPrin urmare intensitatea curentului din circuit este icircn aceste cazuri

1

171AEIr R

= cong+

2p

b

O reprezentare mai accesibilă a circuitului arată astfel

Rezistenţa furnizată de potenţiometrul special este reprezentată alăturat

Rezistenţa echivalentă a potenţiometrului este

22 ( ) 22ABx R xR x x

R Rminus

= = minus

2p

4p

22 ( ) 22ABx R xR x x

R Rminus

= = minus

adică este o funcţie de gradul al doilea cu argumentul xGraficul acestei funcţii este o parabolă cu valorile importante prezen-tate icircn figura alăturată

2p


185

c

Curentul indicat de ampermetru este minim cacircnd rezistenţa totală din circuitul exterior este maximă Acest aspect este realizat cacircnd rezistenţa potenţiometrului este maximă conform graficului această valoare este

max 2ABRR =

2p

4pRezultă min

1 2

EI Rr R=

+ + 1p

Rezultă min 1I A= 1p

dsc

EIr

= 2p3p

12AscI = 1p



a

Intensitatea curentului electric icircntr-un circuit simplu este EI

R r=

+1p

4p

Puterea debitată pe rezistenţa externă este 2

22( )

REP RIR r

= =+

1p

Se vede că puterea este o funcţie de R Aceasta are un extrem dacă derivata de ordin I are valoarea zero adică

2 2

2 4

( ) 2 ( )0 0( ) ( )

d RE R r R R r R rdR R r R r

+ minus += rArr = rArr = + +

1p

Icircnlocuind valoarea aflată a lui R icircn expresia puterii se obţine 2

4mEP

r= De aici

2

14 m

ErP

= = Ω 1p

b

Rezistenţa echivalentă a circuitului exterior se poate deduce urmacircnd curbele icircnchise de la cea mai mică la cea mai mare Pentru curba din interior rezistenţa echivalentă este R Pentru curba mai mare rezistenţa echivalentă este 3R Icircn final se obţine valoarea rezistenţei exterioare echivalente 2eR R=

2p3p

Deoarece aceasta trebuie să fie egală cu rezistenţa internă a sursei se obţine icircn final 2 052rR r R= rArr = = Ω 1p

c

Condiţia impusă se scrie mP fP= Adică 2 2

02

0( ) 4R E Ef

R r r=

+ 2p

4pRezolvacircnd se obţine

2

02 4(1 )

4 m

f fERP f

minus plusmn minus= sdot

Rezultă două valori 01 018R = Ω şi 02 582R = Ω

2p


186

d

1

2

22

2

22

RR r

RrR

η =+

η =+

2p

4p1

2

1223

η =

η =1p

2

1

43

η=

η 1p



187

DOPTICAtilde

Subiectul I Punctaj1 a 32 d 33 a 34 d 35 c 3



a

Considerăm o lentilă cu grosimea a iar imaginea intermediară a obiectului aflat la ndashx1 faţă de primul diop-tru se formează la distanţa x icircn interiorul lentilei Aceasta constituie obiect pentru al doilea dioptru faţă de

care se află la 1( )x x a+ minus =

1p

5p

2 1 2 1

1 1

n n n nx x R

minusminus = 1p

3 2 3 2

2 2

n n n nx x a R

minusminus =

minus 1p

Se rezolvă şi se impune condiţia ca a să tindă spre zero (lentilă subţire) şi se obţine3 1 2 1 3 2

2 1 1 2

n n n n n nx x R R

minus minusminus = + 2p

b

Pentru primul dioptru 1 1

1 2 1

y n xy n x

= 1p

4pPentru al doilea dioptru 2 2 2

1 3 1

y n xy n x

= 1p

Pentru lentila subţire 1x x a x= minus cong 1p

Rezultă mărirea liniară transversală 1 2 2

3 1 1

n x yn x y

β = = 1p

c

Pentru focarul imagine 31

2 1 3 2

1 2

imnx f n n n n

R R

rarr minusinfin =minus minus

+

15p

3p

Pentru focarul obiect 12

2 1 3 2

1 2

obnx f n n n n

R R

minusrarr infin =

minus minus+ 15p

d

Indicii de refracţie ai mediilor din lateral devin ambii de valoare 1 iar razele vor fi identice icircn modul Se obţine o lentilă simetrică situată icircn aer 1p

3pRezultă

22( 1)Rf

n=

minus 2 2

1 1

y xy x

β = = 2 1

1 1 1x x f

minus = 2p



188


a

Unda cu parcurs mai mare 2 22 2( )a l b rδ = + + + 1p

3pUnda cu parcurs mai mic 2 21 1( )a l b rδ = minus + + 1p

Rezultă 2 2 2 22 1( ) ( )a l b r a l b rδ = + + + minus minus + minus 1p

b

Putem scrie 2

kx rD l

∆asymp pentru maxime de interferenţă diferenţa de drum optic trebuie să fie multiplu par de

semilungimi de undă iar pentru minime multiplu impar1p

4pPentru maxime rezultă 2 2 2 2

max 2 ( ) ( )2 2kDx k a l b a l bl

λ = minus + + + minus + 1p

Pentru minime rezultă 2 2 2 2min (2 1) ( ) ( )

2 2kDx k a l b a l bl

λ = + minus + + + minus + 1p

Pentru maximul central se obţine 2 2 2 20max ( ) ( )

2Dx a l b a l bl

= minus + + + minus + 1p

c

Interfranja luminoasă se calculează ca diferenţa dintre maximul luminos de ordinul k + 1 şi acela de ordinul k 1p

4pRezultă

2lDi

lλ

= 1p

Interfranja icircntunecoasă se calculează ca diferenţa dintre minimul luminos de ordinul k + 1 şi acela de ordi-nul k 1p

Rezultă 2icircDi

lλ

= 1p

d

Interfranja va creşte 2 (1 ) 2 1

D D fi il f l f

λ λ∆ = minus =

minus minus2p

4p

Şi icircn acest caz interfranja va creşte (1 )

2 2D g Di gi

l l+ λ λ

∆ = minus = 2p


TESTE DE NIVEL AVANSAT


190

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj1 a 32 b 33 c 34 d 35 c 3



a

Pe direcţia mişcării sistemului saniendashsportiv avem 1 2( ) t fG F m m aminus = + sdot 1p

4punde 1 2 1 2( ) sin ( )t

HG m m g m m gL

= + sdot sdot α = + sdot sdot 1p

Deci 1 2( )fHF m m g aL

= + sdot sdot minus

1p

Rezultă 60 NfF = 1p

b

Pentru sistemul saniendashsportiv pe o direcţie perpendiculară cu direcţia mişcării avem 0nN Gminus = 1p

5p

unde 2 2

1 2 1 2( ) cos ( )nL HG m m g m m g

Lminus


Dar fF N= micro sdot 1p

După efectuarea calculelor obţinem 2 2

1 2( )fF L

m m g L H

sdotmicro =

+ sdot sdot minus1p


c

Pe direcţia mişcării saniei avem 2 2 2 t fG F m aprimeminus = sdot 1p

4p

unde 2 2 2sin tHG m g m gL

= sdot sdot α = sdot sdot2 2

2 2 2 2 2cosf nL HF N G m g m g

Lminus

= micro sdot = micro sdot = micro sdot sdot sdot α = micro sdot sdot sdot1p

Obţinem 2 2H L Ha g

Lminus micro sdot minusprime = sdot 1p

Rezultă 25 ms aprime = 1p

d

Pe direcţia mişcării saniei avem 2 2 0t fF G Fminus minus = 1p

3pDeci 2 2

2 2 2t fH L HF G F m g

L+ micro sdot minus

= + = sdot sdot 1p

Rezultă 28 NF = 1p


Bareme teste nivel avansat

191


a

Aplicăm legea de conservare a energiei mecanice M NE E= 1p

4p

Unde 2 2

0 2 2

M MM cM pM

m v m vE E E sdot sdot= + = + =

max max0N cN pNE E E m g h m g h= + = + sdot sdot = sdot sdot1p

Obţinem 2

max 2Mvhg

= 1p

Rezultă max 5 mh = 1p

b

La coboracircre icircn punctul P icircn care energia cinetică a pietrei este o pătrime din energia potenţială a sistemului

piatrăndashpămacircnt avem 25

4 52

QQ cQ pQ cQ cQ cQ

m vE E E E E E

sdot= + = + = =

1p

4pDar M QE E= 1p

După efectuarea calculelor obţinem 5

5Q Mv v= 1p

Rezultă 447 msQv cong 1p

c

Aplicăm teorema de variaţie a energiei mecanice neconservativR ME E Lminus = 1p

4pUnde 0 ( )R cR pRE E E m g d m g d= + = + minus sdot sdot = minus sdot sdot neconservativ rL F d= minus sdot 1p

Deci 2

2M

rm vF m g

dsdot

= sdot + 1p

Rezultă 5010 NrF = 1p

d

Lucrul mecanic efectuat de forţa de greutate din punctul de icircnălţime maximă a pietrei şi pacircnă la oprirea acesteia icircn sol este max( )GL m g h d= sdot sdot +

2p3p

Rezultă 501 JGL = 1p



192


Subiectul I Punctaj1 b 32 a 33 a 34 d 35 c 3



a

Pentru primul compartiment din ecuaţia termică de stare 1 1 1 1p V R Tsdot = ν sdot sdot 1p

3pObţinem 1 11

1

p VR T

sdotν =

sdot1p

Rezultă 1 1 molν = 1p

b

Pentru al doilea compartiment din ecuaţia termică de stare 2 2 2 2p V R Tsdot = ν sdot sdot 1p

4pObţinem 2 2 2

22 A

p V NR T N

sdotν = =

sdot1p

Dar 22

A

NN

ν = 1p

Rezultă 232 301 10 N = sdot 1p

c

Icircn acest caz 1 2Q Q= 1p

4pUnde 1 1 V 1( )Q C T T= ν sdot sdot minus 2 2 V 2( )Q C T T= ν sdot sdot minus 1p

După efectuarea calculelor obţinem 1 1 2 2

1 2

T TT ν sdot + ν sdot=

ν + ν1p

Rezultă 3333 KT cong 1p

d

Pentru primul compartiment 1 1 1p V R Tprime primesdot = ν sdot sdot 1p

4p

Pentru al doilea compartiment 2 2 2p V R Tprime primesdot = ν sdot sdot 1p

Dar 1 2p pprime prime= 1p

Rezultă 2 2

1 1

05VV

prime ν= =

prime ν1p



193


a

Variaţia energiei interne icircntr-un ciclu este 12341 12 23 34 41U U U U U∆ = ∆ + ∆ + ∆ + ∆ 1p

4pDeci 12 23 34 41 1234U U U U U∆ = ∆ + ∆ + ∆ minus ∆ 1p

Dar 12341 0U∆ = 1p


b

Căldura cedată de gaz mediului exterior pentru parcurgerea a 10N = cicluri termodinamice este cedat cedat12341NQ N Q= sdot 1p

4pUnde cedat12341 34 41Q Q Q= + 1p

Şi 34 34 34 ( 1100 J) 0 J 1100 JQ U L= ∆ + = minus + = minus

41 41 41 ( 450 J) ( 100 J) 550 JQ U L= ∆ + = minus + minus = minus1p

Rezultă cedat- 16500 JNQ = minus 1p

c

Lucrul mecanic total schimbat de gaz cu mediului exterior icircntr-un ciclu este12341 12 23 34 41L L L L L= + + + 1p

3punde 12 0 JL = 23 23 23 1150 J 750 J 400 JL Q U= minus ∆ = minus = 34 0 JL = 41 100 JL = minus 1p


d

Pentru un ciclu termodinamic 12341 primit 12341 cedat 12341L Q Q= minus 1p

4p

Căldura primită de gaz pentru parcurgerea a 10N = cicluri termodinamice esteprimit N primit 12341Q N Q= sdot 1p

Deci ( )primit N 12341 cedat 12341Q N L Q= sdot + 1p

Rezultă primit N 19500 JQ = 1p



194


Subiectul I Punctaj1 a 32 c 33 c 34 b 35 d 3



a

Rezistenţa echivalentă a grupării de rezistoare poate fi exprimată din relaţia1 1 1 1 1

PR R R R R= + + + 1p

3p

Obţinem 4 PR R= sdot 1p


b

Tensiunea electromotoare a acumulatorului este SE N E= sdot 1p

4pIntensitatea curentului prin circuit este S

P S

EIR r

=+

1p

Unde Sr N r= sdot 1p

Rezultă 02 AI = 1p

c

Tensiunea internă dintr-un element de acumulator este 0u I r= sdot 1p

4pRezultă 0 02 Vu = 1p

Tensiunea internă din baterie este 0u N u= sdot 1p

Rezultă 4 Vu = 1p

d

Intensitatea curentului icircn cazul bateriei scurtcircuitate este SSC

S

EIr

= 2p

4pObţinem SC

EIr

= 1p




195


a

Intensitatea curentului prin sursă este BD

EIR r

=+

1p

4pUnde 12 34

12 34BD

R RRR R

sdot=

+ 1p

Iar 12 1 2R R R= + 34 3 4R R R= + 1p

Rezultă 1 AI = 1p

b

Tensiunea măsurată de voltmetru icircntre punctele A şi C este 4 1ACU I R I Rprimeprime prime= sdot minus sdot 1p

5p

Dar 1I I Iprime primeprime= + 1p

Şi 12 34I R I Rprime primeprimesdot = sdot 1p

Deci 2II Iprime primeprime= = 1p

Rezultă 135 VACU = 1p

cEnergia electrică consumată de circuitul exterior icircn 1 mint∆ = este 2

BDW R I t= sdot sdot ∆ 2p3p

Rezultă 525 JW = 1p

dIntensitatea curentului icircn cazul scurtcircuitării sursei este SC

EIr

= 2p3p




196

DOPTICAtilde

Subiectul I Punctaj1 c 32 d 33 a 34 b 35 a 3



a

Convergenţa sistemului optic format de cele două lentile subţiri este 1 2C C C= + 2p

4pDistanţa focală a sistemului optic format de cele două lentile subţiri este 1fC

= 1p

Rezultă f = 20 cm = 02 m 1p

b


1 1 1x x f

minus = 1p

3pObţinem 1

21

x fxx f

sdot=

+1p

Rezultă 2 20 cmx = minus 1p

c


1 1 1x x f

minus =prime prime

1p

4pDin a doua formulă fundamentală a lentilelor 2

1

12

xxprime

primeβ = = minusprime

1p

Obţinem 11x f minus βprime = sdot

β1p

Rezultă 1 60 cm 06 mxprime = minus = minus 1p

d


1 1 1x x f

minus =primeprime primeprime

1p


1

xxprimeprime

primeprimeβ =primeprime

1p

Obţinem 1

fx f

primeprimeβ =primeprime+

1p

Rezultă 2primeprimeβ = minus 1p



a


Dil

λ sdot= 1p

3pObţinem

2l iD

sdotλ = 1p



197

b

De o parte şi de alta a franjei centrale se află maximele de ordinul al II-lea iar distanţa dintre aceste maxime este 2 22x x∆ =

1p

4pFiecare maxim de ordinul al II-lea se află faţă de franja centrală de pe ecran la distanţa 2 2x i= 1p

Deschiderea unghiulară α măsurată din centrul paravanului cu fante pacircnă la maximele de ordinul al II-lea

obţinute pe ecran se exprimă din relaţia 2 4tg x iD D

∆α = = 1p

Rezultă 32 10 radminusα = sdot 1p

c

Interfranja este dată de relaţia 11 2 2

D iil

λ sdot= = 1p

4pDeci 1 2

DD = 1p

Icircn acest caz ecranul se apropie pe distanţa 1 2Dd D D= minus = 1p

Rezultă 1 md = 1p

d

Icircn acest caz interfranja este dată de relaţia 2 12 2

Dil

λ sdot= 1p

4pUnde 2

1 cn n

λλ = sdot =

ν1p

Deci 12 2 2

D iil n n

λ sdot= =

sdot1p

Rezultă 2 033 mmi cong 1p



198

TESTUL 2

A MECANICAtilde


b 2[ ] kg msminus∆= rArr = sdot

∆ SIpF Ft 3

2 c 15 mgiT G F= + = 3

3 a 2

0max2 2 10 m 0

2vd h r

g= = = ∆ = 3

4

b 2 2

5 45 4

4 5

452 2

4 s 5 s

at atx x x a

t t

∆ = minus = minus =

= =22 msarArr =

2 210 9

10 9

9 10

19 m2 2

9 s 10 s

at atx x x x

t t

∆ = minus = minus rArr ∆ =

= =

3

5

c Se realizează graficul F = F(t) şi se calculează Δp ca arie a suprafeţei evidenţiate 180 kg ms 36 msp vrArr ∆ = sdot rArr =

3



a


1 1

2

0

0f

f

F T FT F

minus minus =

minus =

2p

4p

1 1 1 2( )F m m g= micro + 1p

2 2 kgm = 1p

b

Deoarece doar m1 icircncepe să se mişte2 1fF F=

1p

3p2 2 1F m g= micro 1p

2 025micro = 1p


199

c

Studiem mişcarea fiecărui corp separat icircn condiţiile unei mişcări uniforme a sistemului Reprezentăm forţele care acţionează asupra fiecărui corp şi aplicăm principiul al doilea al dinamicii

3 1

2

0

0e f

e f

F F FF F

minus minus =

minus =

2p

4p

3 2 1 2( )F m m g= micro + 1p

3 15 NF = 1p

d

Analizăm mişcarea fiecărui corp separat icircn condiţiile unei mişcări accelerate a sistemului Reprezentăm forţe-le care acţionează asupra fiecărui corp şi aplicăm principiul al doilea al dinamicii

4 1 1

2 2

e f

e f

F F F m aF F m a

minus minus =

minus =

1p

4p24 2 1 2

1 2

( ) 75 msF m m gam m

minus micro += =

+ 1p

eF k l= ∆ 1p

2 2( ) 1333 Nmm a gkl

+ micro= =

∆1p



a

00 50 mspv

m= =

1p

3p0u

vtg

= 1p

5 sut = 1p

b

Conservăm energia icircntre punctele O şi A0c cA pAE E E= + 1p

4p0

32 2pA pA

cA c

E EE E= rArr = 1p

20

3vh

g= 1p

833 mh = 1p

c

0 3c cAE E= 1p

4p0

3vvrArr = 1p

29 msv = 1p

29 kg msp mv= = sdot 1p

d

20

max 125 m2vh

g= = 1p

4pc c G fE L E L L∆ = rArr ∆ = + 1p

0 104 mc

f

Ehmg F

= =+ 1p

max 12hh

= 1p



200


Subiectul I Punctaj1 d 1 2

1

2

V V p pQ C T Q C TTT

= ν ∆ = ν ∆

∆rArr = γ

∆

3

2c

11

1 1 ii i f f f i

f

TTV T V V VT

γminusγminus γminus

= rArr =

44

if f i

VV = rArr ρ = ρ

3

3

b Din V = ap rezultă 1 1

2 2

V pV p

=

Din ecuaţia de stare rezultă 1 2 1

2 1 2

V p TV p T

=

Din cele două ecuaţii rezultă 1 1

2 2

3p Tp T

= =

3

4c

11

1 21 1 2 2

2 1

03V pp V p VV p

γγ γ γ

= rArr = =

2 1

1 2

N n VnV n V

= rArr =

3

5b 2 2

1 1

22 1

1

1 1

12 kJ

Q TQ T

TQ QT

η = minus = minus

rArr = =

3



a

1 2

1 2

m m+micro =

ν + ν 1p

4p

1 21 2

1 2

pV pVm mRT RT

micro micro= = 1p

1 2 2 1

1 2

T TT T

micro + micromicro =

+ 1p

165 gmolmicro = 1p

b

1 1 1 2 2 2 1 1 2 2( )i f

V V V V

U UC T C T T C C= rArr

ν + ν = ν + ν2p

4p1 2

2 1

83 5

TTTT T

=+ 1p

2844 KT = 1p

c

pRT

microρ = 2p

3p30699 kgmρ = 1p


201

d

1 21 2

22 fpp pT T T

ν + ν = ν rArr + =

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν

2p

4p1 2

2 1

116 5

TTTT T

=+ 1p

5 22 1

2 1

11 (2 ) 153 10 Nm2(6 5 )f

p T TpT T

+= = sdot

+ 1p



a

2 3T T= 1p

4p

11 1 1

1 1 2 2 2 12

VTV T V T TV

γminus

γminus γminus = rArr =

1p

1 11 300 Kp VT

R=

ν 1p

2 14 1200 KT T= = 1p

b

323 2

2

ln VL RTV

= ν 1p

4pDin legea transformării izoterme 3 2

2 3

V pV p

= 1p

22 1

2

32RTp pV

ν= = 1p

23 2441 JL = 1p

c

31 31 31Q L U= + ∆ 1p

4p

1 3 1 331

( )( )2

p p V VL + minus= cu 3 1

83

V V= 1p

31 1 3( )VU C T T∆ = ν minus 1p

31 1314 JQ = minus 1p

d

min

max

1CTT

η = minus 1p

3p1

2

1CTT

η = minus 1p

75Cη = 1p



202


Subiectul I Punctaj1 b [ρ]SI = Ω m = J m Aminus1 sminus1 3

2 b 1 2 1 2( ) 3ABU E E I R r r V= minus + + + + = 3

3 c bec fire 170 VU U U= + = 3

4 a 2

1 2max

1 2

( ) 4 W4( )

+= =

+E EP

r r 3

5 c 00

0 0

90 mA(1 ) (1 )

U U II IR R t t

= = = =+ α + α

3



a

Se lucrează icircn condiţiile lui K deschis e

e e

EIR r

=+

1p

4p1 3 1 3e eE E E r r r= + = + 1p

2 31

2 3e

R RR RR R

= ++

1p

1 AI = 1p

b

1 1 1U I R= 1p

3p1

1

e

e

EIR r

=+ 1p

1 154 VU = 1p

c

2V pU I R=

1p

4p

2 1

e

p e

EIR R r

=+ + 1p

2 3

1 1 1 1

p VR R R R= + + 1p

11 VVU = 1p

d

Se lucrează icircn condiţiile lui K icircnchis păstracircndu-se caracteristicile circuitului de la punctul (a)e

e e

E IR r

=+

1p

4p

1 23 3 3

1 2

e p e pr rE E E r r r r

r r= + = + = +

+ 1p

1 2

1 2

1 2

8 V1 1 3p

E Er rE

r r

minus= =

+1p

061 AI = 1p



203


a

22 2

2

PRI

=

2p3p

2 10R = Ω 1p

b

1 2e e eP P r R R= rArr = 1p

5p

1 21 3

1 2e

R RR RR R

= ++

pentru K deschis 1p

1 22

1 2e

R RRR R

=+

pentru K icircnchis 1p

1 2er r r= + 1p

2 45r Ω 1p

c

22 int 2P r I= 1p

4p1 2I I I= + 1p

1 1 2 2I R I R= 1p

22 6 AI I= = 2 int 162 WP = 1p

d2

2

e

e e

RR r

η =+ 2p

3p

037η = = 37 1p



204

DOPTICAtilde


b 2

1

ff

β = 3

2 a 2 2 22

1 1 1

1 45 cma

x n x xx n x n

= rArr = rArr = 3

3 c 15 mmaa

iin

= = 3

4

Prin introducerea lentilei icircn lichid distanţa focală devine

c

1 2

1 1 11

l l

l

f n nnn R R

= =

minus minus

lfrArr rarr infin 3

5a 1 1 2 2

192 61 10 J

ex c ex c

ex

L E L E

L minus

ε = + = +

rArr = sdot

3



a2p

3p

La trecerea printr-o lentilă convergentă un fascicul paralel de lumină converge icircn focarul imagine al acesteia deci 1 30 cmf = 1p

b

2p

4p

Din desen se observă că 1 2d f f= + 1p

2 25 cmf = 1p

c

1 12

2 1 1 1 1

1 1 1 75 cmf xxx x f f x

minus = rArr = =+ 1p

4p1 2 3 3 35 cmd x x x= + rArr =

1p

4 3 2

1 1 1x x f

minus = 1p

4 875 cmx = 1p


205

d

3

1

yy

β = 1p

4p1 2β = β β 1p

21

1

15xx

β = = minus 42

3

25xx

β = = minus

1p

3 1 5625 cmy y= β = 1p



a

W N= ε 1p

4p

hcε =

λ1p

W Pt= 1p

17euml 434 10 fotoniPtNhc

= = sdot 1p

b

ex cL Eε = + 1p

4p

c sE eU= 1p

ex shcL eU= minusλ

1p

1914 10 JexL minus= sdot 1p

c

1s

N eIt

=

2p

3p18

1 075 10 fotonisI tNe

= = sdot 1p

d1p

4p

1914 10 J 0875 eVexL minus= sdot = 3p



206

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj1 d 32 b 33 b 34 a 35 b 3



a

2 0yG F N+ + =

1p

3p2 sinN G F= minus θ 1p

Rezultat final 5135 N=N 1p

b

1 2 0x x fF F F+ + =

la limita alunecării 1p

4p

1 2 cos mF F N+ θ = micro 1p

1 2

2

cossinm

F Fmg F

+ θmicro =

minus θ 1p

Rezultat final 068mmicro = 1p

c

1 2x fF F F ma+ + =

1p

4p1 2 cos+ θ minus micro =F F N ma 1p

1 2 2cos ( sin )+ θ minus micro minus θ +=

F F F mgam

1p

2412 msa 1p

d

2 cosm mP F v= θ 1p

4p0 = +v v at =v at 2m

vv = 1p

2 cos2matP F= θ 1p

1854 W mP 1p



207


a

2

2PkxE = 2p

3p

Rezultat final 05 J=PE 1p

b

legea conservării energiei pentru sistem i fE E= 1p

4p2

2

= =i pkxE E sistem

1 2= + =f c c cE E E E

2p

Rezultat final sistem 05 JcE = 1p

c

2 2 21 1 2 2

sistem 2 2 2cm v m v kxE = + =

1p

4plegea conservării impulsului 1 1 2 20i fp p m v m v= hArr = minus

1p

21

1 1 2

( )

=+

kmv xm m m

12

2 1 2( )kmv x

m m m=

+1p

Rezultat final 1 282 ms=v 2 071 msv = 1p

d

fc FE L∆ = 1p

4p

21 1

12m v m gd= micro 1p

21

2vd

g=

micro1p

Rezultat final 2 m=d 1p



208


Subiectul I Punctaj1 d 32 d 33 b 34 d 35 c 3



a1 0 2

lp p g= + ρ 2p3p

51 15 atm 15∙10 Pa p = 1p

b

1 1 2 2p V p V= 1p

4p

1 0 2

= + ρlp p g 1

2=

lV S2 0+ ρ =p gx p 2 ( )= minusV S l x 1p

2 00( ) 0

4p lgx x p glρ minus + ρ + =

1p

10 cm x 1p

c

1 1p V RT= ν 1p

4p1p ShRT

ν = 1p

ndash322∙10 mol ν 2p

d

2iU RT= ν 1p

4p2 2aer N O ArU U U U= + + 1p

2

5 078 2

= νNU RT2

5 02 2

= νOU RT 3 0022ArU RT= ν 1p

aer 138 J U 1p


Subiectul al III-lea Parţial Punctaja reprezentare corectă 3p 3p

b

1 1 2 2 amestec

1 2

V VV

C CC ν + ν=

ν + ν 2p

4p2

=ViC R 1

3 2

=VC R 252VC R= 1p

amestec 3 2493 Jmol K VC R= sdot 1p


209

c

min

max

1 ndashCTT

η = 1p

3pmin 1=T T max 14T T= 1p

Rezultat final 3 754Cη = = 1p

d

cedat

primit

1 ndashQQ

η =

1p

5p

primit 1ndash 2=Q Q 1ndash 2 2 1 1( ndash ) 3p pQ C T T C T= ν = ν 1p

cedat 2ndash3 3ndash1= +Q Q Q 2ndash3 3 2 1( ndash ) ndash 3 = ν = νV VQ C T T C T 1

3ndash1 1 13

ln ndash2 ln 2VQ RT RTV

= ν = ν 1p

p VC C R= + 1p

3 2 ln 2 3 ndash 2ln 21 ndash 3 12+

η = =V

p

C RC

Rezultat final 215η = 1p



210


Subiectul I Punctaj1 a 32 b 33 a 34 c 35 a 3



a2

12

Ve A

V

R RR R RR R

= + ++ 2p

3p

Rezultat final Re = 312 Ω 1p

b

Teoremele lui Kirchhoff 1 1 2 2

2 2 3

1 2 3

( )A

V

E I R R r I RI R I RI I I

= + + + = = +

111 16

=I A

25 8

=I A

31

16I A=

2p

4p

1AI I= 3V VU R I= 1p

069 AAI 6875 V VU 1p

c 1

AA

EI R r R

=+ +

0VU = 2p4p

Rezultat final 1 AAI = 0VU = 2p

d

er R = 2p

4p1e AR R R= + 1p

Rezultat final rprime = 212 Ω 1pTOTAL pentru Subiectul al II-lea 15p


a

Din grafic max 20 W=P 10 V2

=E

2p

4p2

max 4EP

r= 1p

Rezultat final 20 V=E 5= Ωr 1p

b

2middotP R I= 2p

4pEIR r

=+

1p

Rezultat final 875 W=P 1p

c

ext

totala

p

p

RPP R r

η = =+

2p

4p2pRR = 1p

Rezultat final 05 50η = = 1p


211

d

lRSρ

= 1p

3p2

4dS π

= 1p

Rezultat final 86 10 mminusρ = sdot Ω 1p



212

DOPTICAtilde

Subiectul I Punctaj1 d 32 d 33 c 34 b 35 a 3



a2 1 2

1 1 1 minus =x x f

2 22

1 1

y xy x

β = =

2p

4preprezentarea grafică a imaginii 1p

Rezultat final 2 20 cm= minusx 2 5 cm=y 1p

b2 1 1

1 1 1ndash x x f

=

2 2

1 1

y x y x

β = = 2p

4preprezentare grafică 1p

Rezultat final 2 2240 cm ndash30 cmx = y = 1p

c

2 1 1

1 1 1 minus =x x f 2 120 cm=x 1p

4p1 2 0= minus =x d x 1p

2 1 2

1 1 1ndash x x f

=

1p

Rezultat final 2 = 0 x 1p

d 1 2

1 1Cf f

= + 2p3p

Rezultat final ndash083 dioptriiC = 1p



a

11

2λ

=Dil

2

2Dil

λ=

2p

4p1

1

cλ = rArr

ν1 2 1

12id

cλ ν

=

1p

Rezultat final 41 24 10 mminus= sdotd 1p

b

1 1 2 2k i k i= rArr 1 1 2 2k kλ = λ 2p

4p1 25 6k krArr = 1 26 5k krArr = = 16y i=

1p

Rezultat final 6 mm=y 1p


213

c2 1( )δ = minusn e n e 2p

3pRezultat final 4( ) 3 10 mmminusδ = sdot 1p

d

Pentru dispozitivul Young icircn aer diferenţa de drum este 02 sdot

δ =l yD

2p

4pCondiţia de maxim central 2 02

k λδ = = 1p

Rezultat final 1 1 10

( ndash1) 36 mme n iyc

ν= = 1p



214

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj1 b

1

1

01 kgpm mv

= rArr =

m m

vF m a m

t∆

= sdot = sdot∆

m 2 2 2 s

v v∆ = sdot = sdot

2 NF =

3

2

a 0N F G+ + =

0 cos sin 0 (1)x F m gsdot α minus sdot sdot α = 0 sin cos 0 (2)y N F m gminus sdot α minus sdot sdot α = Din (1) şi (2) rezultă

cosm gN sdot

=α

3

3c Pentru a nu aluneca niciunul dintre corpuri de pe pupitru acesta trebuie icircnclinat cu un unghi maxim egal cu unghiul de frecare cel

mai mic din tabelul de date tg 03 sin 029 17θ = micro = rArr θ = rArr θ = deg

4b 12 1 2v v v= minus

12 2 1 v v v= minus 12km 216 h

v = icircn sens contrar mişcării avionului

5

b Lanţul alunecă de pe masă din momentul icircn care greutatea porţiunii care atacircrnă devine puţin mai mare decacirct forţa de frecare dintre porţiunea de lanţ rămasă pe masă şi masă1 3 1 4 4 3

m g m gsdot sdot ge sdotmicro sdot sdot rArr micro =



a

Folosind metoda grafică de determinare a distanţei parcurse de cutie pe plan

obţinem = 45 md 3p 3p

bDin grafic viteza cu care se icircntoarce cutia la baza planului este egală ca mărime cu viteza cu care este lansat corpul icircn sus de la baza planului Aplicacircnd teorema de variaţie a energiei cinetice a cutiei icircntre stările iniţială şi cea de la momentul 3 st = obţinem 0micro =

3p 3p


215

c

Pentru a putea reprezenta grafic variaţia vitezei cutiei icircn funcţie de timp icircn cazul icircn care mişcarea pe plan se realizează cu frecare trebuie determinată distanţa parcursă de cutie pe plan icircn urcare d1 şi timpul de urcare Δtu respectiv timpul de coboracircre Δtc al cutiei pe planLa urcare Aplicăm teorema de variaţie a energiei cinetice a cutiei icircntre stările iniţială şi cea corespunzătoare opririi cutiei pe plan c RE L∆ =

( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d

sdotminus = minus sdot sdot minus micro sdot sdot sdot sdot α

= sdot α

rArr = sdot sdot sdot α + micro sdot α

( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α


1h este icircnălţimea pacircnă la care urcă cutia pe planObţinem ( )

20

1 1 307 m 2 sin cos

vd dg

= rArr =sdot sdot α + micro sdot α

2p

5p

Icircn timpul urcării pe plan acceleraţia cutiei se menţine constantă Icircn aceste condiţii viteza medie a cutiei icircn

timpul urcării poate fi scrisă 1 0 0 102 s2um u

u

d vv tt

+= = rArr ∆ =

∆

La coboracircre Distanţa parcursă pe plan este aceeaşi 1d

1p

Aplicăm teorema de variaţie a energiei cinetice a cutiei icircntre stările cea icircn care cutia este icircn repaus la icircnălţimea 1h şi cea icircn care cutia ajunge la baza planului c RE L∆ =

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus

sdotminus = sdot sdot minus micro sdot sdot sdot sdot α rArr

rArr = sdot sdot sdot α micro sdot α

= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

Icircn timpul coboracircrii cutiei pe plan acceleraţia acesteia se menţine constantă Icircn aceste condiţii viteza medie a

cutiei icircn timpul coboracircrii poate fi scrisă 1 2 0 137 s2cm c

c

d vv tt

+= = rArr ∆ =

∆

Durata totală a mişcării cutiei pe plan este 24 s u ct t t t∆ = ∆ + ∆ rArr ∆ cong

1p

După intrarea pe planului icircnclinat viteza cutiei rămacircne constantă 2v Icircn reprezentarea grafică semnul negativ al vitezei corespunde schimbării sensului mişcării cutiei

1p

d

Notăm cu R

forţa cu care corpul acţionează asupra planului icircnclinat şi cu Rprime

forţa cu care planul acţionează asupra corpului R R R Rprime prime= minus =

1p

4p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG

prime prime= + = + = sdot + micro rArr

rArr = sdot + micro = sdot sdot α sdot + micro rArr

rArr = α sdot + micro = =

2p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




1p



a ( )1 1 1 cos 05Jp pE m g h m g l E= sdot sdot = sdot sdot sdot minus α rArr = 3p 3p

b

Aplicacircnd legea conservării energiei mecanice icircntre stările iniţială şi cea corespunzătoare poziţiei verticale a firului obţinem

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus

sdotsdot sdot = rArr = sdot sdot = sdot sdot sdot minus α

rArr = sdot

3p4p

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot1p


216

c

Pentru a răspunde la cerinţele subpunctelor c şi dNotăm cu 1d porţiunea pe care coeficientul de frecare variază uniform cu distanţa şi aplicăm teorema variaţiei energiei cinetice a bilei pe această porţiune Datorită variaţiei uniforme cu distanţa forţa de frecare medie care

acţionează asupra bilei poate fi scrisă 1 21 1

2medief mediuF m g m gmicro + micro= micro sdot sdot = sdot sdot

( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d

sdot sdot micro + micro∆ = rArr minus = minus sdot sdot sdot rArr

= minus micro + micro sdot sdot

( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) 2 21 2

1 2

2 2

m m v k x

kv xm m

prime+ sdot sdot= rArr

prime = sdot+

( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

2p

4p

Impulsul sistemului de corpuri va fi ( ) ( )1 2 1 2 014 N sp m m v x k m m pprime prime prime= + sdot = sdot sdot + rArr = sdot

2p

d

Scriem legea conservării impulsului sistemului de bile icircn ciocirea lor plastică ( )1 1 1 2

11 1 1 1

1

= 14 m s

m v m m vpm v p v vm

minus

primesdot = + sdot rArr

primeprimerArr sdot = rArr = rArr sdot

1p

4p

( )1 1 1 2

11 1 1 1

1

= 14 m s


minus


primeprimerArr sdot = rArr = rArr sdot 1p

( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr =

1p( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr = 1p



217


Subiectul I Punctaj1 c

2 2CO C Og = + 44

molmicro micro micro =

5

29 10 Pa = 29 atm

m R Tp V R T pV

p

sdot sdotsdot = ν sdot sdot rArr = rArr

micro sdot

rArr = sdot

3

2

a

12

1213

13

12 13

cu 22

5 cu 082

V

p pp

RQ C T C C R

Q CQ C T C RQ C

T T T

= ν sdot sdot ∆ = + = sdot = ν sdot sdot ∆ = sdot rArr = minus =

∆ = ∆ = ∆

3

3

d-1 -1

1 1 2 2 T V T Vγ γsdot = sdot

1 1 1

2 2 2 1 2

2

V S hV S h S h

= sdot= sdot = sdot sdot

2 75

p

v

C iC i

+γ = = =

1

12 1

2

VT TV

γminus

= sdot

Icircnlocuind numeric 2 21213 K -1517 CT t= rArr = deg

3

4

b2 1 1 2 3 4 0 0 0 0i i i iV V L L L Lgt rArr lt lt lt =

Folosind interpretarea grafică a lucrului mecanic obţinem3 2 1 4i i i iL L L Llt lt lt

3

5 c 3TOTAL pentru Subiectul I 15p


a 5 53 10 PaR Tp V R T p pV


b

1 11 1 2 2 T V T Vγminus γminussdot = sdot

1

12 1

2

VT TV

γminus

= sdot

(1) 1p

5p

p

v

CC

γ =

Folosind expresia energiei interne a amestecului de gaze obţinem

1 2 31 2 3v v v vC C C Cν sdot = ν sdot + ν sdot + ν sdot

1 2 31 2 3v v v vC c C c C c C= sdot + sdot + sdot

249 25vC R R= sdot cong sdot 35p v pC C R C R= + rArr cong sdot

Observăm că datorită cantităţii mici de gaz monoatomic valorile căldurilor molare izocoră şi izobară ale ames-tecului de gaze sunt foarte apropiate de cele ale gazului biatomic

2p

Exponentul adiabatic va avea valoarea 7 145

γ = = 1p

Icircnlocuind icircn expresia (1) obţinem 2 2 240 K 33 CT t= rArr = minus deg 1p


218

c1 1

111

1

1344 g

cmm

ν = sdot ν rArr =ν = micro

3p 3p

d

2 2

222

2

4368 g

cmm


1p

4p

3 3

333

3

08 g

cmm


1p

1 2 3 5792 gm m m m m= + + rArr = 1p

32

kg 29 m

mV

ρ = rArr ρ = 1p



a

m mν = rArr micro =

micro ν 1p

3p1 2 3 4m m m m m= + + +

1 2 3 4ν = ν + ν + ν + ν1p

Calculacircnd valorile maselor gazelor din amestec şi icircnlocuind numeric obţinem g 92

molmicro = 1p

b

4p 4p

c

( )12 2 1ced pQ Q C T T= = ν sdot sdot minus 1p

5p

1 2 3 41 2 3 4v v v v vC C C C Cν sdot = ν sdot + ν sdot + ν sdot + ν sdot

1 2 3 41 2 3 4v v v vv

C C C CC

ν sdot + ν sdot + ν sdot + ν sdot=

ν

1p

Icircnlocuind valorile date icircn textul problemei obţinem21 31 v pC R C R= sdot rArr = sdot 1p

1 1 11 1

2121 2

22

2

p V R TT TTV Tp R T

sdot = ν sdot sdot rArr = rArr =

sdot = ν sdot sdot

1p

112 1

1 1 15

31 2

- 155 - 155

155 10 J

ced

ced

TQ Q R T

R T p VQ

= = sdot ν sdot sdot minus =

= sdot ν sdot sdot = sdot sdot

= minus sdot

1p


219

d

23 31 primitQ Q Q= +

1 1 1 1 123 2

2 1

2ln ln ln 22 2

V T V p VQ R T RV V

sdot sdot= ν sdot sdot sdot = ν sdot sdot sdot = sdot

1p

3p5

23 035 10 JQ = sdot

( ) 131 1 2 1 1 121 105

2vTQ C T T R T p V = ν sdot sdot minus = sdotν sdot sdot minus = sdot sdot

1p

531 105 10 JQ = sdot

514 10 JprimitQ = sdot

1p



220


Subiectul I Punctaj1 a

1 1 2 2 3 31 2 3

1 1 2 2 3 31 1 2 2 3 3

1 2 3

U S U S U SI U U Ul l ll l l

S S S S

sdot sdot sdot = = = ρ sdot ρ sdot ρ sdot rArr = = ρ sdot ρ sdot ρ sdot= = = Din grafic 1 2 3 1 V 2 V 15 VU U U= = = Icircnlocuind numeric obţinem

1

2

2775 n m222 n m

ρ = Ω sdotρ = Ω sdot

3

2

b

( ) ( )1

1 10 1

1 0 1

1 1

UI UR IR

R R

= rArr = sdot + α sdotθ= sdot + α sdotθ

(1)

Analog ( )2

0 2

1

UIR

=sdot + α sdotθ

(2)

Raportacircnd relaţiile (1) şi (2) şi icircnlocuind numeric obţinem2 1280 Cθ = deg

3

3b Alegacircnd convenabil un ochi şi aplicacircnd teorema a II-a a lui Kirchhoff

1 2 3 1 15 AE E E I R I+ + = sdot rArr = 3

4

bCei trei rezistori din circuitul exterior sunt grupaţi icircn paralel

1 2 3

1 1 1 1 pR R R R

= + + Icircnlocuind numeric 20 7pR = Ω

7 ApE I R I= sdot rArr =

140 WP E I P= sdot rArr =

3

5

a2

U N SW tl

sdot sdot= sdot ∆

ρ sdot Icircnlocuind numeric = 10 mW hW sdot

3



a

1K rarr

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

2p

4p1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

1p

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω 1p

b

2K rarr

2

2

2

154 A

EIR r

I

=sdot +

rArr =

1p

4p2

2

2

154 A

EIR r

I

=sdot +

rArr =1p

2 1 034 AI I I I∆ = minus rArr ∆ = 2p


221

c2 2

3e eRR Rsdot

= rArr = Ω 3p 3p

d3 3 36 A2

3

EI IR r= rArr =

sdot+ 2p

4p

Intensitatea curentului icircntre punctele 2 şi B este egală cu 2 3 36 ABI I= = 2p



a

deschisK rarr = 075 A

400 = 3

b b b b

bb b

b

P U I IUR RI

= sdot rArr

= rArr Ω2p

4p

bb c

EIR R r

=+ + 1p

485 1616 3cRrArr = Ω = Ω 1p

b ( ) 2 = 995625 Jext b c extW R R I t W= + sdot sdot ∆ rArr 3p 3p

c

icircnchisK rarr b

t

PP

η = Cursorul icircmparte reostatul icircn două părţi avacircnd rezistenţele electrice R1 şi R2 Circuitul exterior sursei conţine rezistorul R1 icircn serie cu rezistorii R2 şi Rb legaţi icircn paralel

Becul funcţionează normal deci 2

100 VR bU U= = 1 2R R R+ =

( )1 bE I R r U= sdot + +

2 b RI I I= +

2 2b RU I R= sdot

2p

4p

Rezolvacircnd sistemul de ecuaţii şi ţinacircnd cont de valoarea lui R obţinem 1 AI cong 1p

Icircnlocuind obţinem 75 333225

η = = 1p

d

Pentru ca sursa să transfere putere maximă circuitului exterior rezistenţa acestuia trebuie să fie egală cu rezis-tenţa interioară a generatorului eR r= 1p

4p

Notăm cu Rprime rezistenţa rezistorului conectat icircn paralel cu reostatul şi cu Rs rezistenţa porţiunii de circuit pe care se află becul şi reostatul cursorul fiind la mijlocul acestuia

1 1 1

e sR R R= +

prime

1p

Intensitatea curentului prin ramura principală a circuitului va fi 225 A2EI I

r= rArr =

sdot

Aplicacircnd teorema a doua a lui Kirchhoff ochiului care conţine sursa de tensiune şi porţiunea de rezistenţă Rs

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

29 V 2s b bRE I r I U Uprime prime= sdot + sdot + rArr = bU prime fiind tensiunea electrică la bornele becului icircn aceste condiţii

Puterea consumată de bec va fi 2

632 Wbb b

b

UP PR

= rArr =

2p



222

DOPTICAtilde


( )( )1

22 2max 2 2 2 5d a a a= sdot + sdot = sdot sdot

3

2 a 3

3

a1

2

50 cm1 80 cm

fC

ff=

= rArr = Observăm că 1 2 D f f= + deci sistemul celor două lentile este afocal

Notacircnd cu 1h icircnălţimea obiectului şi cu 2h icircnălţimea imaginii acestuia din asemănarea triunghiurilor haşurate se obţine

1 2

1 2

1 22

1

2

16 cm

h hf f

h fhf

h

= rArr

sdot= rArr

=

3

4c

39 Vextrextr s s s

LL e U U Ue

ε minusε = + sdot rArr = rArr = 3

5

a

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus

sdot θ = rArr θ = sdot rArr θ = sdot rArrλ sdot =sdot

primerArr θ = rArr θ = θ = = degπ π

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3



a 3p 3p

b

60 30rα = deg rArr = deg 1p

4p0 sin sin an i n rsdot = sdot 1p

Icircnlocuind obţinem 2 2sin arcsin3 3

i i = rArr =

2p

c

2 5 2sin cos tg = 3 3 5

r r r= rArr = rArr

2p

4p1

1

1

tg i = = 50 cmtg = = 775 cmtg

tg

x hh i H H

r hxrH

rArr rArr=

2p


223

d

2p

4p

0

2

sin 90 sin 3 sin = 075 tg 7

tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p



a

19min min

max

= 262 10 J = 163 eVR

h c h c minussdot sdotε = rArr ε = sdot

λ λ 2p

4p19max min 2 = 522 10 J = 326 eVminusε = sdot ε sdot 1p

Pentru ca celula să funcţioneze icircn vizibil trebui ca max extr VL lt ε = ε astfel icircncacirct pot fi folosite Li Ba Cs K

1p

b

19 398 10 J = 249 eVh c minussdotε = rArr ε = sdot

λ2p

4p max

min c

extrextr c

EL im

L E=

rArr = rArrε = + trebuie folosit ca material Cs 2p

c 049 V extrCs s sL e U Uε = + sdot rArr = 3p 3p

d max max min extrv L= rArr ε = = rArr folosim Cs şi radiaţie cu λ = 380 nm 1p

4p252 m = 66 10

2 sc

cm v EE v v

msdot sdot

= rArr = rArr sdot 3p



224

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj1 b 32 a 33 c 34 a 35 d 3



a

Aplicăm principiul I al dinamicii 0F N Gn Gt Ff+ + + + =

1p

4pFăcacircnd proiecţia pe axe obţinem Gn ndash N = F ndash Gt ndash Ff = 0 1pRezolvacircnd sistemul obţinem F = mg(sin q + m cos q) 1p

Rezultă F = 20 N 1p

b

Aplicăm principiul II al dinamicii Ft N Gn Gt Ff m a+ + + + = sdot 1p

4pFăcacircnd proiecţia pe axe obţinem Gn ndash N = 0Ft ndash Gt ndash Ff = m middot a 1p

Rezolvacircnd sistemul obţinem Ft = m (a + g sin θ + μg cos θ) 1pRezultă Ft = 204 N 1p

cAplicăm legea mişcării d = v t 2p

3pRezultă t = 05 s 1p

d

Corpul icircncepe să alunece pe cărucior atunci cacircnd F ge Ff unde F este forţa aplicată corpului iar Ff este forţa de frecare care apare la suprafaţa de contact dintre corp şi cărucior 2p

4pF ge μN F ge μmg 1p

μ le 01 1pTOTAL pentru Subiectul al II-lea 15p


a

Aplicăm teorema de variaţie a energiei cinetice 2 2

0

2 2mv mv Fr Dminus = minus sdot 2p

4p2 20

2G gFr v v df f

= rArr = minus sdot 1p

V = 9 ms 1p

b

Aplicăm legea conservării impulsului (icircn timpul ciocnirii rezultanta forţelor exterioare este 0) 1 1 2 2 1 2( )m v m v m m vprime+ = +

2p

4pPentru ciocnirea unidimensională şi frontală şi ţinacircnd cont de faptul că v2 = 0 obţinem

1 1 1 2( )m v m m vprime= + 1p

Rezultă 1 1

1 1

3 msm vvm v

prime = =+ 1p


225

c

Scriem teorema variaţiei energiei cinetice ΔEc = L unde L reprezintă lucru mecanic efectuat de rezultanta forţelor exterioare ce acţionează asupra sistemului 1p

4p

1 2( )0 m m g Fr Df

+minus = sdot 1p

1 2( )m m gFrf

+= 1p

Rezultă d = 45 m 1p

d

21 2

21 1

( )cs

cl

E m m vxE m v

prime+= = unde x reprezintă fracţiunea din energia cinetică pe care o posedă primul vagon ime-

diat icircnainte de ciocnire2p

3p

Deci 103

x vprime= 1p



226


Subiectul I Punctaj1 b 32 b 33 a 34 d 35 d 3



a

ν1 = 2Omicro

m 1p

3p1

2

116

νν

= 1p

2

2Hmicrom

ν = 1p

b

Scriem ecuaţia de stare icircn cele două compartimente si punem condiţia de echilibru mecanic pe piston p1= p2 = p 1p

4pp 2

ν= ν1RT1

p 2ν

= ν2RT2

2p

Rezultă 21

12

16TT

νν

= = 1p

c

Deoarece T2 = 16T1 prin icircncălzire hidrogenul icircşi măreşte volumul de la V2 la V2prime iar O2 va avea volumul V1primepV1prime= ν1RT1

pV2prime= ν2RT1

1p

4p

1

22

1VV

=νν

V1prime + V2prime = VUnde V = L S

1p

V1prime= 2L x S minus

V2prime= 2

L x S +

1p

x = 088 m x gt 0 rArr pistonul se deplasează către compartimentul care conţine O2 1p

d


4p1 2

2m m m+ =


Obţinem 1 2

1 2

2micro micromicro =

micro + micro1p

Rezultă 1 2

1 2

2micro micromicro =

micro + micro 376 gmol 1p



227


a

Reprezentarea grafică icircn coordonate P-V1rarr2 (comprimare izotermă)2rarr3 (icircncălzire izobară)3rarr1 (răcire izocoră)

4p 4p

b

Gazul cedează căldură pe procesele 1rarr2 şi 3rarr1Qc = Q12+ Q31

1p

4p

Ţinacircnd cont de expresiile căldurilor icircn procesele respective obţinem

12

1

lnc T VQ RV

= ν + νCv(T1 minus T3) 1p

Pentru a calcula temperatura T3 scriem ecuaţia de stare icircn stările I şi III obţinem1 1 1

1 1 3

13 1

3

4

1 44

PV RTPV RT

T T TT

= ν= ν

= rArr =

1p

Rezultă 12 31 293475 JcQ Q Q= + = minus 1p

c

Expresia randamentului unei transformări ciclice biterme | |1 c

p

QQ

η = minus 1p

4pSistemul primeşte căldură pe procesul 2rarr3Qp = O23 = νCp(T3 ndash T1) = 37395 J 2p

215η = 1p

d

minCarnot

max

1 TT

η = minus unde Tmin = T1 şi Tmax = T3 2p

3pCarnot

3 754

η = = 1p



228





a

Cacircnd la bornele generatorului avem un singur voltmetru de rezistenţă RV (1)U1 = I1Rv= v

V

ERR r+

1p

4pCacircnd sunt montate cele două voltmetre icircn paralel rezistenţa lor echivalenta va fi

2vR (2) 2 2 2 2

v v

v

R ERU IR r

= =+

2p

Rezolvacircnd sistemul format din relaţiile (1) (2) obţinem E = 12 V 1p

b

Aplicacircnd legea lui Ohm pentru un circuit simplu cacircnd avem un singur ampermetru 1A

EIR r

=+ 1p

4pCacircnd cele două ampermetre sunt legate icircn serie 2 2 A

EIR r

=+

2p

Rezolvacircnd sistemul obţinem r = 2 Ω 1p

c2 1

1 1VrR

EU U

=

minus

1p

4p

2 1

1 1AR E

I I

= minus

2p

Obţinem RV= 4 Ω RA= 2 Ω 1p

d

RV rarr infin 1p

3p1V

VV

V

ER EU ErR rR

= = =+ + 1p

Uv = 12 V 1pTOTAL pentru Subiectul al II-lea 15p


a

Cacircnd k este deschis rezistenţele R1 R3 sunt legate icircn serie R2 R4 serie 1p

4p

1 1 3

2 2 4

S

S

R R RR R R

= += +

1p

1 3 2 4

1 2 1 2 3 4

1 1 1 ( )( )tot

tot S S

R R R RRR R R R R R R

+ += + =gt =

+ + + 1p

Din legea lui Ohm pentru un circuit simplu obţinem 1 3 2 44

1 2 3 4

( )( ) 4

E R R R Rr RI R R R R

+ += + rArr = Ω + + +

1p

bW = (E minus rI prime)I primet 2p

3pW = 504 J 1p


229

c

Cacircnd k este icircnchis R1 R2 legate icircn paralel R3 R4 legate icircn paralel Rtot= Rp1 + Rp21p

4p1 2 3 4

1 2 3 4

EI R R R RrR R R R

=+ +

+ +2p

Rezultă I = 104 A 1p

d

Cacircnd k este deschis 1 807tot

tot

RR r

η = =+

2p

4p


tot

RR r

η = =+ 2p



230

DOPTICAtilde

Subiectul I Punctaj1 d 32 a 33 a 34 a 35 d 3



a

Pentru expresia distanţei focale a unei lentile subţiri

1 2

11 1( 1)

fn

R R

=

minus minus

2p

4p1

2 0RR

rarr infinlt

1p

R = 15 cm 1p

b

2

1 2 1f f f= + unde f2 reprezintă distanţa focală a lentilei biconcavă formată prin introducerea lichidului f2 = minus20 cm 2p

4p2

1 2

1 1 1( 1)l

fn

R R

=

minus minus

1p

|R1| = |R2| = 15 cmRezultă nl=137 1p

c

2 2

1 1

1y xy x

β = = = minus 1p

3p

2 1

1 1 1x x f

minus = 1p

Rezultă x1 = minus120 cm 1p

d

Vom arăta că pentru un sistem afocal d = f1 + | f2 | mărirea liniară a sistemului nu depinde de poziţia obiectului şi a imaginii Considerăm obiectul luminos aflat la distanţa x1 de prima lentilă

21

1

xx

β = 2 1 1

1 1 1x x f

minus =

1p

4p

Imaginea obţinută icircn prima lentilă devine obiect pentru cea de-a doua lentilă

2 1 2

1 1 1x x fminus =

2

21

xx

β =

Unde x1prime = x2 minus d d = f1 + | f2 |

1 21 1 2

1 1

f xx f ff x

= minus minus+

1p

Mărimea dată de sistem 1 21 2 2

11 1

1 1

f fff x

f x

β = β sdotβ = sdotminus+

+

1p

2

1

ff

β = minus

d = f1 + | f2 |Rezultă d = 60 cm

1p



231


aScriem formula poziţiei maximului de ordin k

2kk Dx

lλ

=k = 5

2p3p

Rezultă xk=10 mm 1p

bFormula interfranjei

2lDi

l= 2p

3pRezultă I = 2 mm 1p

c

Introducacircnd o lamă de sticlă icircn dreptul fantei S1 drumul optic parcurs de unda provenită de la S1 creşte cu (n ndash 1) middot er1prime = r1 + e(n ndash 1)

1p

5p

Diferenţa de drum opticδprime = r2 ndash r1prime = r2 ndash r1 ndash e (n ndash 1) = k ∙ λ 2p

Pentru maximul central k = 0 sistemul de franje se deplasează pe ecran cu 1

2knx eD

lminus

=

xk = 04 m

1p

Sistemul de franje se deplasează spre fanta acoperită de lamă 1p

d

Condiţia de suprapunere a maximelor de interferenţă xk1 = xk2 1p

4pDin formula poziţiei maximului de ordin k 1 1 2 2

2 2k D k

l lλ λ

= 1p

Obţinem k1 = 13 şi k2 = 10 1p

Rezultă xcomun = 26 mm 1p



232

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj1 d 32 b 33 a 34 d 35 d 3



a

2

2f

BG F cB cA

mvL L E E mgh Nl+ = minus hArr minus micro = 2p

4p2 2 (1 )Bv gh= minus micro 1p

4 msBv = 1p

b

0cE∆ = 1p

3p20J

20JpE mgh

E

∆ = =

∆ =2p

c

2 0

2f

BF cB

mvL E mgd= minus hArr minusmicro = minus 2p

4p2 2

2 2B Bv vgd d

gmicro = rArr =

micro1p

4md = 1p

d

( ) 0 0

fG F c final c initialL de readucere L L E E+ + = minus = minus 1p

4p

cos 0L mgh mgd mglminus minus micro minus micro α = 1p

cos ( )sin

hL mgh mgd mg mg h d h= + micro + micro α = + micro + microα

1p

40JL = 1p



233


a

1 1F t psdot ∆ = ∆

1p

4p1 2F t mv mv mvsdot ∆ = minus minus = minus

1p

12mvF

t=

∆1p

1 2kNF = 1p

b

Triunghiul impulsurilor fiind echilateral se poate scrie v v v= =

1p

4p

2

2 2

22

p p mv

F t pp mvFt t

∆ = =

sdot ∆ = ∆

∆= =

∆ ∆

2p

2 1 kNF = 1p

c

H F t= sdot ∆

H F t= sdot ∆1p

3p

1 1

2 2

2 N s1 N s

H F tH F t

= sdot ∆ = sdot= sdot ∆ = sdot

2p

d

2p

4p

3 2 2 2N sp p mv∆ = = = sdot

2p



234


Subiectul I Punctaj1 a 32 c 33 b 34 a 35 d 3



a 4p 4p

b

A BU U U∆ = ∆ = ∆ 1p

4p2 1 2 1 2 2 1 1

5 5( ) ( ) ( )2 2VU C T T RT RT p V p V∆ = ν minus = ν minus ν = minus 2p

250 JU∆ = 1p

c

13 32 3 1 2 3 3 1 2 35 7( ) ( ) ( ) ( )2 2A V pQ Q Q C T T C T T RT RT RT RT= + = ν minus + ν minus = ν minus ν + ν minus ν 1p

5p

2 1 1 1 2 2 2 1 1 2 1 2 2 15 7 5 7( ) ( ) ( ) ( )2 2 2 2AQ p V p V p V p V V p p p V V= minus + minus = minus + minus 1p

650 JAQ = minus 1p

14 42 1 2 1 2 2 17 5( ) ( )2 2BQ Q Q p V V V p p= + = minus + minus 1p

50 JBQ = minus 1p

d- 650 250 900JA A AL Q U= ∆ = minus minus = minus 1p

2p- 50 250 300 JB B BL Q U= ∆ = minus minus = minus 1p



a

12 34 0Q Q= =

123 3 2 3 2 3 2 2 2 2 3 2 3 2

5 5 5 5( ) ( ) ( ) ( ) ( )2 2 2 2V

VQ C T T RT RT p V p V V p p p p= ν minus = ν minus ν = minus = minus = sdot minusε

2p

5p1 1 2 2 2 1

4 1 3 2 3 4

1 2

3 4

p V p V p pp V p V p p

γ γ γ

γ γ γ

rarr = rArr = sdotε


-11

23 4 1 4 1 15 5( ) ( )2 2primit

VQ Q p p p p Vγ γ= = ε minus = minus ε sdotε

3p


235

b41 1 4

5 ( )2cedatQ Q RT RT= = ν minus ν 2p

4p

1 1 4 1 1 4 15 5( ) ( )2 2cedatQ p V p V p p V= minus = minus sdot 2p

c

1 cedat

primit

QQ

η = minus 1p

3p4 1 1

1-1

4 1 1

5 ( ) 121 15 ( )2

p p V

p p Vγminus

γ

minus sdotη = minus = minus

εminus ε sdot2p

d

pL Q= ηsdot = 1p

3p-1 -1

4 1 1 4 1 1-1

1 5 5(1 ) ( ) ( 1) ( )2 2

L p p V p p Vγ γγ= minus minus ε sdot = ε minus minus sdot

ε2p



236





a

11

1 1 2 1 1 2 1 48 16 8 8 8 4 p

p

RR

= + = + = = = Ω 1p

5p

1 4 20 24sR = + = Ω 1p

22

1 1 1 3 1 69 18 18 6 p

p

RR

= + = = = Ω 1p

2 6 6 12sR = + = Ω 1p

1 1 1 3 1 824 12 24 8 AB

AB

RR

= + = = = Ω 1p

b

1 1 3

1 1 05 8 4 VABU I R IR

I R= += sdot =

2p

4p

1 11 1 2 2 2

2

4 1 025 A16 4

I RI R I R IR

= rArr = = = =

1 22 1 A4 1 20 4 20 24 VAB

I I IU

= + == + sdot = + =

2p

c

ABe

AB

UIR

= 2p3p

3AeI = 1p

d

ABAB e e

e

E UE U I R rI

minus= + rArr = 2p

3p

1r = Ω 1p



237


a

1 2

( )

( )

bE I R r UP PE I R r

I

= + ++

= + +

3p

5p

2

6024 24

2 20 50 05A

II

I II

= +

minus + ==

2p

b( )e e e

E EE I r R R r R rI I

= + rArr + = rArr = minus 2p3p

43eR = Ω 1p

c

( ) 24 5 24 24 12 12VbU E I R r= minus + = minus sdot = minus = 2p

4p1

1

21

24 2A1236 3A12

b

b

PIUPIU

= = =

= = =2p

d

11

12 62

bURI

= = = Ω

22

12 43

bURI

= = = Ω2p

3p

2 25 43 3600 387kJext eW I R t= sdot sdot ∆ = sdot sdot = 1p



238

DOPTICAtilde

Subiectul I Punctaj1 a 32 d 33 a 34 a 35 c 3



a1 1 1 1

22 1 1 2 1 1 1 1

1 1 1 1 f x f xxx x f x f x f x

+= + hArr = hArr =

+ 2p3p

2 15cmx = poziţia imaginii dată de prima lentilă 1p

b

1 2( ) (20 15) 5cmx d x= minus minus = minus minus = minus poziţia imaginii dată de prima lentilă icircn raport cu a doua lentilă pentru

care joacă rolul de obiect1p

4p 2 1 2 12

2 1 2 2 2 1 2 1


+= + hArr = hArr =

+ poziţia imaginii finale icircn raport cu a doua lentilă 2p

2 25cmx = minus poziţia imaginii finale icircn raport cu a doua lentilă 1p

c

2 2

1 2 1 1

sistemx xx x

β = β sdotβ = sdot 2p

3p12 12sistemβ = minus sdot = minus 1p

d 5p 5p



239


aIntroducerea unui strat subţire (lamă film peliculă) icircn calea unuia din fasciculele luminoase care interferă conduce la deplasarea figurii de interferenţă spre acel fascicul dar nu modifică interfranja 2p

3pAşadar raportul cerut este 1 1p

b

Icircn cazul acoperirii cu acelaşi film a ambelor fante figura de interferenţă rămacircne aceeaşi ca şi icircn cazul absenţei filmului atacirct ca poziţie cacirct şi ca interfranjă icircntrucacirct fiecare fantă acoperită introduce acelaşi drum optic supli-mentar deci diferenţa de drum optic icircntre razele care interferă nu se schimbă

2p3p

Aşadar şi icircn acest caz raportul cerut este 1 1p

c

Icircn acest caz analizacircnd distribuţia maximelor şi minimelor din figura de interferenţă reiese că deplasarea figurii

de interferenţă este 45x i∆ = sdot unde i este interfranja 2Dil

λ= 2p

5pIcircntre deplasarea figurii de interferenţă x∆ şi diferenţa de drum optic suplimentară δ introdusă de film există

relaţia 2

xD l∆ δ

= care devine ( 1) 45 ( 1) 45 ( 1)2 2 2 2

x e n i e n D e nD l D l D l l∆ minus minus λ minus

= hArr = hArr = de unde rezultă 451

en

λ=

minus

2p

8522 10 m 522 me minus= sdot = micro 1p

d

( 1) noue n kminus = sdotλ 2p

4p

( 1)nouke

nλ

=minus

1p

eprime = 232 10minus8 m = 232 ηm 1pTOTAL pentru Subiectul al III-lea 15p


240

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj1 c 32 b 3

3

d Din grafic 16 JpE = şi h=4m

Expresia energiei potenţiacuteale este 04kgpp

EE mgh m

gh= rArr = =

Aplicacircnd legea conservării energiei mecanice 2 2

892 ms2

poco po c p p o

EmvE E E E E vm

+ = + rArr = rArr = =

3

4 c 3

5

d

0

0 sin 0 cos

f

x

f y f

F N G FOx N F N FOy F F G F mg F

+ + + =

minus = rArr = sdot α+ minus = rArr = minus sdot α

cos 20 10 3 027

sin 10f

f

F mg FF NN F

minus sdot α minus= micro sdot rArr micro = = = =

sdot α

3


Subiectul al II-lea Parţial Punctaja Reprezentarea corectă a forţelor care acţionează asupra corpurilor 4p 4p

b

Pentru corpul (1) avem 1 1 1 1eF m g k l m g= rArr ∆ = 1p

4p

Pentru corpul (2) avem 2 12 2 1 2( )e eF m g F k l g m m= + rArr ∆ = + 1p

1 1

2 1 2

l ml m m

∆=

∆ + 1p

1

2

25

ll

∆=

∆ 1p

c

Pentru corpul (3) avem 23 3 eN G F= + 1p

4p2 2 1eF m g m g= + 3 3G m g= 1p

3 3 1 2 3 1 2 3 + ( ) 9 NN m g m g m g N g m m m= + rArr = + + = 2p

d

Din condiţia de echilibru rezultă1

1m gm g kx k

x= rArr = 2p

3p

100 Nkm

= 1p



241


a

Forţele ce acţionează asupra corpurilor

2p

4p

1 1 1 10N m g N m gminus = rArr = 1p

2 1 2 2 1 2

2

0 ( )80 N

N N G N m m gN

minus minus = rArr = += 1p

b

Pentru corpul (1) 11 1 1 1fT F G N m a+ + + =

Ox 11 1fT F m aminus =

Oy 1 1 1 10N m g N m gminus = rArr =

11 1 1 1 1 1 1 1f fF N F m g T m g m a= micro rArr = micro rArr minus micro =

1p

5p

Pentru corpul (2) 1 21 2 1 2 2 2f fT T F F N N G m a+ + + + + + =

1 22 1 2

2 1 2 2 1 2

0 ( )f fOx T T F F m a

Oy N N G N m m g

minus minus minus =

minus minus = rArr = +

2 22 2 2 1 2( )f fF N F m m g= micro rArr = micro +

2 1 1 1 2 1 2 2( )T T m g m m g m arArr minus minus micro minus micro + =

1p

Pentru corpul (3) 33 2 3 3fG T F N m a+ + + =

33 2 3

3 3 3 3

sin

cos 0 cosfOx m g F T m a

Oy N m g N m g

α minus minus =

minus α = rArr = α

3 32 3 2 3 cosf fF N F m g= micro rArr = micro α

3 2 3 2 3sin cosm g m g T m aα minus micro α minus =

2p

Rezultă 3 2 1 1 2 1 2 1 2 3

3 2 1 1 2 1 22

1 2 3

(sin cos ) 2 ( ) ( )(sin cos ) 2 ( ) 342

m g m g m m g a m m mm m m m ma g

m m m s

α minus micro α minus micro minus micro + = + +α minus micro α minus micro minus micro +

= =+ +

1p

c

1 1 1 1( ) 1626 NT m a g T= + micro = 1p

3p2 1 2 1 1 2 1 2

2

( ) 2 ( )4736

T a m m m g m m gT N

= + + micro + micro += 2p

d

1

1 1

22 ( )

R TR m a g

== + micro 2p

3p3252 NR = 1p



242


Subiectul I Punctaj1 c 32 c 33 d 3

4

c Aplicăm primul principiu al termodinamicii 1 2 1 2 1 2Q U Lrarr rarr rarr= ∆ +1 2 2 1( )( )

2vp p V VC T C T + minus

υ ∆ = υ ∆ + 2 1

32vT T T C R∆ = minus =

1 2 1 1 2 2 2 12 1 2 1

3( ) ( )2 2

p V p V p V p VC T T R T T minus + minusυ minus = υ minus +

Cum procesul 1rarr2 este de forma p = aV rezultă 1 1

2 2

p aVp aV

= =

1 11 2 2 1

2 2

p V p V p Vp V

rArr = rArr =

Obţinem 2 2 1 12 1 2 1

3( ) ( )2 2

p V p VC T T R T T minusυ minus = υ minus +

Din ecuaţia de stare obţinem 1 1 1

2 2 2

p V RTp V RT

= υ= υ

Rezultă 2 3 2C T R T R T C Rυ ∆ = υ ∆ + υ ∆ rArr =

3

5

c Aplicăm legile celor două procese

izoterm 1 1 2 2p V p V= dar 12 1 22

2pV V p= rArr =

adiabat 11 1 2 2 2 2

pp V p V pγ γγ= rArr = Rezultă raportul 12

2

2pp

γminus= 3



a

Legea conservării numărului de moli 1 2ν = ν + ν 1p

4p

Ecuaţiile de stare pentru heliu şi azot icircn starea iniţială şi pentru amestecul de gaze icircn starea finală 1 1

1 1 1 1

2 22 2 2 2

1 21 2

( )( )

p Vp V RTRTp Vp V RTRT

p V Vp V V RTRT

= ν rArr ν =

= ν rArr ν =

++ = ν rArr ν =

1p

Rezultă 51 2 1 11 2 1 1 2 2 2 2

2

( )( ) 25 10p V V p V Np V V p V p V pV m

+ minus+ = + rArr = = sdot 2p

b

Raportul dintre numerele de moli de gaz este 1 2 2

2 1 1

p Vp V

ν=

ν 2p3p

1

2

5ν=

ν 1p

c

Din legea conservării numărului de moli rezultă 1 2

1 2

m m m= +

micro micro micro1p

4p1 2

1 21 2

1 2

2m m

m m m

=micro micro

= + rArr micro =micro + micro

2p

7gmolmicro = 1p

d

Variaţia energiei interne este VU C T∆ = ν ∆ VC este căldura molară la volum constant a amestecului 2p

4p

Energia internă a amestecului este 1 2U U U= +

1 2 1 2

1 2 1 2

1 21 2

1 2

2 1

1 2 1 2

( )22

V V V V V V

V V V VVV

m m mC T C T C T C C C

C C C CC C

rArr ν =ν + ν rArr = + rArrmicro micro micro

micro + micro micro= + rArr =

micro micro micro micro micro

1p

1 22 1

1 2

134

7716

V VV

V

C CC R

mU C T KJ

micro + micro= =

micro + micro

∆ = ∆ =micro

1p



243


a

Reprezentarea icircn coordonate V şi T este

3p 3p

b

1rarr2 este un proces izocor 2 1V VrArr = 2rarr3 este un proces izobar 2 3p prArr =

1rarr3 este un proces de forma p = aV Procesul 3rarr4 este un proces izoterm1p

4p

1 1 1 33 1 2 1

3 3 1

3 3p aV p Vp p p pp aV V

= rArr = = rArr = =

1p

Parametrii stării 2 sunt 2 1V V= şi 2 13p p=

Parametrii stării 3 sunt 3 13V V= şi 3 13p p=1p

3 3 4 4p V p V= 4 1p p= 1 1 1 49 p V p VrArr =

4 19 45lV VrArr = =1p

c

Din ecuaţia termică de stare determinăm temperatura pentru cele 4 stări 1 11 1 1 1

p Vp V RT TR

= υ rArr =υ

1 12 2 2 1 1 2 2 1

33 3p Vp V RT p V RT T TR

= υ rArr = υ rArr = =υ

1 13 3 3 1 1 3 3 1

93 3 9p Vp V RT p V RT T TR


4 3 19T T T= =

2p

5p

32vC R= rArr 1 1

1 3 3 1 1 13 8( ) 122v

p VU C T T R p VRrarr∆ = υ minus = υ =

υ2p

1 3 6000 J = 6 kJU rarr∆ = 1p

dmin

max

1cTT

η = minus

1

3

1cTT

rArr η = minus 2p3p

88cη = 1p



244


Subiectul I Punctaj1 b 32 c 3

3

a Din grafic avem U = 20 V şi I = 5 A 4URI

rArr = = Ω

Intensitatea de scurtcircuit 2scEI rr

= rArr = Ω

Randamentul circuitului este 066 66RR r

η = = rArr η =+

3

4

b Calculăm parametrii sursei echivalente2 3

182 31 1 1 1 11

2 3

k

ke

k

E E E Er Er r rE

r r r r

sum + += = =

sum + +

1 1 61 1 1 1 11

2 3

e

k

rr

r r r r

= = =sum + +

3

5

d Puterea maximă 2

max 4EP

r= Din relaţia max max05 2P P P P= sdot rArr =

2 22 2

2

2 6 04 ( )E E R R Rr r

r R r= rArr minus + =

+

Rezolvacircnd ecuaţia de gradul doi găsim valorile rezistenţei R 12 3 2 2R r r= plusmn1 582 291R r= = Ω 2 018 09R r= = Ω

3



a

K1 K2 deschise rezistenţa echivalentă este 1

1

1 1 1 22 3e

e

RRR R R

= + rArr = 1p

4p1 18V 2AU I= = 11 1 eU I R= sdot 1p

11 1

1

2 33 2R UU I R

I= sdot rArr = 1p

6R = Ω 1p

b

K1 este icircnchis iar K2 deschis rezistenţa echivalentă2

2

1 1 1 32e

e

RRR R R

= + rArr = = Ω 1p

5p

1 1 1 1

2 2 2 2

U E I r E U I rU E I r E U I r

= minus = += minus = + 1p

1 1 2 2U I r U I rrArr + = + 1p

1 2

2 1

1U UrI I

minus= = Ω

minus E = 10 V 2p

csc

EIr

= 2p3p

10AscI = 1p

d

2

max 4EP

r= 2p

3p

max 25 WPrArr = 1p



245


a

1R şi

2R rezistenţele electrice ale celor două porţiuni de fir MC şi CN de lungimi 1l şi 2l 11

lRS

= ρ

22

lRS

= ρ 1p

3p

Prin 3R nu trece curent electric rezultă o punte Wheastone echilibrată 1 2

1 2 2 1 1 2 1 1 2 2l lR R R R R R R l R l

S Sρ ρ

sdot = sdot rArr = rArr =

1p

1 2 2 12l l l l l= + rArr = 1 20 cml = şi 2 40 cml =1p

bRezistenţa echivalentă a circuitului exterior este 1 2

1 2 1 2

1 1 1 ( )e

e

R R RRR R R R R R R

+ sdot= + rArr =

+ + + 3p4p

5eR = Ω 1p

c

legea lui Ohm pe icircntregul circuit3 45A

3e e

nE EIR nr R r

= = =+ +

2p4p

Tensiunea la borne este 225 VeU IR= = 2p

dParametrii generatorului echivalent sunt 3 3e eE E r r= = 2p

4pIntensitatea de scurtcircuit este 12Aesc

e

EIr

= = 2p



246

DOPTICAtilde

Subiectul I Punctaj1 b 3

2

a

Scriem ecuaţia lui Einstein pentru cele două frecvenţe 1 1 1 12 1

2 2 2 2

2 2 2 1 2

1 1 1

62 4V

(155 )155055

h L eU eU h LundeU V U

h L eU eU h LU h L U hdar LU h L U

ν = + = ν minus rArr = = ν = + = ν minus

ν minus ν minus ν= = rArr =

ν minus

140 0 118 10 HzLL h

h= sdotν rArr ν = = sdot

3

3

b

Aplicăm legea refracţiei ( )1aern = 3sin sin sin8

i n r r= rArr =Icircn triunghiul IPB

2

sintg tg1 sin

PB rr PB h r hh r

= rArr = =minus

Icircn triunghiul 90 60o oAP I iα = minus =

2

sintg tg 3 07m1 sin

x rx PB hPB r

α = rArr = α = =minus

3

4

d Formula interfranjei este 13

2r r

v v

D iil i

λ λ= rArr = =

λ

Relaţia dintre cele două interfranje este (1 ) 1 03rr v v v

v

ii i fi i f fi

= + = + rArr = minus =

f = 30

3

5

cUnghiul limită este 1 3sin

4l

n= =

2

tg tg

sin sin 115cos 1 sin

rl r h lh

l lr h h ml l

= rArr =

= = =minus

3



a

Lentila plan concavă este o lentilă divergentă cu 1R R= minus şi 2R rarr infin

Distanţa focală a lentilei este

1 2

1 20 cm1 1 1( 1)( )

Rfnn

R R

= = minus = minusminusminus minus

1p

4pFormula lentilelor subţiri 12

2 1 1

1 1 1 ( 20)( 30) 12cm20 30

fxxx x f f x

minus minusminus = rArr = = = minus

+ minus minus 2p

Mărirea liniară transversală este 2 2 2 12

1 1 1

( 12) 2 08 cm30

x y x yyx y x

minus sdotβ = = rArr = = =

minus

Imaginea este virtuală dreaptă mai mică decacirct obiectul1p

b

Dacă introducem lentila icircn apă 1 2( )R R R= minus rarr infin

1 2

11 1( 1)( )

a

a

a

n Rf n n nn R R

minus= =

minusminus minus2p

3p

80 cmf = minus 1p


247

c

Alipind cele două lentile se formează o lentilă biconcavă 1 2( )R R R R= minus = cu distanţa focală 1 10 cm1 1 2( 1) 2( 1)( )

R ffnn

R R

minus= = = = minus

minusminus minusminus

2p

4pFormula lentilelor subţiri

1

2 2 1 1

1 1 1 ( 10)( 30) 15 cm10 10

f xxx x f f x


+ minus minus 1p

Din formula pentru mărirea liniară transversală

2 2

1 1

x yx y

β = = rezultă

2 12

1

( 15) 2 1cm30

x yyx

minus sdot= = =

minus 1p

d

Distanţa focală a sistemului de lentile este 1 2 12

a

a a

f fFF f f f f

sdot= + rArr =

+1p

4p

1 15 cm1 1 2( 1)( 1)( )( )

aa

a

Rfnn

R R

= = =minusminus minus

minus1p

( 20) 15 30 cm30 20

F minus sdotrArr = = minus

minus1p

12

2 1 1

1 1 1 ( 30)( 30) 15 cm30 30

F xxx x F F x

sdot minus minusminus = rArr = = = minus

+ minus minus 1p



aUtilizacircnd formula interfranjei 15 mm

2Dil

λ= = 2p

3pDin relaţia 8 franjedd Ni N

i= rArr = = 1p

b

Maximul luminos de ordinul K se formează la distanţa 2MK Dx

lλ

=

pentru K = 5 55 75 mm2

Dxl

λ= =

1p

4pDistanţa la care se formează franje icircntunecoase este (2 1)2 2MDx Kl

λ= + sdot 1p

Pentru a treia franjă icircntunecoasă K = 2 3 5 35 375 mm2 2

Dx x x xl

λrArr = sdot = ∆ = minus Rezultă 375 mmx∆ = 2p

c

Icircn absenţa lamei diferenţa de drum este 2 12 Klxr r

Dδ = minus =

Icircn prezenţa lamei diferenţa de drum este 2 1 2 1 2 1( ) ( ) ( ) ( 1)S P S P r r e ne r r e n∆ = minus = minus minus + = minus minus minus

Rezultă 2 ( 1)Kl x e nDsdot

∆ = minus minus

2p

5p

Din condiţia de maxim K∆ = λ rezultă 2 ( 1)Kl xK e nDsdot

λ = minus minus

Maximul central se obţine pentru K = 0 rArr 0 02 2( 1)( 1)

l x l xe n eD D nsdot sdot

= minus rArr =minus

2p

0 33 45 mm2

Dx xl

λ= = = rArr 18 me = micro 1p

dDin relaţia 1

2KK Dx K

lλ

= = rArr 22l x

Dsdot

λ = 2p3p

2 480nmλ = 1p



248

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj1 b 32 c 33 b 34 b 35 c 3



a

2

2cpEm

= 2p3p


b

pFt

∆=

∆ 1p

4pF ma= 1p

pam t∆

=∆

1p

rezultat final 25msa = 1p

c

amFFF fx

=++ 21 1p

4p1 2 cosF F N ma+ θ minus micro = 1p

1 2 2cos ( sin )F F F mgam

+ θ minus micro minus θ += 1p

rezultat final 2412msa 1p

d

2

2atd = 1p

4p2 1d d∆ = minus

21

1 2atd =

22

2 2atd = 1 1 st = 2 2st = 2p

rezultat final 75m∆ = 1p



249


a

Energia căruciorului se conservă fi EE = 1p

4p1iE m gh= 2

1

2fm vE = 2p

rezultat final 2 2 10 632 msv gh= = 1p

b

Pentru mişcarea bilei 2

2gth = 1p

4pPentru mişcarea căruciorului pe orizontală

2

2atd vt= + 1p

Acceleraţia căruciorului 1 fm a F= minus fF mg= micro a g= minusmicro ( )2d h= minus micro

1p

rezultat final 3 md = 1p

c

Teorema de variaţie a energiei pentru cărucior pe pista orizontală2 2

1 1 112 2

m v m v m gdminus = minusmicro legea conservării impulsului ( )

1 1 1 2i fp p m v m m v= hArr = +

2p

4p( )1

1 2

2m g h dv

m mminus micro

=+

1p

rezultat final 2 10 21 ms3

v = 1p

d

Legea conservării energiei pentru sistemul cărucior-bilă după cionire( ) 2

1 2

2sistem

m m vE

+=

2

2sistemk lE ∆

=

1 2m ml vk+

∆ =

2p

3p

rezultat final 2 47 cm30

l∆ =

1p



250


Subiectul I Punctaj1 b 32 d 33 c 34 a 35 d 3



a

pV RT= ν 1p

4p1

2

2

1

2

1

2

1

micromicro

νν

sdot==mm

pp

2p

rezultat final 47

2

1 =pp

1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p

127gmolamestecmicro = 1p

c

0

2p V RT= ν

1p

4p0 finalp V RT= ν

1p

rezultat final 4fT T= 1200KfT = 2p

d

Icircn decursul transformării gazul parcurge o icircncălzire izocoră pacircnă cacircnd presiunea devine 0p apoi se dilată izobar pacircnă la volumul 2V

2p

4p0 0middot middot 2L p V p V RT= ∆ = = ν 1p

rezultat final 196kJL = 1pTOTAL pentru Subiectul al II-lea 15p


b1 1L p V p V= ∆ sdot ∆ = 2p

3prezultat final 1000JL = 1p

c

min

max

1CTT

η = minus 1p

3pmin 1T T= max 14T T= 1p

rezultat final 3 754Cη = = 1p


251

d

efectuat

primit

LQ

η = 1-2 2-3primitQ Q Q= + 2p

5p

1-2 2 1 1( )V VQ C T T C T= ν minus = ν 2-3 3 2 1( ) 2p pQ C T T C T= ν minus = ν 1p

p

V

CC

γ = p VC C R= + 1p

RC γ=

γ minus 1V

RC =γ minus

1p

12 1γ minus

η =γ +

1p

rezultat final 2 11

19η = = 1p



252


Subiectul I Punctaj1 d 32 c 33 a 34 b 35 d 3



a

4 52

4 5s

R RR RR R

= ++ 2p

4p3

13 3

se

R RR RR R

= ++ 1p


b

1e

EIR

= 1

4ptensiunea la bornele rezistorului 3R se poate scrie 1 2middot middotp sI R I R= unde 3

3 3

sp

R RRR R

=+

2p

rezultat final 2 625 AI = 1p

c 1

EIR

=

2p3p

rezultat finalIprime = 4 A 1p

d3

1 3

EIR R

=+ 3p

4p




alR

Sρ

= 2p3p


b2 middotbU U I R= + 3p

4prezultat final n = 8 1p

c

becuri b

sursatilde

P UP U

η = =

3p4p


η = 1p

d

middotx

xRS

ρ= 1p

4p2 x

UIR

= 2p

rezultat final 864kmx 1p



253

DOPTICAtilde



Subiectul al II-lea Parţial Punctaja reprezentarea grafică a imaginii 3p 3p

b

2 1 2

1 1 1x x f

minus =

2 2

1 1

12

y xy x

β = = =

2p

4p012

21

=+ff 1p

rezultat final 2 20 cmf = minus 1 40 cmf = 1p

c

( )1

1 11nf R

= minus 2p

4p( ) 11R n f= minus 1p

rezultat final 20 cmR = 1p

d

reprezentare grafică a imaginii 1p

4p2 1 1

1 1 1x x f

minus = 2 120 cmx = 1p

2 1 2

1 1 1 x x fminus = 1 2x x dminus = minus 1p

rezultat final 1 22 2

1 2

16 cm

x fx xx f

= = minus+

1p



a0 3x x= maximul de ordinul 0=k se deplaseaza icircn locul maximului de ordinul 3k = 1p

4p 3δ minus δ = λ 2p

rezultat final 7 15∙10 mminusδ minus δ = 1p

b

2 1r rδ = minus 1p

4p( )2 1 1r r e nδ = minus minus minus 1p

( ) 1e nδ minus δ = minus 1

enδ minus δ

=minus

1p

rezultat final 63∙10 me minus= 1p

c 2Dil

λ= 2p

3p

rezultat final 310 mi minus= 1p


254

d

Pentru dispozitivul Young icircn aer diferenţa de drum este 2 middotl yD

δ = 1p

4p

Pentru dispozitivul Young cu sursa deplasată cu distanţa h faţă de axul de simetrie al dispozitivului diferenţa

de drum este 2 middotl hd

δ = δ + 1p

Cele două diferenţe de drum suplimentare trebuie să se anuleze reciproc ( )2 middot 1l h e nd

= minus 1p

rezultat final( )12

ed nh

lminus

= 75 mh = micro 1p



255

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj1 c 32 c 1 J = 1 N ∙ m = 1 kg ∙ m ∙ sndash2 ∙ m = 1 kg ∙ m2 ∙ sndash2 3

3b F t m vsdot ∆ = sdot

5 20 205

F tv vmsdot ∆ sdot

= rArr = = ms 3

4

d [ ]

( )

21

2

0 0

m1s

2

sC

atx t x v t

C a

=

= + +

rarr

3

5

a ( )

1 1 2 2

1 2

1 2

1 2

1 2

1 2

N12m

s

s

s

s

F k F kF k

F Fk F Fk k

k kkk k

k

= ∆ = ∆

= ∆ + ∆

rArr = =∆ + ∆ +

rArr =+

=

3



a

1p

4p

1 1 1

1 2

2 2

1 2

1 2

2

m2s

m g T m aT T Ff ma N G mgT G m a

m m ma gm m m

a

minus =minus minus = = =minus =

minus minus micro= rArr

+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

= 1p

b2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+

2p4p

2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+ 2p


256

c

In această situaţie corpul cu masa m nu se deplasează a = 0 1p

3p( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot =

1p

( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot = 1p

d( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

2p

4p( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

1p

( )1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

= 1p



a 05 JA

A

E mghE

==

2p3p

1p

b

2p

4p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=

1p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=1p

c

Conform teoremei de variaţie a enegiei mecanice

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

4p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

= 1p

d

Conform teoremei de variaţie a enegiei cinetice

2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p

4p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

= 1p



257



b [ ] Jmol KSi

C =sdot

3

2 a 2100 kJ

Q m c tQ

= sdot sdot ∆=

3

3d 1 1 2 2

1 2

g kg3028 3028mol mol

N NN Nmicro + micro

micro =+

micro = =

3

4

c 2

1

11 1

2

ln

ln

280 J

VL RTVpL p Vp

L

= ν

=

= minus

3

5 c Gazul efectuează lucru mecanic asupra mediului exterior atunci cacircnd volumul său creşte deci icircn procesul A rarr B 3TOTAL pentru Subiectul I 15p


a0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p

0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

b

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

4p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot 1p

c

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =

ν = rArr = νmicromicro

=

νmicrorArr ρ =

rArr ρ = sdot

1p

4p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot 1p

d

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

4p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= = 1p



258


a

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

3p

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

b

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

5p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

c( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

=

2p3p( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

= 1p

d

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL

minus minus minus minus minus minus

minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p

4p

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p



259



d 2 WRtU

= [ ]SI1st = 3

2 b nEIne r

=+

3

3a

2

32 2 323 7221320

AB

ABAB

AB

RR RR R RR RR

R R RRR R

sdot sdot= + = +

+

sdot= =

+

3

4

c ( )0

00

02 1

1

10 grad

R R tR RR R A t

Rminus minus

= + α

minusminus = α rArr α = sdot

rArr α =

3

5 c Ampermetrul se conectează icircn serie iar voltmetrul icircn paralel cu rezistorul 3TOTAL pentru Subiectul I 15p


a

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω

2p3p

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω 1p

b

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

2p3p

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

1p

c

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R

V E I r VU V V I r EU

= minusprime+ minus =

minusprime prime= rArr =+ +

prime+ minus =primerArr = minus = minus

rArr = minus

1p

5p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus 1p


260

d

Icircn acord cu sensurile curenţilor indicate icircn schema alăturată legile Kirchhoff devin

( )( )

1 2 23

1 1 1 1 23 23

2 2 2 2 23 23

I I IE I R r I R

E I R r I R

+ =

= + + sdot

= + + sdot

1p

4p

cu 2 323

2 3

23

12

8 A3

R RRR R

I

= = Ω+

rArr =

1p

Din 2 2 3 3

2 3 23

2 16 A

I R I RI I I

I

prime prime=prime prime+ =

primerArr =

1p2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr = 1p



a

Intensitatea curentului prin generator este e

EIr R

=+

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus1p

4p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p

rArr r = 20 Ω 1p

b

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

3p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

= 1p

c

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

2p

4p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

1p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr = 1p

d

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

4p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

şi eR = r 2p



261

DOPTICAtilde

Subiectul I Punctaj1 b 1 d = 1 mndash1 3

2 a 0h L hν minus = ν se măsoară icircn J ca şi max

2

2c

mvE = 3

3b Scafandrul vede pescăruşul mai departe decacirct este icircn realitate

aparent aparent 1h n h n h h= sdot gt rArr gt

3

4

d 2 2 22 1

1 1 1

x y yx xx y y

β = = rarr =

Din grafic 1 2

2

1 2

1 2

40 cm 10 cm40 cm

20 cm

x yx

x xf fx x

rArr = =

rArr =

rArr = rArr =minus

3

5 b 9

63

550 10 1 275 10 m2 2 10

275igrave m

Di ie

i

minusminus

minus

λ sdot sdot= = = sdot

sdot=

i = 275 μm 3



a

Din grafic 1 75 cmxrArr = minus 1p

3pcorespunde cu 1 2

05

= minusβrArr β = minus

1p1 2

05

= minusβrArr β = minus 1p

b

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

4p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr = 1p

c

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

4p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = = 1p

d

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =

rArr β = = minusminus

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

4p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p



262


a 4p 4p

b0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν

= sdot sdot sdot = sdot

2p3p0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν

= sdot sdot sdot = sdot 1p

c

00

86

0 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot

2p

3p0

08

60 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot1p

d

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=

sdot sdot sdot sdot=

sdot

= sdot

2p

5p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot 1p



263

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj1 a 32 a 33 d 34 c 35 b 3



a

Corpul de masă m1 de deplasează cu acceleraţia 21 1 05m sa g= minusmicro = minus

Aplicacircnd legea vitezei pentru corpul de masă m1 se obţine viteza acestuia 1 1 1v a t= 1p

4p

Din conservarea impulsului pentru momentul destinderii resortului se determină viteza corpului de masă m21 1 2 2

1 12

2

m v m vm vvm

=

=

1p

rezultat final 1 2 m sv = 2 15 m sv = 2p

b

Aplicacircnd legea lui Galilei se determină distanţa parcursă de corpul de masă m12 21 1 1

112 2

v gtda

micro= =

2p3p

rezultat final 1 4 md = 1p

c

Aplicacircnd legea vitezei şi legea lui Galilei pentru corpul de masă m2 se obţine acceleraţia acestuia 2

222

2

0375m s2vad

= = 1p

3p

Corpul de masă m2 de deplasează cu acceleraţia 22 2 2

aa gg

= minusmicro rArr micro = 1p

rezultat final 2 00375micro = 1p

d

Distanţa dintre corpuri reprezintă suma distanţelor parcurse de cele două corpuri icircn timp de 2 s determinate aplicacircnd legea spaţiului 1 2D D D= + 1p

5p

21

1 1 2a tD v t= minus 1p

22

2 2 2a tD v t= minus

1p

( )2 2 2

1 21 2 1 2 1 2( )

2 2 2a t a t tD v t v t v v t a a= minus + minus = + minus + 1p

rezultat final 525 mD = 1p



264


a

conservarea energiei tA tDE E= tD cD pDE E E= + 1p

4p(1 cos )tAE mgh mgl= = minus α 1 (1 cos )pDE mgh mgl= = minus β

cD tA pDE E E= minus 1p

(1 cos ) (1 cos )cDE mgl mgl= minus α minus minus β (cos cos )cDE mgl= β minus α 1p

rezultat final 219 JcDE = 1p

b

Aplicacircnd conservarea energiei pentru poziţia iniţială şi poziţia de echilibru se obţine tA BE E= 1p

3p20

0(1 cos ) 2 (1 cos )2

mvmgl v glminus α = rArr = minus α 1p

rezultat final 0 387 m sv = 1p

c

Cacircnd firul face cu verticala unghiul γ = 30deg viteza este dată de relaţia 2 cosv gl= α 2 33 mscDEv

m= = 1p

4p

iar componenta orizontală a vitezei este cos 165 msxv v= γ = 1p

Corpul de masă m are deci viteza vx corpul de masă M fiind icircn repaus şi aplicacircnd legea de conservare a impul-

sului viteza corpului nou format va fi dată de 066 msxmv v

m M= =

+

Acceleraţia corpurilor fiind a g= minusmicro spaţiul parcurs pacircnă la oprire de cele două corpuri se determină din

ecuaţia vitezei 2

2vd

g=

micro

1p


d

Căldura degajată icircn procesul ciocnirii plastice este determinată de variaţia de energie cinetică2 2 2 2

2 2( ) 2( )x x xmv m v mMvQ

m M m M= minus =

+ + 2p

4p

rezultat final 0326 JQ = 2p



265


Subiectul I Punctaj1 b 32 a 33 a 34 c 35 c 3



a

Din ecuaţia de stare şi formula de definiţie a densităţii se obţine

pV RT

mpV RTm= ν

rArr =ν = micromicro

1p

4pm pV RT

microρ = =

1p

rezultat final 30 141 kg mρ = 1p

3128 kg mρ = 1p

b

Din ecuaţia de stare se obţine 1 1 1

1 1 11

p V RTmp V RTm

= ν rArr =ν = micromicro

1p

3p1 1

1

p VmRT

micro=

1p

rezultat final 025 kgm = 1p

c

Se determină numărul de moli rămaşi icircn recipient la presiunea p2= 105 Nm2 din ecuaţia de stare

2 1 2 1

22

1

p V RTp VRT

= ν

ν =

1p

3pNumărul de moli ce trebuie scoşi se calculează ca diferenţă dintre numărul de moli aflaţi iniţial icircn recipient şi

numărul de moli rămaşi 1 2 11 2

1

( ) 722 molip p VRTminus

∆ν = ν minus ν = = 1p

rezultat final 1 2 11 2

1



d

Din conservarea numărului de moli şi aplicacircnd ecuaţia de stare se obţine1 2ν = ν + ν

1p

5p

1 11

1 1 1 2 2

2 2 12

1

( )p VRT p V p Vp V RTRT

ν = + rArr ν =ν =

1p

1 2 1 1 1 2 2( )p V V RT p V p Vν+ = = + 1p

1 1 2 2

1 2final

p V p VpV V

+=

+ 1p

rezultat final 6 216 10 Nmfinalp = sdot 1p



266


a

Reprezentare corectă

3p 3p

b

Pentru trecerea din starea (1) icircn starea (2) lucrul mecanic reprezintă aria porţiunii situate sub grafic Presiunea icircn cele două stări este 1 1p kV=

respectiv 2 2p kV= Astfel lucrul mecanic se poate scrie ca fiind

2 22 1 2 1 2 1 2 12 1

( )( ) ( )( ) ( )2 2 2

p p V V k V V V V kL V V+ minus minus += = = minus

1p

3p

2 22 1

2LkV V

=minus

1p

rezultat final 8 510 Nmk = 1p

c

Stările iniţială şi finală sunt caracterizate de temperaturile T1 respectiv T2 Trecerea din starea (2) icircn starea (3) se realizează printr-un proces izoterm deci T2=T3 şi variaţia de energie internă este nulă Din ecuaţia de stare se determină temperaturile pentru starea (1) respectiv starea (2) apoi se icircnlocuiesc icircn for-mula energiei interne

21 1 1

1 1 1 1p V kVp V RT T

R R= ν rArr = =

ν ν

1p

5p2

2 2 22 2 2 2

p V kVp V RT TR R

= ν rArr = =ν ν

1p

2 22 1 2 1( )kT T T V V

R∆ = minus = minus

ν 1p

2 22 1

3 ( )2VU C T k V V∆ = ν ∆ = minus

1p

rezultat final 300 JU∆ = 1p

d

Aplicacircnd principiul icircntacirci al termodinamicii se calculează căldurile pe fiecare transformarePentru procesul 1rarr2

1 2

1 2

1 2 1 2 1 2

100 J300 J

400 J

LU

Q U L

minus

minus

minus minus minus

=∆ =

= ∆ + =

1p

4p

Pentru procesul 2rarr32 3

23 12 3 2 2

2 2

2 3 2 3

0

ln ln 245 J

245 J

UV VL RT kVV V

L Q

minus

minus

minus minus

∆ =

= ν = = minus

= = minus

1p

Cantitatea de căldură schimbată cu mediul exterior şi lucrul mecanic efectuat de gaz pe icircntregul proces se de-termină ca sumă a căldurilor respectiv lucrului mecanic pe fiecare transformarerezultat final 1 2 2 3 145 JL L Lminus minus= + = minus

1p

1 2 2 3 155 JQ Q Qminus minus= + = 1p



267


Subiectul I Punctaj1 b 32 b 33 b 34 a 35 b 3


Subiectul al II-lea Parţial Punctaja Reprezentare corectă

3p 3p

b

Tensiunea electromotoare a unei grupări de 3n = elemente icircn serie este 45 VsE ne= = iar tensiunea electro-motoare echivalentă a grupării de acumulatoare va fi egală cu sE E=

1p

4p

Rezistenţa internă a grupării serie de acumulatoare este 06sr nr= = Ω

Rezistenţa internă echivalentă a celor patru grupări legate icircn paralel este 1 4 3

4 4s

ee s

r rrr r

= rArr = =1p

rezultat final 45 VE = 1p

015er = Ω 1p

c

Curentul pe ramura principală este I iar fiecare ramură din gruparea de acumulatoare este parcursă de un curent i Legile lui Kirchhoff pentru ochiul format de sursa echivalentă şi cele două rezistoare respectiv pentru nodul A se scriu sub forma

1 23 ( )4

E i r I R RI i

= sdot + + = sdot

Din rezolvarea sistemului se obţine intensitatea curentului principal 1 2

142A34

neIr R R

= =+ +

1p

4p

şi intensitatea curentului prin fiecare acumulator 035A4Ii = = 1p

Intensitatea curentului de scurtcircuit se obţine punacircnd condiţia ca rezistenţa echivalentă a celor două rezis-

toare să fie nulă Astfel 43scEIr

= 1p

rezultat final 30 AscI = 1p

d2R Sl =ρ

3p4p

rezultat final 20 ml = 1p



268


a

Aplicacircnd legea lui Kirchhoff pentru ochiul de reţea se determină curentul1 2 1 2 1

1 2

1 2 1

( )

1

E E I r r RE EI A

r r R

+ = + ++

= =+ +

1p

3p

Din legea lui Ohm se calculează căderea de tensiune icircn circuitul exterior U şi tensiunea la bornele fiecărui generator Ub1 respectiv Ub2

1

1 1

2 2

b

b

U IRU E IrU E Ir

== minus= minus

1p

rezultat final

1

2

8 V55V25V

b

b

UUU

===

1p

b

După introducerea rezistorului de rezistenţă R2 se va modifica intensitatea curentului pe ramura principa-lă Astfel dacă R2 este icircn serie cu R1 rezistenţa echivalentă este 1 2 12sR R R= + = Ω iar curentul devine

1 2

1 2

035Ass

E EIr r R

+= =

+ +

1p

4pSimilar pentru legarea icircn paralel se obţin 1 2

1 2

266pR RR

R R= = Ω

+ respectiv 1 2

1 2

214 App

E EIr r R

+= =

+ +1p

Puterile consumate icircn circuitul exterior vor fi 2 612 Wext s s sP R I= = respectiv 2 1224 Wext p p pP R I= =

1p

rezultat final 612 Wext sP = 1224Wext pP = 1p

c

Puterea consumată de R1 este 21 1 408 WsP R I= = 1p

4pPuterea consumată de R1 reprezintă o fracţiune din puterea totală consumată icircn circuitul exterior 1 ext sP fP= 1p

1

ext s

PfP

=

1p

rezultat final 066f = 1p

d

Energia disipată icircn circuitul exterior icircn condiţiile legării celor doi rezistori icircn paralel icircntr-un interval de timp oarecare este

2ext p pW R I t=

1p

4piar energia totală dezvoltată de cele două surse icircn acelaşi interval de timp este 1 2( )total pW E E I t= + 1p

1 2 1 2

057p pext

total p

R I RWW E E r r R

= = =+ + + 1p

rezultat final 057ext

total

WW

= 1p



269

DOPTICAtilde

Subiectul I Punctaj1 a 32 c 33 b 34 d 35 c 3



a

Convergenţa sistemului se determină ca suma convergenţelor celor două lentileLentila L1 este convergentă imaginea se formează la distanţa x2 determinată din mărirea liniară transversală

2 2

1 1

12

y xy x

β = = = 1p

3p

Pentru determinarea convergenţei se scrie legea lentilelor 11 2 1

1 1 1 125Cf x x

= = minus = δ 1p

rezultat final 1 2 15C C C= + = δ 1p

b


1 1

10 cmf xxf x

= =+

faţă de lentilă 1p

4p

Cum distanţa dintre lentile este de 20 cm faţă de lentila L2 imaginea este situată la distanţa

1 2 10 cmx D x= minus = 1p


1 40 cmfC

= = Aşadar lentila L2 se comportă ca o lupă formacircnd o imagine

finală virtuală situată la distanţa 2 12

2 1

f xxf x

=+

faţă de lentila L2 1p

rezultat final 2 133 cmx = minus 1p

c

Măririle liniare transversale ale lentilelor sunt 21

1

025xx

β = = respectiv 22

1

133

xx

β = = 1p

4pMărirea liniară transversală a sistemului este β = β1β2 = 033 1p

Icircnălțimea imaginii finale va fi 2 1y y= β 1p

rezultat final 2 066 cmy = 1p

d

Dacă lentilele sunt alipite lentila echivalentă are distanţa focală 1 6cmFC

= = 1p

4pUtilizacircnd formula lentilelor 2 1

1 1 1F x x

= minus 1p


1

FxxF x

=+

1p

rezultat final 2 705 mx = 1p



270


a

Cum maximul de ordinul k = 2 se află faţă de maximul central la o distanţă 1x ki= se obţine interfranja

1 0108 mmxik

= =

Din formula interfranjei 11

Did

λ= se obţine 1

1i d x dD k D

λ = =

1p

3p

Din 22 2 2

D di id D

λ= rArr λ = 1p

rezultat final 1 540 nmλ = 2 630 nmλ = 1p

b

Maximele de ordinul 3 situate de aceeaşi parte a axei de simetrie a dispozitivului pentru cele două radiaţii utilizate se formează la distanţele

3 1 13 3 Dx id

= = λ

1p

4prespectiv 3 2 23 3 Dx i

d= = λ 1p

Distanţa dintre cele două maxime de ordinul 3 va fi ( )3 3 2 13 Dx x x

d∆ = minus = λ minus λ 1p

rezultat final 054 mmx∆ = 1p

c

Condiţia ca maximele să se suprapună este ca 1 1 2 2mx k i k i= = unde 1k şi 2k să fie cele mai mici numere icircntregi care să satisfacă egalitatea

1p

4pIcircnlocuind interfranjele se obţine egalitatea 2 2

1 1 2 2 11

kk k k λλ = λ rArr =

λ 1p

Condiţia este icircndeplinită pentru 1 7k = şi 2 6k = Deci distanţa minimă la care se suprapun maximele celor două radiaţii este 1 1mx k i=

1p

rezultat final 0756 mmmx = 1p

d

Dacă dispozitivul se cufundă icircn apă lungimile de undă vor fi 1

1

nλ

λ = respectiv 2

2

nλ

λ =

A treia franjă icircntunecoasă dată de radiaţia cu lungimea de undă λ2 este poziţionată faţă de axa de simetrie a

dispozitivului la distanţa 23min

2 12

k Dxn d

+ λ=

1p

4pPentru k = 3 se obţine 3min 3307 mmx =

Maximul de ordinul 2 dat de radiaţia cu lungimea de undă 1λ se formează la distanţa 12max 162 mmDx k

n dλ

= = 1p

Distanţa dintre a treia franjă icircntunecoasă dată de radiaţia cu lungimea de undă 2λ şi maximul de ordinul 2 dat de radiaţia cu lungimea de undă λ1 dacă dispozitivul se cufundă icircn apă este

3min 2maxx x x∆ = minus 1p




271

TESTUL 11

A MECANICAtilde


c 2 2 2 2

2 2 2 cp m v m v E

m msdot sdot

= = =sdot sdot

1 J = 1 N mc SIE = sdot 3

2 c 2

0 0 2g tx x v t sdot

= + sdot + 2

55 2

g tx sdot=

24

4 2g tx sdot

= 5 4 45 mx x x= minus = 3

3 b v ct= 0a = 0t fG Fminus = sincos

αmicro =

α3

4 b 1N100m

Fkl

= =∆

2N200m

k = 1

2

12

kk

= 3

5 b p F t∆ = ∆ F G= m100 kgs

p m g t∆ = sdot sdot ∆ = sdot 3



a

1 1 t fm T G F m aminus minus = sdot

2 2 2m G T m aminus = sdot 2p

4pPentru expresia acceleraţiei

2 2 1

1 2 1 2

( sin cos )t fG G F m g m g mam m m mminus minus sdot minus sdot sdot sdot α + micro sdot α

= =+ +

1p

Rezultă 2

m375s

a = 1p

b

Tensiunea icircn fir 2 2 3125 NT G m a= minus sdot = 1p

4pForta din scripete 2 2 22 cos( )2

F T T T π= + + sdot sdot minus α

23 3F T T= sdot =

2p

Rezultă 54 NF asymp 1p

c

Pentru viteza constantă1 11 t fT G F= + 1 2T G= 1p

4pRezultă

2 (sin cos )m g M gsdot = sdot sdot α + micro sdot α 2 25 kg

sin cosmM = =

α + micro sdot α2p

Obţinem 25 kgM = 1p

d

Pentru (sin cos )a g= sdot α minus micro sdot α

V = a Δt2p

3p

Obţinem m5s

v = 1p



272


a

Din conservarea impulsului se obtine viteza dupa impact a sistemului de corpuri

i fp p=

21 1 2 0

1 2

m v m vu

m msdot + sdot

=+

2p

3p

Rezultăm60s

u = 1p

b

Impulsul sistemului icircn B va fi 1 2( )B Bp m m v= + sdot 1p

4p

Din teorema de variaţie a energiei cinetice obţinem

B A fc c FE E Lminus =

2 m2 40 2 56sBv u g d= minus sdotmicro sdot sdot = = 2p

Obţinem Bp =m56kgs

1p

c

Din condiţiile problemei energia totală a sistemului este

B R Bt c PE E E= + 1p

4pObtinem

21 2

1 2( ) ( )

2B

Bt

m m vE m m g h+ sdot= + + sdot sdot 2p

Rezultă 1700 JBt

E = 1p

d

Din teorema de variatie a energiei cinetice pe portiunea BC se obtine2

2C f

Bc G F

m vE L Lsdotminus = + 1p

4pDeci 2

2sin cos2C

t Bc t t

m vE m g d m g dsdot= + sdot sdot α sdot minus sdotmicro sdot sdot α sdot

cossinC Bc t tE E m g h α

= minus sdotmicro sdot sdot sdotα

2p

Rezultă 1650 JCcE = 1p



273



b 12

Q C T R T= ν sdot sdot ∆ = sdotν sdot sdot 3

2 b 201204 10A

A

V N V NN moleculeV N Vmicro micro

sdot= rArr = = sdot 3

3 c U Q L∆ = minus pQ C T= ν sdot sdot ∆ L R T= ν sdot sdot ∆ 3U L∆ = 3

4 b 1 2 1 2

1 2

m m m m+= +

micro micro micro 3

1 2

5 kg777 101 4 molminusmicro = = sdot

+micro micro

3

5 b mp V R Tsdot = sdot sdotmicro

3 2 11mp R T m m mV

= sdot sdot sdot rArr gt gtmicro

3



a

Din ecuaţia termică de stare 1 1 1A

Np V R TN

sdot = sdot sdot 3 31

11

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro

1p

4pSe obţine 1

11

AN P NnV R T

sdot= =

sdot2p

Rezultă 25

1 3

molec12 10m

n = sdot 1p

b

Numărul de moli poate fi exprimat din relaţia2 2

22

p VR T

sdotν =

sdot2p

3p

Rezultă 2 437 moliν = 1p

c

Pentru starea iniţială din ecuaţia termică de stare1 1

11

m R TpV

sdot= sdot

micro 1p

4p

Putem scrie densitatea pentru fiecare stare1

11

pR T

sdotmicroρ =

sdot

22

2

pR T

sdotmicroρ =

sdot1p

Se obţine 1 1 2

2 2 1

p Tp T

ρ sdot=

ρ sdot1p

Rezultă 1

2

113

ρ=

ρ 1p


274

d

Pentru fiecare balon se poate scrieSI 1 1 1 1

2 2 2 2

p V R Tp V R T

sdot = ν sdot sdotsdot = ν sdot sdot

SF 1 1 1

2 2 2

p V R T

p V R T

primesdot = ν sdot sdot


1p

4pDin conservarea numărului de moli 1 2 1 2prime primeν + ν = ν + ν 1p

Icircn final obţinem

1 1 2 2

1 2

1 2

1 2

p V p VT Tp V V

T T

sdot sdot+

=+

1p

Rezultă 535 10 Pap asymp sdot 1p



a

Pentru fiecare stare se obţine 12 1 2 1 1( ) 2L p v v R T= sdot minus = sdotν sdot sdot 223 2 2

1

ln ln 2vL R T R Tv

= ν sdot sdot sdot = ν sdot sdot sdot

34 3 4 3 3

1( )2

L p v v R T= sdot minus = minus sdotν sdot sdot

141 1 1

4

1ln ln2

vL R T R Tv

= ν sdot sdot sdot = ν sdot sdot sdot1p

4p

Pentru icircntregul proces ciclic 12 23 34 41tL L L L L= + + + 1 2 3 11 1(2 ln 2 ln ) 39057 J2 2tL R T T T T= ν sdot sdot sdot + sdot minus sdot + sdot = 2p

Obţinem 39057 JtL = 1p

bVariaţia energiei interne pe 3rarr4 este 34 4 3 1 2

3( ) ( )2vU C T T R T T∆ = ν sdot minus = sdotν sdot sdot minus 2p

3p

34 12465 JU∆ = minus 1p

c

Căldura cedată de gaz icircntr-un ciclu de funcţionare este 34 41cQ Q Q= + 1p

4pObţinem

1

4 3 14

( ) lnc pVQ C T T R TV

= ν sdot sdot minus + ν sdot sdot sdot1 2 1

5 1( ) ln2 2cQ R T T R T= sdotν sdot sdot minus + ν sdot sdot sdot 2p

Rezultă 38226 JcQ = 1p

d

Randamentul unui motor se calculează t

p

LQ

η = 1p

4pPutem scrie| |t p cL Q Q= minus | |p t cQ L Q= + | |p t cQ L Q= + | |p t cQ L Q= +

2p

Obţinem 505η = 1p



275



c 2

2 UW R I t tR

= sdot sdot ∆ = sdot ∆ 2V sRJsdot

= 3

2 c 3

3 b Sarcina totală reprezintă aria figurii cuprinse icircntre graficul intensităţii şi axa timpului 375 10Q C= sdot 3

4 c eSC

e

EIr

= 65er = Ω 24 V

5eE = 4AeSC

e

EIr

= = 0 VU = 3

5b

2 2U UP RR P

= rArr =

2

10 mS UlP

sdot= =

ρ sdot3



a

pentru K icircnchis e

e e

EIR r

rArr =+

1p

4pParametrii sursei sunt

1 2

1 2

1 2

30 V1 1e

E Er rE

r r

+= =

+ 1

2err = = Ω 1p

Din schema circuitului rezultă 1 8sR = Ω 2 8sR = Ω 1 2

1 2

4s se

s s

R RRR R

sdot= = Ω

+ 1p

Obţinem 6 AI = 1p

b

Pentru nodul A 1 2I I I= + 1p

4pPentru ochiul II 1 1 2 2 1 2s sI R I R I Isdot = sdot rArr = 2p

Obţinem 1

2

1II

rArr = 1p

c

Pentru K deschis rArr 3 3U I Rprime= sdot 1p

4p Noua intensitate va fi34

10 A3

e

e

EIR r

prime = =+

2p

Rezultă 3 10 VU = 1p

dTensiunea indicată de un voltmetru ideal pentru K deschis va fi V eU I Rprime= sdot 2p

3pRezultă 266 VVU = 1p



276


a

Cacircnd cursorul se află icircn AP E I= sdot 1p

3pSe obţine din circuit 8eR = Ω 2 AI = 1p

Rezultă P = 40 W 1p

b

Energia disipată de rezistorul R2 este 22 2 2W R I t= sdot sdot ∆ 1p

4pIntensitatea va fi 2 2I R I r Esdot + sdot = 2 04 AI = 2p

Rezultă 2 768 JW = 1p

c

Randamentul va fi e

e

RR r

η =+

1p

4p

Rezistenţa echivalentă va fi15AC ABl l= sdot

1 325AC AB BCR R R= sdot rArr = Ω

16021P E P ACR R R R= rArr = +

2p

Rezultă 886η = 1p

d

Energia disipată icircn interiorul sursei 2intW r I t= sdot sdot ∆ 1p

4p

Pentru calculul intensităţii prin sursă2

12

21

8022 3

2

e

RR RR RR

sdot= + =

+

30 A43e

EIR r

= =+

2p

Rezultă int 60 JW asymp 1p



277

DOPTICAtilde




a

Din prima formulă fundamentală a lentilelor 2 1 2 2

1 1 1 1 1 12x x f x x f

minus = rArr minus =minus

2p

4pRezultă 2

1 32

cf x

minus= =

minus1p

Obţinem 2

3 2015 10

c minus= = δsdot

1p

b

2 1

1 1 1x x f

minus = 1p

4pobţinem 2

1 21

1 22

x x xx

= minus rArr = minus 2p


c

Pentru a doua lentilă f = f primeprime

2 1

1 1 1x x f

minus = 1p

4pObţinem 2 1 1 2 75 cmx x d x d xprimeprime prime+ = = minus = 1p

Icircnlocuim icircn prima formulă şi obţinem 12

1

x fxx f

primeprime =

prime +1p

Rezultă 275 5 15 cm75 5

x minus sdotprime = =minus +

1p

d Realizarea corectă a desenului 3p 3pTOTAL pentru Subiectul al II-lea 15p


aDin formula interfranjei obţinem

D a ii Da

λ sdot= rArr =

λ2p

3p

Rezultă 3 3

9

2 10 1 10 5 m400 10

Dminus minus

minus

sdot sdot sdot= =

sdot1p

b

Pentru noua poziţie a ecranului avem 4 m 16 mmD xprime = = 1p

4pRezultă

axD

δ = 2p

Obţinem 3 3

72 10 16 10 8 10 m4

minus minusminussdot sdot sdot

δ = = sdot 1p


278

c

Pentru maximul de ordinul 4 avem maxDx ka

primeλ= unde k = 4 2p

4pObţinem

9

max 3

400 10 442 10

xminus

minus

sdot sdot= sdot

sdot1p

Rezultă max 32 mmx = 1p

d

Dacă dispozitivul se introduce icircn apă avem Di

naλprime =

1p

4pDin Dia

λ= rezultă ii

n= 2p

Rezultă 3

3 31 10 3 10 075 10 m4 43

iminus

minus minussdot= = sdot = sdot 1p



279

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj1 d 32 b 33 b 34 d 35 b 3



a

Pentru mişcarea sistemului ca un singur corp se scrie0 0

00

( )

05

kt M m m akta t

M m m

= + +

= =+ +

2p

4pDupă desprindere corpul de pe scacircndură este tractat doar de forţa de frecare dintre corp şi scacircndură

2

2 2

m8 02s

Mg ma

a

micro =

= micro =

2p

b

Pentru sistemul de corpuri desprinse se poate scrie2

0 1( )Mg ma

kt Mg m m amicro =

minus micro = +

2p

3p

De aici rezultă pe de o parte 1 20

m( 2)s

kt Mga tm m

minus micro= = minus

+ 1p

c

Impunacircnd ca la momentul desprinderii acceleraţiile să fie egale 1p

4p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + =

2p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + = 1p

d

Pentru momentul t = 2 s forţa de tracţiune are valoarea (2) 20 NF = 1p

4p

Pentru fracţiunea de tijă precizată se poate scrie( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus = 1p



a

Energia cinetică a şnurului imediat după momentul pornirii este2

30 5 10 J

2cmvE minus= = sdot 1p

3pDeoarece şnurul este tras cu viteză constantă energia lui cinetică nu se schimbă 1p

Deci 0cE∆ = 1p


280

b

Lucrul mecanic necesar urcării şnurului integral pe planul icircnclinat este

1 1 2sin cos2 2 2l l lL mg mg mg= micro + α + micro α

Primul termen reprezintă lucrul cheltuit pentru aducerea centrului de masă al şnurului la baza planului icircnclinat al doilea reprezintă lucrul mecanic necesar ridicării centrului de masă al şnurului la jumătatea icircnălţimii planului icircnclinat iar al treilea termen este lucrul efectuat icircmpotriva forţei de frecare pentru transportarea centrului de masă de-a lungul planului icircnclinat pe jumătate din lungimea lui Această abordare permite tratarea rapidă clară şi explicită a unui proces care altfel ar fi presupus o analiză grafică sau un calcul integral

3p4p

Numeric 1 15 JL = 1p

c

Icircn conformitate cu cele precizate la punctul b se scrie

2 2 3sin cos2 2 2l l lL mg mg mg= α + micro α + micro 3p

4p


d

Deoarece puterea momentană este egală cu produsul dintre forţa momentană şi viteza momentană (icircn cazul nostru aceasta are aceeaşi valoare pe toată durata procesului) rezultă că puterea maximă dezvoltată corespunde momentului icircn care forţa de tracţiune este maximă ceea ce corespunde momentului icircn care icircntreg şnurul se află pe planul icircnclinat

1p

4pmax maxP F v= sdot max 2sin cosF mg mg= α + micro α

2p

Numericmax 15 WP = 1p



281


Subiectul I Punctaj1 d 32 a 33 c 34 a 3

5

b

1 11

1

abs ced

abs

Q QQ

minusη = respectiv

2 22

2

abs ced

abs

Q QQ

minusη =

Pe de altă parte 1 21 2

1 2

abs abs

L LQ Q

η = η = iar

2 1abs cedQ Q=

Din aceste relaţii rezultă 1 2 1 2 092η = η + η minus η sdotη =

3



a

Ecuaţia de stare 0mp V RT=micro

1p

3p0p VmRT

micro=

1p


b

Numărul total de molecule icircn urma disocierii este ( )0 0 02 (1 ) 1N N N N= α + minus α = + α 1p

4p

Pentru starea iniţială 00

A

Np V RTN

= 1p

Pentru starea finală 0 (1 )

A

N RTpVN+ α

= 1p

Rezultă 0

1 14pp

= + α = 1p

c

Icircn recipient sunt două tipuri de molecule 1 02N N= α molecule monoatomice cu masa molară 1kg16

kmolmicro = şi

2 0(1 )N N= minus α cu masa molară 2kg32

kmolmicro =

1p

4p

012i ii

A i

N m iN

ν = = =micro

1p

1 2

1 2

m m m= +

micro micro micro

1p

0

1 2 1 2

1 2

32 kmol 2286kmol1 14

A

A A

Nm N

m m N NN N

micromicro

micro = = = = cong+ α+ +

micro micro

1p


282

d

Sistemul este alcătuit din două componente o cantitate 1 02ν = αν de molecule monoatomice şi 2 0(1ν = minus α)ν molecule biatomice 1p

4p

Pentru starea iniţială 0 0 pVpV RTRT

= ν ν = 1p

Căldura absorbită de sistem este suma căldurilor absorbite de către fiecare componentă a amestecului la volum constant şi pentru acelaşi interval de temperatură

1 2Q Q Q= + 1 1 1 2 2V VQ C T C T= ν ∆ + ν ∆1p

( 5) 450 J2

pV TQT

α + ∆= sdot =

1p



a

Procesul 1rarr2 este izocor 2 11 2

2 1

p pT T T nTT T

= = = 1p

3p

Procesul 2rarr3 este izobar 3 23

3 2

V V T nkTT T

= = 1p


1 4

V V T kTT T

= =

1p

b

( ) ( )12 23 2 1 3 2abs V pQ Q Q C T T C T T= + = ν minus + ν minus ( ) ( )34 41 3 4 4 1ced V pQ Q Q C T T C T T= + = ν minus + ν minus

1p

4p

(5 2 3)2absRTQ nk nν

= minus minus

(3 2 5)2cedRTQ nk kν

= + minus 1p

Cercetăm să vedem relaţia de ordine dintre căldura primită şi modulul căldurii cedate presupunem căabs cedQ Qgt rezultă 2( 1)( 1) 0n kminus minus gt ceea ce este evident icircn condiţiile icircn care n şi k sunt supraunitare 1p

Rezultă că maşina termică va funcţiona ca motor adică ab-soarbe căldură de la sursa caldă cedează energie sursei reci şi efectuează lucru mecanic cu respectarea principiilor ter-modinamicii Icircn această situaţie are sens să calculăm randamentul motorului termic reprezentacircnd raportul dintre energia utilă şi cea cheltuită

1p


283

c

Randamentul ciclului Carnot care ar funcţiona icircntre aceleaşi limite extreme de temperatură este11 1rece

Ccald

T T nkT nkT nk

minusη = minus = minus = Presupunem că

Cη gt η adică 1 2 2 2 25 2 3

nk nk n knk nk n

minus minus minus +gt

minus minus se obţine 2 23( 1) 2 ( 1) 0nk n kminus + minus gt ceea ce este

adevărat

1p

4p

Pentru k rarr infin avem 1lim 1C k k

nknkrarrinfin rarrinfin

minusη = =

Pentru n rarr infin avem n1lim 1C k


minusη = =

1p

Pentru ciclul dat k rarr infin 2 2 2 2 2 2lim5 2 3 5k k

nk n k nnk n nrarrinfin rarrinfin

minus minus + minusη = =

minus minus

Acest randament este mai mic decacirct 1 (aceeaşi limită a ciclului Carnot) 2 2 1 3 2

5n n DA q e d

nminus

lt gt minus

Pentru ciclul dat n rarr infin 2 2 2 2 2 2lim5 2 3 5 2n n

nk n k knk n krarrinfin rarrinfin


minus minus minus

1p

Şi acest randament este mai mic decacirct 1 (aceeaşi limită a ciclului Carnot)

Pentru n rarr infin şi k rarr infin simultan se obţine 2 15n

krarrinfinrarrinfin

η = lt

1p

d

Dacă un ciclu termodinamic este parcurs icircn sens trigonometric maşina care funcţionează după acest ciclu absoarbe căldură de la sursa rece pe baza lucrului mecanic primit din exterior şi cedează căldură sursei caldeDacă sursa caldă este un spaţiu icircnchis pe care dorim să icircl icircncălzim iar sursa rece este un izvor maşina se nu-meşte pompă de căldură şi are eficienţa

cedQL

ϕ =

1p

4p

Valoarea eficienţei ca pompă de căldură ne arată cacircţi jouli de căldură se furnizează sursei calde pentru un joule de lucru mecanic cheltuit de mediu

1 5 2 32 2 2 2

cedQ nk nL nk n k

minus minusϕ = = =

η minus minus +

Dacă sursa rece este un spaţiu icircnchis pe care vrem să-l răcim iar sursa caldă este izvor sistemul se numeşte maşină frigorifică (frigider) şi eficienţa ei este

absQL

ε =

1p

Valoarea eficienţei ca maşină frigorifică ne arată cacircţi jouli de căldură sunt preluaţi de la spaţiul refrigerat pentru un joule de lucru mecanic cheltuit din exterior Vom avea

1 3 2 512 2 2 2

nk knk n k

+ minusε = minus =

η minus minus +

Se observă că icircntotdeauna1 1 1 1

ϕ minus ε = minus minus = η η

1p

Ceea ce se verifică şi pentru maşina noastră 5 2 3 3 2 5 12 2 2 2 2 2 2 2

nk n nk knk n k nk n k

minus minus + minusϕ minus ε = minus =

minus minus + minus minus + 1p



284


Subiectul I Punctaj1 a 3

2

d O metodă care se poate aplica icircn situaţii similare este aceea de a introduce mintal o rezistenţă x icircntre punctele a şi b icircn locul firului Pentru notaţiile din figură scriem teoremele lui Kirchhoff

1E I r Ix= + 2 ( )E I r R Ix= + minus 1 2I I I+ +

Rezolvacircnd se obţine ( )( )

ERIR r r x xr

=+ + +

Icircn acest moment pentru x se atribuie valoarea zero şi se obţine rezultatul solicitat 6 A( )

ERIr R r

= =+

3

3 c 3

4

d Pornind de la dreapta către stacircnga calculăm rezistenţa echivalentă din aproape icircn aproape şi aflăm valoarea 2R

dintre punctele a şi b

Intensitatea curentului principal injectat icircn circuit este

2 2

U UI R R R= =

+ Icircn aceste condiţii avem 2 2

R UUab I= sdot =

Curentul prin rezistorul vertical dintre a şi b este 2

abR

U UIR R

= = Puterea debitată de acest rezistor este 2

2

4RUP RI

R= =

3

5

c Circuitul se deschide cacircnd tensiunea de intrare este mai mare decacirct suma dintre tensiunea de deschidere diodei (2V) şi tensiunea electromotoare a sursei (3V) Intensitatea curentului este

0 i D

i D

U U EI U U E

r R

le += minus minus +

Tensiunea de ieşire este eU IR= Numeric se obţine 0 5V

1 5 V2

i

e ii

UU U U

le= +

gt

Se vede că ( 1) 0 (0) 05 (5) 3 (5) 9Ve e e eU U V U V Uminus = = = =

3



a

Pentru circuitul dat sunt valabile relaţiile 1

n

ki

k k k

I I

U E I rU IR

=

=

minus = minus =

sum

Eliminacircnd U icircntre ultimele două relații obţinem kk

k k

E RI Ir r

= minus

2p

4p

Utilizacircnd şi prima relaţie din sistem se obţine 1

1

11

nk

i kn

i k

ErI

Rr

=

=

=+

sum

sum și respectiv 1

1

11

nk

i kn

i k

ERrU IR

Rr

=

=

= =+

sum

sum2p

b

La bdquomers icircn golrdquo se obţine 1 1

1 1

lim1 11

n nk k

i ik kgol n nR

i ik k

E ERr rU

Rr r

= =

rarrinfin

= =

= =+

sum sum

sum sum 1p

4prespectiv 1

1

lim 011

nk

i kgol nR

i k

ErI

Rr

=

rarrinfin

=

= =+

sum

sum 1p

La bdquoscurtcircuitrdquo se obţine 1 1

0 0n n

ksc sc k sc

i ik

ER I I Ur= =

rarr = = =sum sum 2p


285

c

Intensitatea curentului electric se mai poate scrie sub forma

1

1

1

1

11

nk

i kn

i k echivalent

echivalentn

i k

Er

r EIR rR

r

=

=

=

= =++

sum

sum

sum

2p

4p

adică 1

1

1

nk

i kechivalent n

i k

ErE

r

=

=

=sum

sum 1p

respectiv

1

11echivalent n

i k

r

r=

=

sum

1p

d

echivalentE E= 1p

3pechivalent

rrn

= 1p

EI rRn

=+

1p



a

Pentru circuitul dat se scriu teoremele lui Kirchhoff

0 2

A B

A

B A

I I IE Ir RI

RI RI

= + = + = minus

1p

3pRezultă

3 22

3 23

3 2

B

A

EIr R

EIr R

EIr R

= + = +

= +

2p


286

b

( 2 )3 2CD B A

E R xU RI xIr R

minus= minus =

+ este funcţie de gradul I deci are ca grafic o dreaptă 1p

5p

Pentru max

min

03 2

3 2

2

CD

CD

CD

ERx Ur R

ERx R Ur R

RU o x

= = + = = minus +

= =

1p

Graficul

1p

Panta dreptei este derivata de ordin I a tensiunii UCD icircn funcţie de x 2tg 3 2

CDdU Edx r R

α = = minus+

1p

Combinacircnd această relaţie cu expresia tensiunii maxCDU se obţine max26

tg CDU

R = minus = Ωα

respectiv 3 tg 2 15 V

2 mrE Uα

= minus + = 1p

c

Scriem teoremele lui Kirchhoff pentru circuit 1

2

1 2

( )E I R x IRE I x IRI I I

= minus + = minus = +

1p

3p

Rezultă

2 2

(2 )E x RIRx R x

minus=

+ minus1p

Pentru

0 2A

2A

02

Ex IR

Ex R IR

Rx I

= = minus = minus = = =

= = 2

max max 20 WP RI= =

1p

d Plecăm de la presupunerea că numărul de surse are o valoare fixată iar x şi y sunt variabile cu condiţia ca produsul lor să dea n O linie serie de x surse echivalează cu o sursă echivalentă cu tensiunea electromotoare echivalentă xE şi rezistenţa internă xr Un număr de y astfel de surse grupate icircn paralel echivalează cu o sursă

cu tensiunea electromotoare Eechivalent = xE şi rezistenţa internă echivalentxrry

=

Avacircnd icircn vedere că xy = n rezultă că intensitatea curentului prin rezistorul R este 2

xEI cu xy nx rRn

= =+

1p

4p

După prelucrare se obţine 2 ( )nxEI I xnR rx

= =+

1p

Această funcţie are un extrem pentru 0dIdx

= După derivare rezultă 2rxn

R= 1p

Icircn final rezultă 1000n = 1p



287

DOPTICAtilde

Subiectul I Punctaj1 b 32 b 33 c 34 a 35 b 3



a

Sfera este un ansamblu de doi dioptri sferici La prima trecere se poate scrie

2 1

1 1n nx x R

minusminus = 1p

5p

Icircn interiorul sferei 2 1( ) 2x x R+ minus = 1p

La a doua trecere 2 1

1 1n nx x R

minusminus =

minus 1p

Se deduce

12

1( 1)Rxx

nR n x=

+ minus 1p

Rezultă 2 7cmx = 1p

b

La prima trecere 2 2

1 1

y nxy x

= 15p

4pLa a doua trecere

2 2 1 1

y xy x

= 15p

Dar 1 2y y= deci

2 03 cmy = minus 1p

c

Prima imagine este realizată de oglinda sferică convexă formată de sfera de sticlă (o parte a luminii de la

lumacircnare este reflectată) 2 1

1 1 2x x R

+ = Rezultă 12

1

2 214 cm2

xxx R

= =minus

1p

3pDimensiunea verticală 2

21

007 cmxyx

minus= = 1p

Imagine dreaptă virtuală şi mai mică decacirct obiectul 1p

d

Următoarea imagine este realizată de a doua oglindire (oglindă concavă) la ieşirea din sferă Pentru studierea ei vom considera c este obiect virtual pentru imaginea de pe paravan

12 1

1

5 cm 7cm2Rxx R xx R

= = minus = Rezultă 2 184 cmx = minus faţă de al doilea dioptru1p

3p

Pentru dimensiunea imaginii 12 1 225 cm

2 x

Ryy yx R

= minus = minusminus

1p

Rezultă 2 059 cmy = Imaginea este dreaptă mai mare decacirct obiectul iniţial şi virtuală 1p



288


a

Drumul optic al primei radiaţii este 2 21 0 1 1 1 1( ) ( ) ( )n a l b n r e n eδ = minus + + minus + 1p

4pDrumul optic al primei radiaţii este ( )2 2

2 0 2 2 2 2( ) ( ) ( )

nn a l b n r e n e

r n rδ = + + + minus +

minus 1p

Rezultă ( )2 2 2 22 1 0 2 2 1 1( ) ( ) ( ) ( ) ( ) ( )n a l b a l b n r e n n e n nδ + δ minus δ = + + minus minus + + ∆ + minus minus minus 2p

b

Se poate scrie 2kx r

D l∆

= 1p

4pCondiţia de maxime de interferenţă ( ) kδ = λ 1p

Rezultă ( )2 2 2 20 2 2 1 1( ) ( ) ( ) ( )

2kDx k n a l b a l b e n n e n n nnl

= λ minus + + minus minus + minus minus + minus 2p

c

Interfranja (luminoasă sau icircntunecoasă) se calculează ca diferenţa coordonatelor a două franje consecutive de acelaşi tip 2p

4pRezultă

2Dinlλ

= 2p

d

= 0 b nu mai contează 1p

3pDimensiunile lamelelor să fie reduse la zero sau indicii de refracţie să fie ca ai aerului 1p

0 1 1n n= = 1p



289

Cuprins

Teste de nivel minimal 3Testul 1 4Testul 2 12Testul 3 18Testul 4 26Testul 5 34Testul 6 40Testul 7 48Testul 8 55Testul 9 62Testul 10 70Testul 11 78Testul 12 86

Teste de nivel mediu 93Testul 1 94Testul 2 101Testul 3 109Testul 4 116Testul 5 126Testul 6 134Testul 7 142Testul 8 150Testul 9 157Testul 10 165Testul 11 173Testul 12 181

Teste de nivel avansat 189Testul 1 190Testul 2 198Testul 3 206Testul 4 214Testul 5 224Testul 6 232Testul 7 240Testul 8 248Testul 9 255Testul 10 263Testul 11 271Testul 12 279

Această variantă electronică cu rezolvările testelor de fizică icircnsoţeşte lucrarea

TESTE DE FIZICĂ PENTRU BACALAUREAT (ISBN 978-606-38-0224-9)

copy Editura NICULESCU 2018

Bd Regiei 6D 060204 ndash Bucureşti Romacircnia

Telefon 021 312 97 82 Fax 021 312 97 83

E-mail edituraniculescuro

Internet wwwniculescuro

Comenzi online wwwniculescuro

Comenzi e-mail vanzariniculescuro

Comenzi telefonice 0724 505 385 021 312 97 82

Redactor Geta Vicircrtic

Tehnoredactor Carmen Birta Şerban-Alexandru Popină

ISBN 978-606-38-0224-9

Toate drepturile rezervate Nicio parte a acestei catilderthorni nu poate fi reprodusatilde sau transmisatilde sub nicio formatilde ordmi prin niciun mijloc electronic sau mecanic inclusiv prin fotocopiere icircnregistrare sau prin orice sistem de stocare ordmi accesare a datelor fatilderatilde permisiunea Editurii NICULESCU Orice nerespectare a acestor prevederi conduce icircn mod automat la ratildespunderea penalatilde fathornatilde de legile nathornionale ordmi internathornionale privind proprietatea intelectualatilde ____________________________________________________________________________________________________________________________________________________________________________________________

Editura NICULESCU este partener ordmi distribuitor oficial OXFORD UNIVERSITY PRESS icircn Romacircnia E-mail oxfordniculescuro Internet wwwoxford-niculescuro



4

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 d 34 c 35 a 3



a



= 1p


b




deci 0fF F =- 2p3p

Rezultă F = 4 N 1p

d



2p

4pObţinem 1

2 fF F Fam m

= =- 1p






b






5

c



Obţinem 2Ahh = 1p


d





Rezultă 2 JGL = 1p



6


Subiectul I Punctaj

1 b 32 b 33 a 34 c 35 d 3




11

mν =

micro2p

3p


b


A

NN

ν = 1p


22

mν =

micro 1p


2

mN N= sdotmicro

1p


c


1 1

m R TpVsdot

= sdotmicro 1p

4pRezultă 5 2



2 2

m R TpVsdot

= sdotmicro

1p


d


4p

unde 1 2m m+ν =

micro 1p


1 2 1 2 2 1

1 2

( )m m m mm m m m



1p




7


a


1 0

311

1

300 K

831 831 10 Pa

= 3 dm

Tp p

R TVp

=


=

1p


32 1

32 1

2 600 K

831 10 Pa

=2 6 dm

T Tp pV V

= =

= = sdot

=

2p


313

33 2

600 K

4155 10 Pa2

=2 12 dm

T Tpp

V V

= =

= = sdot

=

2p

b


12 2 1 V 2 1 17( ) ( ) ( )2


4p


23 2 12




12 23 17 2ln 22


Q Q Q R T 1p


c


12 V 2 1 15( )2



d


4p


23 2 12







8


Subiectul I Punctaj

1 c 32 a 33 d 34 d 35 b 3




3p


b



S

EIR r

=+

2p

Rezultă 1 AI = 1p

c


torul R este 1

1

EIR r

prime =+

2p




1SC

EIr

= 3p4p




9


a


5p


1 1 1

PR R R= + 1p


S PS P

E ER RR r R r

sdot = sdot

+ + 1p



b


S

RR r

η =+

1p



P

RR r

η =+

1p


c


1 11

EP RR r

= sdot

+ 1p

3pDeci 1

11

( ) PE R rR

= + sdot 1p

Rezultă 6 VE = 1p


EIr

= 2p3p




10

DOPTICAtilde

Subiectul I Punctaj

1 b 32 a 33 d 34 a 35 c 3




f= 2p

3p


b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+1p


c


1

xx

β = 1p

5pRezultă 12

β = minus 1p




3p




11



1

cλ =

ν2p

3p


b




2 1

( )

S S


=minus

1p


c



Obţinem 10 1

Se Uh


0 2Se U

hsdot

ν = ν minus 1p


d



4pRezultă 22







12

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1a [ ] J Nm= =SIL 3

2 c 2

2 2 00 2 270 m

2vv v ad d

a= + rArr = = 3


2

10 6

6 ms

v v at v t

a

= + = +

rArr =

854 kgFma

= = 3

4 b 2

2

2

2

C

C

mvE p mv

pEm

= =

rArr =

2

2

2

2

C

C

mvE p mv

pEm

= =

rArr = 3

5 b 2

0max

0

20 m 10 m2

5 JC C p C

vh hg

E E E E

= = rArr =

= + rArr =

3



a


1 1

1

t f

n

G F ma

N G

minus =

= 2p

4p


21 25 msa = 1p

b


2 20 1 02 0v v a l v= + = 1p

3p1 12v a l= 1p

1 10 316 msv = = 1p

c


2 2

2

fF ma

G N

minus =

=

1p



22 01a

gmicro = minus =

1p


13

d


1t f

n

G F FN G

= +

=

2p

4p


5 NF = 1p



a

1 1f fL F d= minus

1p

4p1f

y

F N

N G F

= micro

= minus

1p


1p

1

75 JfL = minus 1p

bcosL Fd= α 2p

3p

255 JL = 1p

c

1C fE L L∆ = + 1p

4p

2

2CmvE =

1p

2 CEvm

=

1p

2 3 34 msv =

1p

d

2C fE L∆ =

2 2f fL F D= minus

1p

4p2f

F N

G N

= micro

=1p

C

f

EDF

=

1p

D = 6 m 1p



14


Subiectul I Punctaj

1b mpV RT=

micro

30 kPamp RT

V= =

micro

3



4c

3127 kgmpRT

microρ = = 3

5

b 1 1 2 23 32 2

U RT U RT= ν = ν

2 1 2 12 2U U T T= rArr =

11

2 200 J3UL R T RT= ν ∆ = ν = =

3



a


1 2


= = 1p

4p


1 2

1 2

V p pmR T T

micro∆ = minus

1p

1 2

1 2 2 1( )mRTTV

p T p T∆

=micro minus

33324 mV =

1p

b

11

1

p VRT

ν =

2p3p

1 026 kmolν = 1p

c

22 2

mp V RT=micro

2p

4p22

2

p VmRT

micro=

1p

2 512 kgm = 1p

d

11

1

pRT

microρ =

3p

4p

31 25 kgmρ = 1p



15


a


3p 3p

b


1 2

3V V V VT T

= rArr = 1p

4p12 1 2 1( )L p V V= minus 1p

121

2 1( )Lp

V V=

minus 1p

5 21 2 10 Nmp = sdot

1p

c


23 31cedQ Q Q= + 1p


1p

131 1 1 1

3

ln ln3VQ RT p VV

= ν = minus

1p


d


4pp VC C R= + 1p

1 1 15 5absQ RT p V= ν = 1p




16


Subiectul I Punctaj

1b 1 Wh = 3600 J 6720 10 JW = sdot

3

2 c 1 1 2

2

E I EIR r R R r

= =+ + +

1 1 2

2

E I EIR r R R r

= =+ + +

1 22

1

2 8R R r RR r+ +

rArr = rArr = Ω+

3

3 b 23ABRR = = Ω 3

4 c V1sc

EIr

= =Ω

3

5c

2

bec 90n

n

URP

= = Ω


3



a

1e pR R R= + 1p

3p2 3

2 3p

R RRR R

=+

1p

44eR = Ω 1p

b

1e

EIR r

=+

1p


88VU V= 1p

c

1 2 3I I I= + 1p

4p2 2 3 3I R I R= 1p

1 23

2 3

I RIR R

=+

1p

3 08I A= 1p


17

d


1 23

1 2e

R RR RR R

= ++

1p

4p125 A

e

EIR r

=+

1p

1 2

1 1 2 2

I I I I R I R

= +=

1p

12 083 A3II = =

1p



a


1 3 3E Ir U I R= + +

2p

4p

13

3

E Ir UIR

minus minus=

1p

3 15 AI = 1p

b


2 2 3 3I R I R= 1p

4p3 3

22

75I RRI

= = Ω

1p

22 2 2P R I= 1p

2 1875 WP = 1p

c

1e pR R R= + 1p

4p

11 30UR

I= = Ω

1p

2 3

2 3p

R RRR R

=+

1p

4875eR = Ω 1p

d

1 1

s

P UP E

η = =

2p3p

05 50η = = 1p



18

DOPTICAtilde

Subiectul I Punctaj

1 c 3

2 a 2

1

sin 3sin 60sin 2

i n r rr n

= rArr = rArr = 3

3 b 1 2 3 3 4C C C C C= + + rArr = δ 3


3

5b 2 2

1 1

2 1

3

3 150 cm

y xy x

x x

β = = = minus

rArr = minus =

2 1

1 1 1 375 cmff x x

= minus rArr =

3



a

2 21

1 1

3y xy x

β = = = minus

1p

4p2 13x x= minus 1p

11

43fx = minus

1p

1 40 cmx = minus 1p

b

32

1

3xx

β = =

1p

4p3 13x x= 1p

1 32

1 3

x xfx x

=minus

1p

2 60 cmf = 1p

c 3p 3p


19

d

1 2

1 1 1F f f

= + 1p

4p

20 cmF = 1p

14

1

FxxF x

=+

1p

43

1

1xx

β = = minus 1p





b

0L h= ν 1p

4p

150 055 10 Hzν = sdot

1p

151

1


λ 1p


c2

2

hcε =

λ 2p

3p


d

2 ex CL Eε = + 2p

4pCS

EUe

=

1p

079 VSU = 1p



20

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 a 34 a 35 d 3



a


0eF G+ =

1p


11

m glk

∆ = 1p


b

fF F= 1p

4pN G= 1p

fF N= micro


∆ =m glk 2 125 cm∆ =l 1p

c

e fF F ma+ =

1p

4peF F= 1p

2

2

minus micro=

F m gam 1p


d

22

2pk lE ∆

=

2p

4p2

Flk

∆ =

1p




21


a

F G ma+ =

1p



b

vat

∆=

∆1p

3p0 = +v v at 0 0v = 1p


c

total∆ =cE L 2p

4p2

2cmvE∆ = 1p


d


4p

2

1 2

= +mvE mgh

2

2 2PmvE =

1p

2( )2∆

=a th

1p


1p



22


Subiectul I Punctaj

1 a 32 d 33 b 34 a 35 c 3



a A

N mN

=micro

2p4p


b


4p0

0

mV

ρ =

mV

ρ = 1p

0

0middotmVV

V m=

∆ρ +1p



3p


d


4p0 p VRT

ν =


2p



a


2 2

1 1

=p Vp V

2p

3p

Rezultat final 2

1

2pp

= 1p


23

b



2 14T T= 1p


U RT∆ = ν rArr 12 10800U J∆ = 1p

c


( )=L A p V 2p

4p1 1 1

2 2p V RTL ν

= =


2p

d



2 14 =T T 3 12T T= 1p





24


Subiectul I Punctaj

1 d 32 b 33 a 34 c 35 c 3



aRS= R1+ R2 2p


b

UAB = R2 I 1p

4ps

EIR r

=+ 2p


c s A

EIR R r

=+ +

3p4p

I primecong 12 A 1p

d 1

EIR r

=+

3p4p




a


4pISC = 10 A 1p

ISC = scEIr

= 1p


b


4pEIR r

=+

1p


c

echivalent

echivalent

RR r

η =+

2p


91η 1p





25


1 a 32 b 33 b 34 d 35 a 3




b

2 1

1 1 1x x f

minus = 1p

4p1fC

= 1p


c

2 2

1 1

β = =x yx y 2p

3p


d

1 1 60 cm= + = minusx d x 1p

4p12

1

x fx x f

=+

1p





3p


bfW h= ν 2p

4p








26

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1c

5

8

km = 3 10 uas = 012

1ua min1 km = 15 10

cc

sdot rArrsdot

3


m

m

km = 200 m 48 h

= 15 s

dvt

d vt

= ∆ rArr =∆

3

3c

25

3

= 75 10 J2 = 208 kW h

11 J = 1 W s = 10 kW h3600

c c

c

m vE EE

minus


sdot sdot

3

4



= 75 Nm m

m v vF F

tsdot +

= rArr∆



5


0 0 02

0 0 2

0

4 1 3828 cm4

F l F lE lS l E S

d FS l l ld E

l l l


∆ = minus

3



a


4p2 = 2 m sxx

RR m a a am






27

d


5p


1p



1p



2p



a



= 11314 mdrArr

2p

5p2p

2 1 2

= 14 s

= 12 h 24 min 34 s

dv tt

t t t t

= rArr ∆∆

∆ = minus rArr1p

b

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr =

2p3p

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr = 1p

c


4p


2p

( )1 20

879 MJr

r

F

F

m g h h L

L


rArr = minus

1p

d

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

2p

3p

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

1p



28


Subiectul I Punctaj


3



3

b 1 1

2 2

1 1 11

22 2 2

5 cu 80 K 52 1673 3 cu 80 K2

V V

V V



ν

ν3

4

d 23 23 23

2323

40 J0

Q U LU

L= ∆ +

rArr ∆ = minus=


123112

1231 12 23 31

0110 J

UU

U U U U∆ =

rArr ∆ =∆ = ∆ + ∆ + ∆


3

5

c 22 72 10

VA

VA

Q C TQ NN NNC T

N


3



a

1 2 m m m= = 1p

3p1 2

2 1

125ν micro= =

ν micro2p

b1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


1p

4p1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


c

3

kg 157 m

mp V R TpR Tm

V


rArr ρ =

3p

4p

3

kg 157 m

mp V R TpR Tm

V


rArr ρ = 1p


29

d5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

4p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot 2p





c

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

4p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

2p

d

5 1 3

pp v

v v

p v

CC C R RC C

C C R


γ minus= +

1p

5p


( )( )

1 2 1 2

1 2

1 2

1 2

1

1v v v v v v


f f



2p

2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = =

1p2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = = 1p



30


Subiectul I Punctaj

1a 2

2

514 nC

U S qI U r tl t ql

S rq


=

3

2

a ( )( )

1 0 1 2 1

1 2 2 12 0 2

3 1

1 - 1

4 10


Kminus minus


α = sdot

10 0

1

9 1

RR Rt


3

3




3

4b

=12800 C

PIP tU q q

q UIt


3

5


= rArr = Ω


= rArr = =


3





1

9 E EI R r R


+ 2p

b



5p


1817 18

E EI R R rR R r I

R


rArr = Ω cong Ω1p




1p

2 05 mml lR S SS R



31

c


1p

4p1

1

105 A 11 s

EIR R r

I t


primeprime = rArr =

2p

1

1

105 A 11 s

EIR R r

I t



dI = 0 1p

3pU = E 1pU = 20 V 1p



a = 2400 mA h = 8640 C q sdot 2p


b1

06 AUI IR

= rArr = 3p 3p

c 14400 s = 4 hq qI t tt I


3p 3p

d


1p

5p


UI IR



2 22

6480 s = 18 hq qI t tt I


1p

2 22

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆

1p2 2

2

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆ 1p



32


1


3

2 c 3

3




3



D iil n n


3


0

2 4

s

s

h c h c e U

h c h c e U



3



a

22 1

12

1

2 4 cm

2 2 cm

xx x

xx

x


2p3p


b

2 1

1 1 1

1

25 4 cm

x x f

Cf

Cf

minus = =

rArr = δrArr =

2p4p


c

2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus

2p

4p2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus 1p



33

d


4p

2

1 2 1

1 1

2 1

22

8 cm 1875 1 1

xx x x

x x C

C Cx x


primeminus = +



3p




b

1

1

22

kDxD li

l


sdot= sdot

1p


2

R V Dx

lλ minus λ sdot

∆ =sdot

2p

076 mmx∆ = 1p

c2

2

2 2 192 mm

2

k x ixDi

l


= sdot 3p 3p

d


4p

1 21 22 2

D Dk kl l


sdot sdot1p


Soluţia 1

2

32

kk

= =

1p


0Cprime lt rArr


34

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj

1 a 32 b 3

3


0 sin cos 0sincos

Gt Ff mg mg

tg


micro = = αα

3



5



= + = + =

3



a


4p 4p

b


şi 2a


1p


2 2

T m g m aT m g m a

minus sdot = sdot



1 2



+ 1p

c T = m1(a + g) = 2222 N 3p 3p

d


1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p

4p

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p



35


a

2p

3p

cos5640 J

L F dL

= sdot α=

1p

b

NFfL Ff d

Ff= minus sdot

= micro

1p

4pN

FfL Ff dFf

= minus sdot

= micro 1p




4p4504 JfEc = 2p

d564 W

med

med

LPt

P

=∆

=

2p4p

564 W

med

med

LPt

P

=∆

= 2p



36


Subiectul I Punctaj

1 c 3

2 a 33 b 34 a 35 c 3



a3096 kgm

pRT

microρ =

ρ =

2p3p

3096 kgm

pRT

microρ =

ρ = 1p

b

a

a

N m N mNN

sdot= rArr =

micro micro1p

4p



a

a

N p VNp V RT NN RT

= rArr = 1p


RT= = sdot 1p

c

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus=

1p

4p

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus= 1p

1

2

11

pp f

=minus

1p

p2 = p1(1 ndash f ) = 96 ∙ 105 Pa 1p

d


4p


1p

21 O 2

1 2


ν + micro ν

ν + ν2p



37




3p 3p

b


2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= =

2p

4p2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= = 2p

c9kJ

p

p

LO

LO

η = rArr

= =η

2p

4p

9kJ

p

p

LO

LO

η = rArr

= =η

2p

d


4pT2 = T3 1p

ΔU = 0 1p



38


Subiectul I Punctaj

1 a 32 c 33 c 34 a 35 b 3



a


1 1 1

PR R R= + 1p



R = 11 Ω 1p

b




cE = I3r + UAB 2p

3pUAB = 22 V 1p

d


1

UIR

= 1p



2

UR


U1 = U2 = 96 V 1p




3pUb = 120 V 1p

b



Uf = 4 V 1p




W = 4464 kJ 1p



39


1 c 32 a 33 d 34 c 35 c 3





3p

Rezultă cvn

= v = 212 ∙108 ms 1p

b



r = 450 1p090 45rθ = minus = 1p

c




1p

4p1sin sin2

n l n π= 1p

1

2

sin nln

= 1p

l = 450 1p


3p 3p




1Cf

= 2p3p


c

2 1

1 1 1x x f

minus = 2p



1 1

x yx y

β = = 1p

y2= y1=10 cm 1p


2p3p




40

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj

1 c 32 a 33 c 34 c 35 b 3



a

2p

4p

2p

b





M m M m


= =+ +

1p

21msa = 1p


09 NT = 1p

d



17 NF = 1p



a


4pfc u FL L L= + 2p

2500JcL = 1p


41

b

u

c

LL

η = 2p

3p2000 08 802500

η = = = 1p

c

cossin sinfF f


1p

4p

u GL L mgh= = 1p

f

f

FF u

u

LL ctg L

L ctg


α1p

1 0254

micro = = 1p

d

2 60 40 s3

t∆ = sdot = 1p

4p

u cu c

L LP Pt t

= =∆ ∆

1p

2000 50 W40

2500 625 W40

u

c

P

P

= =

= =2p



42


Subiectul I Punctaj

1 d 32 d 33 b 34 c 35 c 3



a

1 1 1 1 1

2 2 2 2 2

( ) ( 2 )2 2 2

( - ) ( - 2 )2 2 - 2




+sdot


4p5

51

55

2

10 1 5 10 Pa18 9

10 1 5 10 Pa02

p

p

sdot= = sdot

sdot= = sdot

2p

b

F



133(3) NF = 1p

c

1 1 1

2 2 2

( 2 ) ( 2 )





2p

4p

51

52

510 Pa6625= 10 Pa

6

p

p

prime =

prime2p

d

1

2 p ll hpsdot

+ = 1p

4p1

1 12

p lhp

sdot= minus prime

1p

01m = 10 cmh = 1p




43



3p4p



c

212 1 1 1

1


= ν = = sdot =


1p

5p

5 51 11 1 2 2 2 1

2 1



3 523

5 2 10 3 10 (1 27) 1500 17 2550J2



1620 2550 1530 600 JQ = minus + = 1p

d

U Q L∆ = minus 1p

3pL Q= 1p

600JL = 1p



44


Subiectul I Punctaj

1 a 32 a 33 d 34 b 35 d 3



a

1 1 1 1 1AU I R I= rArr =

2 2 2 2 075AU I R I= rArr =2p

4p( )1 1 2 1 1 2

2 2 1 2

( )6V

( )E I R r R R I I

EE I R r I I


2p

b

1 11 1 2 2

2 2

( )( ) ( )

( )E I R r

I R r I R rE I R r

= +rArr + = + = +

2p

4p2 2 1 1 2 1

1 2 1 2

I R I R U UrI I I I

minus minus= =

minus minus 1p

2r = Ω 1p

c

66

15012 10 603 10fir

lRS


sdot2p

4p6 075A 075 60 45C8fir

EI q I tR r

= = = = sdot ∆ = sdot =+ 2p

d0

00 0

1(1 ) 1 1

fir

fir firfir



α2p

3p

40 Ct = 1p



a

ss

s

EIR r

=+ 1p

4p10sE E= 1p

10sr r= 1p

4 133 A3sI = = 1p


45

b

pp

p

EI

R r=

+ 1p

4p10prr = 1p

10

110p

ErE E

r

= = 1p

05ApI = 1p

c


4p2 pU I R= 1p

1 1 VU

1p

2 35VU = 1p

d1

s

RR r

η =+

2p

RR r

η =+

1p

3p

17 259227

η = 235 972236

η =

2p



46


1 d 32 c 33 c 34 c 35 a 3



a 4p 4p

b

sin sini n r= 1p

4psin sin n r i= 1p



c


4p

sin sin

3sin 12sin

23

i n r

irn

=

= = =1p

30r = deg 1p



d cosADAB

r= 2p

3p

115cmAB 1p



47


a


1

2

1

2

s

s

ch L eU

ch L eU

= +λ

= +λ

2p

5p


2

2 1

1 1 2

1 2


( )( )

s s

s s

hc e U U

e U Uh

c

minus = minusλ λ

minus λ λ=

λ minus λ

2p


b1

1s

cL h eU= minusλ 2p


c0 0

Lh Lh

ν = hArr ν = 2p3p

150 0924 10 Hzν = sdot 1p

d

2max

2smveU = 1p

4pmax2 seUv

m= 1p





48

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj


0 m 8 N4 m 0 N

x Fx F

= rArr == rArr =


(6 2) 2 8 J

2L + sdot

= =

3


22 2i f

kx mv kxE E vm

= rArr = rArr = = 3

3

a Viteza medie mdvt

= (1)





= şi 2 2dtv



1 22 2

mm m


+= rArr = rArr = =

++km h


3

4 a 35 d 3



a


1km20h

v =

2km40h

v =

1130min h2

t = =

2115min h4

t = = 2p4p


b

Viteza medie mdvt

= 2p


t = = 1p




d


21

2cmvE =

2

22

2cmvE =

Rezultă 2

1

222

1

c

c

E vE v

=2p

3p

2

1

4c

c

EE

= 1p



49


a




hl =α

iar 1

0NL =


2p

ECB = 32 J 1p

bLFf

= LFf1 + LFf2



c


1p

4p2




= = 1p

d



CDEE = 1 1

15





50


Subiectul I Punctaj

1 b 32 c 33 b 3





3

5 c 3



a



1 2

12p p p pT T

= rArr = 1p

2p = 5

2

N12 10m

sdot 1p


max1 max 1

p p p TTT T p

= rArr = 2p4p

max 12 600KT T= = 2p

c


1 00 1

91p Tp T

ρρ = = ρ 2p

3p

31 0819kgmρ = 1p

d


mp V RT=micro

şi 21 2



12 112





a


3p

2p


51

b


VVrArr = 1p


2 3

V VT T



2 12 2888KT T= = 1p


3

1 2p

vv

C RC RC


4p

1 3 1 1 1 1 13 3 3(2 ) 1800 J2 2 2


d

2

1

| |1 QQ

η = minus 1p

4p

1 2 3 3 2( )5

1 2p

Q Q Cp T TRC R

rarr= = υ minusγ

= =γ minus

Rezultă 1 1 1 15 252

Q RT p V= υ =

1p

12 1 2 3 1 1 3 1 1 1 1 1 1 1

2

3| | | | ln ( ) ln 2 (ln 2 15) 21932v



122η = 1p



52


Subiectul I Punctaj

1 c 32 c 33 d 34 d 3




a



1 21

PRI

= 2p

R1 = 5 Ω 1p

b



EIR r

= =+

1p



e

RR r

η =+

2p3p

80η = 1p

d


4pPuterea maximă

2

max 125 W4 EPr

= = 2p



a


5p


2

V

p

V

RRR RR

= = Ω+

32e pRR R= + = Ω 2p


EIR r

= =+

1p



3pP = 75 W 1p


3pPE = 100 W 1p


e

RR r

η =+ 2p

4pη = 75 2p



53



2max

max max2

2S

C S SmV eUE eU eU V

m= rArr = rArr = 3

2 b 33 a 34 c 3

5


1

2

R RR R

= minus= +

1

1 2

1 1 1 1 2( 1)( 1)( ) ( 1)( ) 5mnC n nR R R R R







3pDin 2

2 1 21

4 20cmy y y yy


c


4pCum 2

2 1 11

4 4 5x x x d xx


1

2 1

16cm54 64cm

dx

x x

= minus = minus

= minus =1p

d


2 1 1 2

1 1 1 ( 16) 64 128 cm16 64

x xfx x f x x


minus minus minus2p


1 1 2( 1)( 1) ( )

Rfnn

R R


2p

2( 1) 128 cmR n f= minus = 1p



54


a


3p 3p

b

302



sinsin irn

rArr =

3sin8

r = 1p

c


5p

tg hOI

α = rArr 1038mtg 3

h hOI = = =α 1p


= rArr2

sintg 0809m1 sin

rAB H r Hr

= sdot = =minus

2p


d


4p



53

i n r i n i

i i i n

i

= rArr = minus


rArr =

2p



55

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj

1 a 32 d 33 a 34 b 35 a 3



aG1 = m1g 1p


b




1 2

g m ma

m mminus

=+

1p


c


4p1 2

1 2

2m m gTm m

=+

2p


d

eF k l= sdot ∆ 2p

4p2Tlk

∆ = 1p




a





fFL = minus 1p

b

totalcE L∆ = 1p

4p

2

2cmvE∆ =




1p


56

c

util

consumat

LL



11 ctg

η =+ micro α

ctg xh

α = 1p


d

c totalE L∆ = 1p

4p

2

2cmvE∆ = minus

total fFL L= 1p

2

2mv mgd= micro

2

2vd

g=

micro 1p




57


Subiectul I Punctaj

1 d 32 d 33 d 34 a 35 c 3



a

1 1 1p V RT= ν 1p

3p

11

1

RTVp

ν=

1p


b


1 2

V VT T

= 2p

4p1 22

1

TVTV

=

1p


c


4p11

mV

ρ =

22

mV

ρ = 2p


d

( )12 1 2 1L p V R T T= ∆ = υ minus 1p


52VC R= 1p





58


a

11

NnV

= 1p

3p1 1 1p V RT= ν A

NN

υ = 1p

rezultat final 11

1

Ap NnRT


b

123 3 1U U U∆ = minus 1p

4p1 1VU C T= ν

3 3VU C T= ν 1p

( )123 3 1VU C T T∆ = ν minus 1p


c

43-4 3

3

ln VL RTV

= ν 1p


33-4 3

4

ln pL RTp

= ν 34 3 ln 2L RT= ν 1p


1p

d

cedat 4-1Q Q= 1p


1p

4p VC C R R= + = 1p




59


Subiectul I Punctaj

1 d 32 a 33 d 34 a 35 d 3



a

0

0p

R RRR R

=+

2p

4pe pR R R= + 1p


b


EIR r

=+

1p


( )0

A

E I r RI

Rminus +

= 1p


c

V eU R I= 1p

4pu r I= sdot 1p


d

A

EIr

= 1p

3p 0VU = 1p




a

21 1 1P R I= sdot


4p1

1

EIR r

=+

22

EIR r

=+

1p




60

b scEIr

=

2p3p


c

ext s

totala s

P RP R r

η = =+

2p

4p1 2sR R R= + 1p


η = = 1p

d rEP4

2

max =

2p




61


1 a 32 d 33 c 34 b 35 b 3



a1

1

1Cf

=

11

1fC

=

2p3p




c

112

111fxx

=minus

1p

4p2 2

1 1

y xy x

β = =

1p


d

1 2

1 1 1

sf f f= + 1p

4p2 1

1 1 1

sx x fminus = 1p







c0extL h= ν 2p

4prezultat final 14


dc sE e U= sdot 2p

4p




62

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj

1 a 0

0

FllE s

∆ =sdot

3

2 c 2

m4 ms 22 s s

va at

∆= rArr = =


3

3 b


= sdot =

3

4d

355 10 2750 N20

PP F v Fv

F

= sdot rArr =

sdot= =

3


cos ( sin )cos sin


α + micro α 3



a


3px24 = 3 ∙ (4 minus 2) 1p

x24 = 6 m 1p

b

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

=

2p

4p

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

= 2p

c

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

5p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1ptotal 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


d

total

total

9 m2254 s

m

m

xvA

v

=

= =

2p

3ptotal

total

9 m2254 s

m

m

xvA

v

=

= = 1p



63


a


3p2

0 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

=

1p20 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

= 1p

b


0

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus = 1p

4p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

c 056B

B

p

p

E mgh

E J

=

=

2p4p

056B

B

p

p

E mgh

E J

=

= 2p

d


4p024 J

f AB

f AB

pF B A

pF

L E E

L

= minus

= minus

2p

024 Jf AB

f AB

pF B A

pF

L E E

L

= minus

= minus 1p



64


Subiectul I Punctaj


2 d [ ]SI

J1mol K

C =sdot

3

3c 2

1

ln

1929 J

VL RTV

L

= ν

= minus

3


5 a A

NN

ν = 3



a

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

2p

4p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

1p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot 1p

b0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

c

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

2p

4p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

1p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr = 1p

d

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

4p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

= 1p



65


a

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

4p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

= 1p

b

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

2p

4p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

1p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr = 1p

c

L12341 = Aria12341 1p

4p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot 1p

d( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot

2p3p( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot 1p



66


Subiectul I Punctaj


2 a [ρ]SI = 1 Ω m 3

3 c q = I Δt 3

4 b 0 0 00

112 3 6eR R RR R= + + = 3

5 d 6 2003

UR RI

= rArr = = Ω 3



a2 3

12 3

11

e

e

R RR RR R

R

= ++

= Ω

2p3p

2 31

2 3

11

e

e

R RR RR R

R

= ++

= Ω 1p

b


4p


2

3

RR

1p

rArr 22

3

1 R

I RI =

+

1p

rArr 32

2 3

IRIR R

=+ rArr I = 04 A 1p

c

e

EIr R

= rArr+ 2p

4pe

Er RI

= minus 1p


d

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

2p

4p

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

1p2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =1p



67


a

Din legea Joule

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

3p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr = 1p

b

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1p

4p

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


c

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

2p

4p

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

1p2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω 1p

d

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

2p

4p

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr = 2p



68


1 c 1

SI

m ms s

cminus

= = ν sdot 3


3 c Negativă 3


3

5 a sin 2 141sin

in nr

= rArr = = 3



a 3p 3p

b

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

2p

4p

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p

c

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


2p

4p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

d

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

4p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1psistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr = 1p



69


a

00

814

0 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot

2p

3p0

08

140 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot1p

b 8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

2p

3p8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

1p

c

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

4p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ



d


34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

2p

5p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr = 1p



70

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj

1 a 32 a 33 b 34 c 35 d 3



a


2p4p


b


1 1 1 1

1

(2) ( ) ( )

G T m a G T m a


+ = rArr minus + =

+ + = + rArr + minus = +

1p

4p



1 2 1

( )2


minus += =

+ + +

1p


2 1 1 21 2

1 ( ) ( ) ( )2



+ +rArr 1 1

1

2 ( )2

m m m gTm m

+=

+ 1p

rezultat final 2

m02s

a = 25NT = 1p


1

4 ( )22

m m m gF Tm m

+= =

+ 2p

3p


d




2p

4p1

1

22

m mgNm m

=+

1p


1p


71



a




2p

4p2 2

90 m2 2

at F th gm

= = minus =

1p


1p

b


22

2

45 m2

2

FF g tv g t mm h hg

v gh


2p

4p



c



3pmax2v gh= 1p


d



4p2 maxL mgh= 1p





72


Subiectul I Punctaj

1 c 32 a 33 d 34 c 35 b 3



a


11

1

AA

A

pV RTNpV RTm NN

NN RTVpN


=

3p4p


b


11 A A

m N NmN N

microν = = rArr =

micro

1p2p


c


1A

NN

ν = 2 12 1

2 2A A

N NN N


3 3A A

N NN N

ν = = = ν 3p


1 2 3

2 36

amestecamestec

total


ν ν + ν + ν 1p


d



pν

= 1p


ν + ∆ + ∆= = 1p




73


a


3p 3p

b


4p


1

p Vp V RTRT

= ν rArr ν = 1p


2 11 2

3p p T TT T

= rArr = 1p


c


5p


2 22 2 3 3 3 2

3

32

p Vp V p V V Vp

= rArr = =

1p


2 3 2 1 12

3ln 3 ln 4800 J2



3

32

p TT Tp



1p


1p


1 2 2 3primit

L LQ Q Qminus minus

η = =+

1p3p




74


Subiectul I Punctaj

1 a 32 b 33 b 34 c 35 d 3




sc

E EI rr I

= rArr = 2p3p


b


4p


2 3 4

( )R R RRR R R

+=

+ + 1p



2 3 4

( )e

R R RR R RR R R

+= + +

+ +1p


c


EIR r

=+


eU IR= 1p

4p


1p

2 2 3 4 ( )U I R I R R= = +

1 2 5 1 5 2( ) U U U U I R R I R= + + = + + 1p

1 5

2

( ) eI R R RIR


d






75


a




2p

4p


W WIUt Ut

= = şi 22

23

W WIUt Ut

= =


2p

b


18URI

= Ω 22

9URI

= Ω 2p


1 2p

R RRR R

=+

1p



1 1 1WP R It

= = 2 21 2 2

WP R It

= = 2p4p



p

p

RR r

η =+ fie cu putil

consumat

R EPP EI

η = = 2p3p




76


1 d 32 c 33 a 34 d 35 a 3



a


1 1 1

0 4 4y x x x xy x x



2p4p


b



1 1 1 5Cf x x

= = minus = δ 1p

3p



c


= = 1p


2 1

1 1 1F x x

= minus 1p


1

FxxF x

=+

1p


d


FC

= = 1p


1

25 cm

F xXF x

= =+

1p





a



ν = = sdot 1p


1p


77

b



2extmvLε = + 1p


mε minus


c



LUe

ε minus= 1p


d



5p


1 4λ

λ = (1) 1p


exthcL h= ν =λ


hcL

λ = (2)


hcL

λ =

1p

1 1 11






78

TESTUL 11

A MECANICAtilde

Subiectul I Punctaj

1 b 2 2

2 22

m1kg kg m s2 sSI


2 b 3





m2 sin 10s



sdot sdot= = =

∆ ∆ 3



a


1p

4p



1 2 1 2



+ +

2p




c


1 2 0T Gminus + = 12 fG F= 2p


=micro

1p


d


1p

4p





79




Rezultă 42500 JAt

E = 1p

b


4p

2 2

2 2 f

B AG F


1p

Obţinem 2

2f

BF



c



sinhd =

α1p



1p


d




2p

4pObţinem 1 2

AtE

h sm g

=sdot sdot

1p




80


Subiectul I Punctaj

1d 1 1J kg K

SI

Q Jc cm T kg K


3


R Tp ρsdot sdot

=micro 3

kg8m

pR T

sdotmicroρ = =

sdot3



5 b 3



a


A

m NN

=micro

2p

4pRezultă 0A

mNmicro

= 1p



1p VT

Rsdot

=ν sdot

2p3p


c


4p


2

pR T

sdotmicroρ =

sdot

33

3

pR T

sdotmicroρ =

sdot 1p


3 1 1

2T TT T

ρ sdot= =

ρ1p

Rezultă 2

3

2ρ=

ρ 1p

d


Np V R TN

sdot = sdot sdot 1p

4p

Obţinem 1 12 4A

Np V R TN



1

24

Ap NnR T

sdot sdot=

sdot sdot1p

Rezultă 263

mol036 10m

n = sdot 1p



81


a


4pRezultă 3 11

22 R TV Vp



b


12 23tL L L= + 12 0L = 3

23 22

ln VL R TV






d







82


Subiectul I Punctaj


2 1J A sSI



sdot ∆= = = = Ω

∆

3

3 b 2 2

3 3eR R RR

Rsdot sdot sdot

= =sdot

3

2 3e

E EIR r R r

sdot= =

+ sdot + sdot3

4 d 3


WIR t

= =sdot ∆

50 mAI = 3



a





e e

EIR r

=+

= 5A 1p


b


2 32RI R I= sdot 1 2 3I I I= +2p

4p




1lR

Sρsdot

= 2

4dS πsdot

= 2

1 11 4


ρ sdotρ2p



4pJustificare

1

2 666Ate

EI I cresteR r

sdot= = rArr

+2p



83


a


4pRezultă 2bb

PRI

= 1p


b


1p

4pRezultă b

Rb

PU U sI

= minus 1p

RU R I= sdot

R

b

URI

= 1p


c




d

2R bP R I= sdot 2p

3p




84



= rArr = rArr =λ

3

2 d 3

3 c 22 1 1 1

1


2

i n in n n nr n r

= rArr = = sdot = 3

4 c 1 1 1 02 m = 20 cm5

C ff C

= rArr = = = 3

5 b 3




1 125 10

Cf mminus= =

sdot2p

3p

Rezultă 4C = δ 1p

b


2 1 2 1 1

1 1 1 1 1 1 fxxx x f x f x f x



4pobţinem 2

2 22

2fxx x ff xminus

= rArr =minus

1p



d


4pRezultă 1 2

1 1 100 10025 20

Cf f

= + = minus 2p




a


2p

4pRezultă 25 8

199

66 10 3 10 199 10450 10 4

minusminus

minus


sdot1p


1p


85


0

hcL h= ν =λ

2p3p


c

0eU L= ε minus 1p


eε minus

= 1p

Obţinem 19

19

(49 44) 1016 10

Uminus

minus

minus sdot=

sdot1P


d


2cmE eU ν

= =

1p


m mε minus

ν = = 2p




86

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj

1 b 32 c 33 b 34 a 35 a 3






12

m3 s

F mg maFa gm

minus micro =

= minus micro =

2p

4p1

12

m3 s

F mg maFa gm

minus micro =

= minus micro = 2p

c


2 2

2

cos 0

sin 0f

f

F FN F mgF N

α minus =

+ α minus == micro

2p4p


mgF micro= =

α + micro α 2p


4pRezultă 3 kgFm

g∆ = =

micro2p




2

2MvMgh = 2p

4p




2

0 cos1802c


4p


2p


2 2


2p

4p


2p



87


Subiectul I Punctaj

1 c 32 c 33 a 34 d 35 b 3




1p3p


micro= = 2p

b

Din pV RT= ν 2p

3p

rezultă 4molpVRT

ν = = 4 moli 1p

c


5p

1 2 2

1 1

2871 1 004(3) 4(3)300

U U TfU Tminus



2

1 2 2 2

11 1 1

1 1 1 4(3)

RTp p p TVk RTp p T

V

νminus


scade cu k 1p

d


4p1 0


micro= = =

micro2p



88


a


1 1 1

8atm 300K 1200 1200 273 927 degC2atm

p T pT T K tp T p

= = = = = minus = 1p

4p


Graficul

2p

b

1 1 1p V RT= ν 1p

3p1 1

1

p VRT

ν =

1P

v cong 012 moli 1p

c

323 2

2

ln VL RTV

= ν

1p



2 1


= = = = 1p

223 2 1

1

V ln 11088 JpL pp

= = 1P

d

1abs ced ced

abs abs abs

Q Q QLQ Q Q


4p2

12 23 2 1 21




025 25η cong = 1p



89


Subiectul I Punctaj

1 b 3

2


2 3E E EI I I

R r R r R r= = =

+ + +




1 23

1 2

3 3 2727A3 4

E I IIR r I I

= = =+ minus

3

3 d 3

4


RR R r

η =+ +



R rη =

+



3

5


2

8 8( ) 9 9 4m

RE EPR r r

= = sdot+


5 34

2

rr rR r

plusmn = =


3



a

12 3

12

1 AEI R Rr RR R

= =+ +

+1p

4p2 3

12

2 VBAR RU I

R R= =

+ 1p

22

2 A3

BAUIR

= = 1p

33

1 A3

BAUIR

= = 1p

b

22 2 2qI q I t

t= = sdot ∆

∆2p

4p2

2 A 30 s 20 C3

q = sdot = 2p

c



scEI

R r=

+ 1p


d

p

EIR r

=+ 1p

4p1 2 3

1 1 1 1 142pp

RR R R R

= + + rArr = Ω 1p




90


a


2LL

U UI RR R R=

+ + 2p


Randamentul este 2

2

1 05 50( ) I 2L

RIR R

η = = = =+

1p

b


2 290400J3U tW I Rt

R= = =



290400 003 kWh36 10

W = congsdot

1p


kWhC = sdot = = 1p

c


xxrS


lRS

= ρ


=


UI xRl

=+

1p

4p


2(1 )

UP xRl

=+

1p



max0 605Ux P WR

= = =


min 20163Ux l P W

R= = cong

1p


1p





91


1 c 32 d 33 c 34 d 35 c 3



a


2 1

n nx x

=



= minus

2p

3p


nminus

δ = = 1p

b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=



c


1 1

x yx y

β = = 1p

5p

Rezultă 22 1

1

xy yx

= 1p

Rezultă 2 11

fy yx f

=+

1p








lλ

= 2p3p



4kk Dx

l+ λ

= 2p4p



92

c



4pSe obţine

2Di

lλ

= 2p


d


i i∆ minus

ε = = 1p


ε = minus 1p

Rezultă DD

δε = minus 1p





94

TESTUL 1

A MECANICAtilde




aDeoarece N = Gn 1p


b


0

v vat t

minus=

minus 2p



c



d





a




f g= sdot =

sdot1p



max 2 2cm v pE

msdot

= = 2p3p


c




d


4obţinem rF

r

Ld

F= minus 1p




95





a



obţinem 0

0

p L SR Tsdot sdot

ν =sdot

1p


b


1 2

m m=

micro micro1p

4Dar 1 2 ν = ν + ν sau 1 2 1 2

1 2

m m m m+= +


Deci 1 2

2micro + micro

micro = 1p


c


0

m mm R TV

p

ρ = =sdot

sdotmicro

1p

3pDeci 0

0

pR T

sdot microρ =

sdot1p


d


4Deci 0max

0

p TTpsdot

= 1p




96


a


31

1 11

2 10 Pa

= 4 dm

9627 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

4


32

2 22

4 10 Pa

= 4 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p


33

3 33

4 10 Pa

= 8 dm

38508 K

p

Vp VT

R

= sdot

sdot= cong

ν sdot

1p


34

4 44

2 10 Pa

= 8 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

b


4

unde 34 V 4 3 4 35( ) ( )2


şi 41 P 1 4 V 1 4 1 47( ) ( ) ( ) ( )2



c




d


primit

LQ

η = 1p


Deci 12341

12341 cedat

LL Q

η =+ 1p




97





a


1 1 1

Pr r r= + 1p

3pObţinem 1 2

1 2P

r rrr r

sdot=

+1p


b


1 2P

E r E rEr r

sdot + sdot=

+1p


P

S P

EIR r

=+

1p


Rezultă 05 AI = 1p

c


2

P

P

EIR r

prime =+

3p4


d


4Deci 2 22

2

E I RIr

primeminus sdot= 1p




a






98

b


4

unde S

S

EIR r

=+

1p



RP




S

RR r

η =+

2p3p

Rezultă 50 η = 1p

d



P

EIR r

prime =+ 1p





99

DOPTICAtilde




a


1 1 1x x f

minus = 2p

4Obţinem 1 21

1 2

x xfx x

sdot=

minus1p



1

1Cf

= 2p3p


c


1 1 1 x x f

minus =prime 1p

4

cu 1 2

1 1 1 f f f

+ = 1p

obţinem 12

1

f fff f

sdot=

minus sau 2 2

22 2

x xfx x

primesdot=

primeminus 1p


d


1 1

y xy xprime prime

β = = 1p

4


1 1

y xy x

β = = 1p

Deci 2 2

2 2

y xy xprime prime

= 1p

Rezultă 2

2

12yyprime

= 1p



100


a


3Obţinem xiN

= 1p


b


Dil

λ sdot= 2p

4Obţinem

2l iD

sdotλ = 1p


c


2Dx

lλ sdot

=1p


42

Dxl

λ sdot= 1p



2Dx x x

lλ sdot

∆ = minus =1p


d


2Di

lλ sdot

= 1p

4unde 1

1 cn n

λλ = sdot =

ν1p

Deci 1 12

D inl i i

λ sdot= =

sdot1p




101

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj


2gt hh t

g∆

∆ = rArr = = 3p

2 b pF F t pt

∆= rArr ∆ = ∆

∆

3p

3


L = 22 J 3p



α 3p



b



2p

3p

2 2

1 1

G T m aT G m a

minus =minus = 1 2

g mam m

∆rArr =

+ 2p

c



+ = = 1p


2 1

08 kg01 kg

m mm m

+ =minus =

2p



2p

4p

1 2( ) 126 kg ms= + = sdotp m m v 2p



102




2p3p


b

GL mg h= ∆

1p


2

2atx = 1p


45 JGL = 1p

c



4p

21 2 2c f

h xE mg F= minus

1p

21 (sin 03)

2cmgxE = α minus

1p

1 12 JcE = 1p

d

22 2

1 12c

cc

p p EEm p E

= rArr =

1p

4p



2 24 JcE =

1p

2

1

2pp

=

1p



103



1 2

1 2

m m+micro =

ν + ν

1 1 2 21 2

1 2

A A

N Nm mN N

N N

micro micro= =

=

1 2

2micro + micro

micro = b

3p


1

1 1 ii i f f f i

f

VTV T V T TV

γminus

γminus γminus

= rArr =

1

1iV i

f

VL C TV



c

3p

41 1 2 2

1 32 2 5

1 1

3 3

p V p V

V VV V

γ γ

γ

=

= rArr = b

3p

52 2

1 1

22

1

1 1

065 260 K

Q TQ T

T TT

η = minus = minus

rArr = rArr =

b3p



a

1 2

1 1 2 2( )

m m m

m p V p VRT

= +micro

= +3p

4p


b


4pRezultă 1 1 2 2

1 2

p V p VpV V

+=

+2p

5 28 10 Nm3

p = sdot

1p

c

mpV RT=micro 2p

4p1 21 2pV pVm m

RT RTmicro micro

= = 1p



104

d

2 22

2

m RTp V

=micro 2p

3p5 2

2 241 10 Nmp = sdot 1p



a


3p 3p

b

3 31

1


11

186 10 NmmRTpV

= = sdotmicro 1p

4p

3 312 2 12 4 10 m

2V V minusρ

ρ = rArr = = sdot

1p

5 21 12

2

93 10 Nmp VpV

= = sdot 1p

2 1

212 1

1

ln 25915 J

T Tm VL RT

V

=

= =micro

1p

c

13 450 K

2TT = = 1p

5p

11

3 323 2

3

224 10 mTV VT

γminusminus

= = sdot

1p

5 233

3

083 10 NmmRTpV

= = sdotmicro 1p


4 44 40 Kp VT

mRmicro

= = 1p

d

min

max

1CTT

η = minus 1p

3p4

1

1CTT

η = minus 1p

95Cη = 1p



105


Subiectul I Punctaj


21

2

1 A

AA

A

EI Rr R

EIR R r

= rArr = Ω+

= =+ + a

3


c3

4 22

225 W4

ER r Ir

EPr

= rArr =

= = c

3

5( 1) 10 ka V a

V

UR R n n RU

= minus = rArr = Ω


3



a

e

UIR

= 3 4 15sR R R= + = Ω 1p

4p1

1

5sp

s

R RRR R

= = Ω+ 1p

2 10e pR R R= + = Ω 1p

1I A= 1p

be

EIR r

=+ 2p

4p

2eEr RI

= minus = Ω 1p

c

2 3

2 3

25pR RR

R R= = Ω

+ 1p

4p

1 10s pR R R= + = Ω 1p

4

4

5se

s

R RR R R

= = Ω+ 1p

12 A7e

EIR r

= =+

857 VeU IR = = 1p


106

d


1p

4p1 2

2 1 4

1 26 A7

s

I I I I R I R

I I

= +=

rArr = =

2 3 4

3 2 4 3

3 46 A

14

I I I I R I R

I I

= +=

rArr = =1p

1 4AI I I = + 1p

9 A7AI = 1p



a


4p1 3 1 1 2 3( )P P r R R R R= rArr = + + 1p

( )21 2 1 1 2 3( )R R R R R R+ = + + 1p


b

2

max 84EP r

r= = Ω 1p

3pPmax = 450 W 1p

1 2

1 2

05R RR R r

+η = =

+ + 1p

c

2 Ann

n

PIU

= = 1p

4p1( ) 6 AnE I R r U I= + + rArr = 1p


4 15n

R

URI

= = Ω 1p

d

P UI= 1p


432 WP = 1p



107

DOPTICAtilde

Subiectul I Punctaj

1

2 22 1

1 1

12

1

2

53

23 3

y x y yy x

fxx ff x

fy

β = = rArr = β

= =+


3

22 5π

∆ϕ = δ = πλ c 3

3 375 nmaan

λλ = = b 3

4 206 nmexex

hc hcL eUL eU

= + rArr λ = =λ +

a 3

5 00


c 3



a


4

2( 1)C nR


( 1)2

CRnrArr minus = 1p

15n = 1p

b

2 1 2 1

1 1 1 1 1 Cx x f x x


3p12

1

15 m1

xxCx

= =+


2

1

5xx

β = = minus 1p

c



4

12( 1)

l

nCn R

= minus 1p

1 12

lnn C R=

+ 1p

163ln = 1p


108

d


4p

13 1 1 3 1

1 1 1 1 1 Cx x f x x


13

1 1

025 m1

xxC x

= = minus+ 1p

3

1

083xx

β = = 1p



a




2kk ix x iminus

= rArr = 1p


b

2Dil

λ= 1p

3p2liD =λ

1p

142 mD = 1p

c


4p

2 ( 1)klx e n kD




Pentru 33

1kx i en

λ= rArr =

minus 1p

42 me = micro 1p

d

1 2r vk kδ = λ = λ 1p

4p2 12k k= 1p

1 21 2k k= = 1p



109

TESTUL 3

A MECANICAtilde




a

1

2

AI dtgIB d

α = = 1p

4p1 0 =d v t

2

2 2atd = 2p

Rezultat final 2 tg

=sdot α

vta

15s=t 1p

b

2 0v v at at= + = 1p

3p1

21 1 135 kJ2

= =cm vE 1p

2

22 2 90 kJ2

= =cm vE 1p

c1 1= ∆ = minus

fF c cL E E 2p3p


d







a

0T F G+ + = 1p

4pcossin

T mgT F

α =α =

1p


α=

αF mg

3 N 09 N2

= congF 2p

b cosmgT =

α2p

3p



110

c


3p=AE mgh

2

2BmvE = 1p



d


4

p mv= 1p

final2mv mv= 1p


= =vv 1p



111





a

pV RT= ν 1p

4pν =

microm m

Vρ =

1p

p RTmicro = ρ 1p


b

1 2 3 1 2 3( ) ( )p V V V RT+ + = ν + ν + ν 1p

4p

1 1 2 2 3 3

1 2 3

p V p V p VpV V V

+ +=

+ + 1p

173pp = 1p


c

1 1 1 ∆ = minusU U U 1 1 1 15 5 2 2

= ν =U RT p V 1 15 2

U RT= ν 1p

4p1 1 1 1

5 2

= ν =pV RT U pV

1 1 1 15 2

= ν =pV RT U pV 1p

1 1 1 1 15 10( )2 3



d

1 1 2 2 3 3

1 2 3


ν + ν + ν p

3p1 2 34 9


micro = 1p




112


a


4p

2 1 1 1 2 13 3= =p V p V p p 2 1 1 1 2 13 3= =p V p V p p 1p

1 1 1 1 12 2 4 4L p V p V RT= = = ν

14LRT

ν =

1p


b


4p

12 1 13 2 32


12 34LU∆ = 1p


c


3p23 1 15 6 152



= sdot =Q L 1p

d

P

LQ

η = 1p



92PQ L= 1p


η = = 1p



113








c


2 2

23

U rU r

= = 1p

5pK icircnchis 1 11

1 1

=+p

r RRr R

2 22

2 2

=+p

r RRr R

11 2p p

U R RI

= = 2p


d

1

1G

p

U IR

= 2p

4p1 11

1 1

12 k= = Ω+p

r RRr R 1p




a

2

b

UPR

= 2p3p


b


5p

2 A= =bPIU

1p



UnE I r

1p


c4 (4 )= + +b b aE I r R R 2p


d

2

max 4

=EP

r cacircnd R r= 2p

4p




114

DOPTICAtilde



b 2 1 1

1 1 1x x f

minus = 2p3p


c


4p2 1 2

1 1 1x x f

minus = 1p


d



1 12

1 1

x fx x f

=+ 1p





2Dil

λ= 2p

3p


b


4p2 2 4d i i i= + = 2p


c


4p

2 π∆∆ϕ =

λr 2

3sdot λ

∆ = =y lr

D

Rezultă 2 3π




II

= 1p


115

d


21

15 cm= =+

x fxx f




1

22 sdot=

minusl xl

x

2p

4p


λ= 1p




116

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1

b8

3

m 3 10 s

1 al 631 10 ua

1 an

c

dc dt

t


∆ ∆ =

3

2 d 3

3

a

m m

vF m a m

t∆

= sdot = sdot∆

m5s

v v∆ = =

= 25 NF

3

4

c

fR N F= +




θ = = micro

3

5


2

22 2

2 42

m v F dv v

m v F d


3



a

1

1

0const2m

d va vt

+= rArr = =

∆ 2p

4p11

1

2 100 m sdv vt


∆ 2p

b

22

2

50 sdv tt

= rArr ∆ =∆ 2p

4p

1 2 110 st t t t∆ = ∆ + ∆ rArr ∆ = 2p


d11 2 727 m sm m

d dv vt

minus+= rArr = sdot

∆ 4p 4p



117



b


0

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

2p

4p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

1p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p 1p

c


1 1 231 msin

hd d= rArr =α

1p


( )

2

2 2

2

2 2

0 sin - cos 2

237 m2 sin cos

ABAB R

m vEc L m g d m g d

vd dg



2p

1 2 868 mD L d d D= + + rArr = 1p

d


4p


2 2 1p

2 = 09 m 1p

pp

Ek xE xk

x


1p2

= 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr =

1p



118



11

1 1 2

2 2 12

2

1 8

A

A

m NN N

m N NN


micro

3

2

d

0 V

Q U LL U C T

Q= ∆ +



= 12465 JL

3

3

c

12 12

12 12

1213 13 13

13

12 13

31 2 32

5 61 3 80 1202 5

V

V

RC C RQ R T

Q C TQQ C T R T KQ

T T


3


3

5


2

1

1C

p

LQ

TT

η =

η = minus


3



119



3p 3p

b51

1 1 1 1 15 10 PaR Tp V R T p pV


c


1p

5p

0const 0 0 1i f

QV L U U U pQ U L

= = rArr = rArr ∆ = rArr == ∆ +

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p 1p

d ( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

2p

4p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

1p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p 1p



a

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

1p

4p


1

2 3 3 23

53 51 1 1 1

1

2 = 8 = 15

5 3

p

v

p V V pVp V V pVC

C

γγ

= rArr = rArr =

γ = =

1p

13 1 3

1

15 15 374 lR TV V Vp



120

b

2p

4p

( )12 2 1

12

37395 J

VQ C T TQ


=2p

c

1231 12 23 31 1pL L L L= + +

12 0 L =1p

4p

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

( )

( )

23 23 3 2

23 1 1 13 315 22 4

vL U C T T

L R T T R T



1p

23 186975 J 1pL =


31 12465 J 1pL = minus

1231 62325 JL =1p

d


2 3 2 3

2 3 2 3

A A A

B B B

Q Q QQ Q Q

= += +

32 2

2

lnAVQ R TV

= ν sdot sdot sdot


32 2 2

2 3

ln ln BB

V VQ R T R TV V





( )2 3 23

ln 0BVR T R T TV


( )2 3 2 2 3 23

1

2 3 2 3

4ln ln 3

42 ln 15 2 0 3

B

B A

VR T R T T R T T TV

R T

Q Q



rArr gt

3p 3p



121


Subiectul I Punctaj

1

b

1

2

21

2

156

lRl LLR

l lR lR L

= ρ sdotsdot

= ρ sdotsdot

= =

3

2

a U R I= sdot

2

lRmS R

m m d Sd ll S d S


sdot sdot

2 mU Id S

= ρ sdot sdotsdot

= 3 VU

3

3


2 WRI t

=sdot ∆

1 R = Ω

3

4



02 AI =

3

5

b2

max 4

p

p

EP

r=

sdot

1 2

1 2

6 12 5p pr rr r

r rsdot

= rArr = Ω = Ω+ 1 2

1 2

26 Vpp

p

E E E Er r r

= + rArr =

max 14083 WP =

3



a

1

3 1 1

075 A

EI THORN IR R r

= =+ + 1p

3p1

1

1

45 C 1p

R

R

qI p

tq

=∆

=

1p1

1

1

45 C 1p

R

R

qI p

tq

=∆

= 1p

b

2

3 2 2

06 A

EI IR R r


4p2 2 1

84 1U E I r pU V p

prime= minus sdot=

1p2 2 1

84 1U E I r pU V p



122

c


( ) ( )( )

1 1 2 2

1 1 2 2

90 19p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1p

4p1 2

1 1 2 2

18 V 19

pp

p

E E E Er R r R r


3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= =

1p3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= = 1p

d



( )3

3

2 1

1 1 1 1

1

2

115 106

R

R

I I I

E I R r I

I AI A

+ =

= sdot + +

==

2p

4p

1 1 2 2 1 1883 1

AB

AB

U I R I R pU V p


1p1 1 2 2 1

1883 1AB

AB

U I R I R pU V p




a


1 2 1 2

085 A

E EI IR R r r

minus= rArr =

+ + +3p 3p

b


3 2 2

2 A 1p

EI IR R r


1p




123

c

[ ] 015 mint isin ( ) 1 1 1

685 V

AB

AB

U E I R rU

= minus sdot +

=1p

6p


1p


( ) ( )( )

1 1 2 2

1 1 2 2

079

p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1 2

1 1 2 2

7 V

pp

p

E E E Er R r R r

= + rArr =+ +

3

25 A 1p

p

p

EI I


+

1p


2p

d 3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


=

2p3p3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


= 1p



124

DOPTICAtilde

Subiectul I Punctaj

1



02

2

sinsin n irnsdot

=

2sin 04r =


3

2


1 0

1 1ln gf n

= minus sdot

2

1 1l

a

n gf n

= minus sdot


( )2

1

14a l

l a

n nff n n

sdot minus= =

minus

2 1 4f f= sdot

3

3 c 3

4

d


1

22

2

2

2

extr

extr

m vL

m vL

sdotε = +

sdotε = +


3

5

c

2

11 2

2k

Dx kl

λ sdot= sdot sdot

sdot 2

2 2k

Dx kl

λ sdot= sdot

sdot

22 1 k k

d x x= minus

( )1 2 2 - 02k Dd

lsdot


3




b


4p



0 01 1

1

sin 2 cmsin

r rr d dd r

= rArr = rArr =

1p



125

c

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot

3p4p

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot 1p

d


4p





2n = = 1p




1

13 1 3 3

2Dx ilλ


sdot 2p

3px3λ1

= 36 mm 1p

b

x4λ2

= 36 mm 1p

4p

x4λ2

=i2 1p


12

2

3 4

450 nm 1p

sdot λλ =

λ =1p

c

3

1

3 1

34

13 3

4 3

4 2

3 360 nm2

DxlDxl

x x

λ

λ

λ λ


sdot =

3p4p


d1 1

14 4 4 6 mm

08 2Dx x

lλ λ


4p 4p



126

TESTUL 5

A MECANICAtilde




a


0N Gn Gt Ff+ + + =

1p


2p

1p

b




1p




c


N T T= + 1p

4p( ) 306BT m a g N= + = 1p

( )( )2 22 1 cos 90N T= + minus α 1p

5911N N= 1p

d


3p( )2

2B B


072m sBv 1p



127


a




b




2B



c


2

cc

mvEc p mv= = 2p


2 2C B

Ffmv mv Lminus = 1p

Rezultă

6 m sCv = 12 m sCp = 36 JCEc = 1p

d


max

2 2Cmv kx

=


4p

max cmx vk

= 1p




128






HemNamicro

= 2p3p


b



c


1p

4p


1p

Obţinem

1 1 1

2 2 2 1 2

1 2 1 2


ν= ν =gt ν = ν =

ν + ν = ν + ν


d





a


1 2

p pT T

= 1p


3 4

p pT T

= 1p


T2 = T4 = 1 3 360KTT = 1p

b

LTOT = L12 + L23 + L34 + L41 1p



129

c

TOT

p

LQ

η = 1p



d

min

max

1 TT

η = minus 1p

3p1

3

1 TT

η = minus 1p

30η = 1p



130





a


1 2 3

1 2 3

1 2 3

36V1 1 1

E E Er r rEp

r r r

+ += =

+ +

1p

3p

1 2 3

1 1 1 1 1pp

rr r r r

= + + =gt = Ω 1p


p

EI

R r= =

+A 1p

b


1 2

1 1 1

2 2 2


= +


2p

5p


11 2 1 2

E r E E RI


=sdot + sdot + sdot

2 1 2 12

1 2 1 2


+ minus=

+ + 2p


c

Deoarece

ABU I R= sdot

1 2I I I= + 1p

4p1 2

1 2

1 2 3

1 1 1AB

E Er rU

r r r

+=

+ +2p

UAB = 40 V 1p

d

Icircn relaţia

1 2

1 2

1 2

1 1 1AB

E Er rU

r r R

+=

+ +1p

3p


vRrarr 1p




131


a


4p



sdot=

sdot


U tminus sdot

=sdot

2p

5AQIU t

= =sdot

1p

b


4pRezultă 2pQRI t

= 1p

Rp = 22 Ω 1p


3pE = 120 V 1p

d


x p

x p

R Rr

R R=

+

2p

4p

px

p

R rR

R r=

minus 1p



132

DOPTICAtilde




a



2 smv e U= sdot 1p

4p11

22

S ex

S ex

ch eU L

ch eU L

= + λ = + λ

2p


b


11

S exch eU L= +λ

2p

4p

11



c


4p0exLh

υ = 1p


d2

2

cW Nh Nh= υ =λ 2p

3p




a1 2

11 11e

m

fnn R R

=

minus minus

2p3p


b


2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2p


minus x1 = 80 cm 1p


133

c


1 22 1

1 1 1d x xx x f

= minus + minus =

1 22 1

1 1 1d x xx x f



11

1y xy x

β = = = minusminus

2p


d


1 2

2

1

2 1

1 1 1

2

1 1 1

F f fxx

x x F

= +

minusβ = =

minus

minus =

2p4p



134

TESTUL 6

A MECANICAtilde




atg 1

3tg ϕ = micro = 2p

3p

30ϕ = 1p

b

t fG Fa

mminus

= 1p

4psin cosmg mga

mα minus micro α

= 1p


2310 =577ms3

a = 1p

c



1p

4pt fG F F= + 1p


310 3 10 100 N3

F = sdot = 1p

d



1p

4p

t fG F F+ = 1p

sintG mg= α 1p

3 1 1 100 3 100 3 200 N2 23




135


a

sin 5

1000m54 =15ms3600s

hpl

v

= = α =

= sdot1p


cosUcirc α


rArr sintg cos


α1p



4 352 sin 2 10 15 15 10 W=15kW100


b


4psincos


α1p

7500W=75kWP 1p


2p3p



3p25kJpE E∆ = ∆ = 1p



136





a

1

1 1 2 1 21

22 1 2 2 1

2

2 540 2 54 4 083 450 3 45 5

pVV T V TRT

pV T V V TRT


ν2p

3p

1

2

4 085

ν= =

ν 1p

b

1

1 1

22 2

p VVRT

p V VRT

ν= =

ν 2p

3p

1

2

08VV

= 1p

c

1 1 11 2 1 1 2 2

2 2 2

( ) ( )pV RT

p V V R T TpV RT

= ν rArr + = ν + ν= ν

1p

5p

1 1

1 2 1 22 2

( ) ( )

p V RTp V V RT

p V RT

= ν rArr + = ν + ν= ν

1p

1 1 2 2 2 2

1 2 2

5 08 450 540 ( ) 3 18

p T Tp T T


ν + ν sdot ν 2p


d

1 36lV =

12

2

2 3 36 3 18 54l3 2

V VV


1p

4p 1 2 1 2

1

1 22

9

08 08

V V V V lV V VV

+ = + =

= rArr =2p




137


a 4p 4p

b

212 1

1

41 1 4 1 4

ln 0

5( - ) ( ) 02V

VQ RTV

RQ C T T T T

= ν gt


4p

12 41 229356JprimitQ Q Q= + = 2p

c

23 4 1 4 1

134 4

2

5( ) ( ) 02

ln 0

VRQ C T T T T

VQ RTV


= ν = lt2p

4p


d1 cedat

primit

QQ

η = minus 2p3p

3043η = 1p



138





a

2AP R I= sdot 2p

3p25 WP = 1p

b

0

0 0

0

A

A

A

I I IxI R I Rl

l xE I R I Rl

= + = minus

= +

3p5p

05 mx = 2p

c

ee

E EI RR I

= rArr = 1p

4p0

00

05A

A

A

I I IR lI Ix R

= +sdot

= =sdot

2p

151 151eI A R= = = Ω 1p


3p2250Wh 225 kWhW = = 1p



139


a


3p200 4A50

totaltotal

PP U I IU


b



5p



1p

21

1 1 1 2

22

2 2 2 2

40 102 4

2

60 152 4

2

I PP R RI

I PP R RI

= rArr = = = Ω

= rArr = = = Ω

2p

10VABU = 1p



d

1

1 01 1 3

0

191 1800 C

5 10

RR Rt tR minus


α sdot

2p

4p

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot= t2 =

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot

2p



140

DOPTICAtilde




a

1 2


n ff R R


2p

4p

1 2


n ff n R R


2p

b

2 1

1 1 1

aerx x f= + 1p

4p2

24 cm 48cm5


2

1

xx

β = 1p

04β = 1p

c

2 1

1 1 1

apax x f= + 1p

4p

2

96 cm 192cm5


21

xx

β = 1p

04β = 1p

d2p

3p




141


a

412 10 m 12mm2Dil


3p


b

λapăapatilde nλ

λ = 1p

3piapă

12 3 09mm4apatilde

iin

= = sdot = 1p


c



2p




2p

10nouk = 1p

d






142

TESTUL 7

A MECANICAtilde


3


md dvt t t

= =+

Cum 112 6

d dtv v

= = şi 222 2

d dtv v

= =

6 154

6 2

md vv vd d

v v

rArr = = =+

a

3

4


∆= = minus

∆






Icircnlocuind 35

rArr micro =c

3

5


11 2

2mvmgh v gh= rArr =




∆ b

3



a


3p 3p

b


4p



Randamentul este 2

u

c

L mghL Fl

η = = 1p


l= =

η1p


143

c


0fG T N F+ + + =


2p


lα =

Rezultă 2fhF F mgl

= minus

2p

600 NfF = 1p

dcos

cosf

f f

FF N F mg


α 2p3p

065rArr micro = 1p



a


cE L∆ = 1p

4p



1p

Ff = μN = μmg

Rezultă 2 2

200 2 21 ms 458ms


2p

b



1p


1

2 2mv mv mgR= + 1p


1 11 ms 331msv = = 1p

c



max max2 2mv vmgh h

g= rArr =

2p3p

max 105msh = 1p

d


max 081 m2vh fh fg

= = = 1p



222 2


2p

2 204 v m s= 1p



144




1 2 2 1 1 21 2 2 3 2 3 2

2 3

( )( ) 150 J ( ) 200J 0752

p p V V LL L p V VL

rarrrarr rarr

rarr


c

3

5


max

1CTT

η = minus


1 max 1

1L T LQ T Q

η = rArr minus =


3 150 K8

TT = =d

3



a


1 1

0 1

V VT T

= şi

2 2

0 2

V VT T

= 1p

4p

dar 1 2V V V+ = şi 2 1 12


23VV = iar

1 2 2VV V= = 1p


32TT = şi 0

234TT = 1p

Rezultă 1 4095KT = şi 2 20475KT = 1p

b


4p

1 1 ( )

2 2l lV S V x S= = +


2 2 ( )2 2l lV S V x S= = minus 1p

Obţinem 50 00 1 1 2

500 2 2 0 2

2 N( ) 066 102 2 2 3 m

N( ) 2 2 102 2 2 m





1p

51 2

52 2

N066 10m

N2 10m

p

p

= sdot

= sdot1p


536 NF = 1p

d


mp V RT=micro

şi 2 2 2 20 2 2

1 1 1

m V m Tp V RTV m T

= rArr =micro 2p

5p

Dar 0 01 2

3 32 4T TT T= = şi 1 2

23 3V VV V= =

1p

2 2

1 1

2m Vm V

rArr = 1p

2

1

2mm

rArr = 1p



145


a


1p

3p


2p

b


1 2 1 2


= rArr = rArr = 1p


2 3

p pT T

=


31 13 3

p p ppT T

rArr = rArr =

2p

5 23 10 Nmp = 1p

c


4p1 1

32 32



1 2 2700JU rarr∆ = 1p

d

1

LQ

η =



1p



44 1 4 1 3

p p p p TTT T T T

= hArr = rArr =


2p


1 1

4 2 02218 9

p Vp V

η = = = 22rArr η = 1p



146



41 2sR R= 1 1 1 2

2 3pp

RRR R R

= + rArr = 22 53 3sR RR R= + =

1 1 23AB

RR R

= +


b

3



a

1 2R R R R= + = 1p

4p

22 1

1 1 1 3 22 3

RRR R R R

= + = rArr = 1p

3 22 53 3R RR R R R= + = + = 1p

3

1 1 1 3 1 85 5

5 6258

AB

AB

R R R R R RRR

= + = + =

= = Ω1p

b


UU IR IR

= rArr = = 2p


( ) 264ABAB

EI E I R r VR r

= rArr = + =+

2p

c




AB

EIR r

= =+

1p





d

2

max 4EP

r= 2p

3p

max 8712WP = 1p



147


a


2 4 1

3 4 5

00


minus minus =

2p


1 3 1 1 3 1 2 2

2 3 2 4 4 2 2

2 3 2 5 5



3p



4p19MNU V= minus 1p

c

24 4 4P I R= 2p

3p4 8P W= 1p

d2

1 1 1W I R t= 2p3p

1 10800 J 108 kJW = = 1p



148

DOPTICAtilde


680nma a aa

nnλ

λ = rArr λ = λ =

d 3

2 c 3

3



2MO PO AOMOAO PO

= rArr =

2ON PO O BONO B PO

= rArr =


=09 m2 2


b

3

4


k Dxl

λ=


2

1 1 mm222 2 mm2

Dk xl

Dk xl

λ= rArr = =

λ= rArr = =

c3

5


2 1 1 2

1 1 1 x xfx x f x x

minus = rArr =minus


1

x x xx

β = rArr = β 1 1

11 1

x xf βrArr = =

minus β minusβ


20 10 cm1 1

f minusrArr = =

minus minusb

3



a



1 2

1 1 1 1( 1)( ) n nC n RR R R C


R = 10 cm 1p

b

Din 1 1 20C f cmf C

= rArr = = cm 1p


2 1 1

1 1 1 fxxx x f f x

minus = rArr =+

2p


x minus= =

minus2p


149

c1 2d x x= minus +

2p3p


d


1 1

y xy x

β = = 1p

3p22 1

1

xy yx

= 1p

2 2 cmy = minus




a


1 11 1

2 12 2

( )(1 )

c ch L eU h L eU

c ch L eU h L eU f


2p

5p

11

1

2

08

ch LeU

c eUh L

minusλ

=minus

λ

1 2

1 2

0802


λ λ2p



LL hh


1410 HzL = 1p


1

ch hε = ν =λ

2p3p

191 011 10 J 006eVminusε = sdot = 1p

d


11

22

2

2

2

e

e

m v eU

m v eU

=

=

2p

4p1 1 1

2 2 108v U Uv U U

= = 1p

1

2

54

vv

rArr = 1p



150

TESTUL 8

A MECANICAtilde




a


4p2

1 2atx = 2 0x v t= 2p


= = 1p

b0v v at= + 2p

3p16msv = 1p

c


3prezultat final

-=fracircnare

fracircnare

va


d

1 2= +d d d 2p

5p1 0=d v t 1 1 2= = = 64d x x m 64 m 2p


2 2 fracircnare

vdaminus

= 2 = 32d m




aE mgh= 2p


b

c totalE L∆ = 1p

4p

2

2cmvE∆ = = +

ftotal G FL L L 1p



1p



151

c

c totalE L∆ =

1p

4p 0cE∆ = GL mgh=

f


rezultat final = -f

totalFL mgh = -4500

f

totalFL J 1p

d

= +f f f


4p2f


F oprire

F FL d

+= minus 1p







152





a

pV RT= ν 1p

4p1 1 1 2

2 2 2 1

p mp m

ν micro= = sdot

ν micro2p

rezultat final 1

2

74

pp

= 1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0


4pVf = 2 V 1p


d




4p 4p



a1 1

1 1 11

p Vp V RTRT

= ν rArr ν = 2p3p


b


4p

1 2

1 2 12 1

2 1

34

V

p pV ctT T U C T

T T


=2p




153

c

min

max

1 TT

η = minus 1p

4pmin 1=T T 1p

max 2 1= = 4T T T 1p


η = = 1p

d

P

LQ

η = 1p

4p


1


1-2 2-3 3-1L L L L= + + 1-2 0L = 32-3 2

1

ln VL RTV

= ν 3-1 1 3( )L R T T= ν minus 1p


minusη =

+ 25η 1p



154





a


( )1 1 1 3E I R R= + de unde 1 3AI =2p


b



2422

131 )()( 1p


c


2p4p


d


2p3p




a



bP1 + P2 = Rp I

2 2p3p


c

echivalent

echivalent

RR r

η =+ 2p



d


4p


EIR r

=+

1p




155

DOPTICAtilde




b

2 1

1 1 1x x f

minus = 1p

4p2

1

xx

β = 1p


c

21 CCCsistem += 1p


2 1

1 1sistemC

x xminus = 1p


d

1 2= -x d x 2p

4p

1 22

1 2

x fxx f

=+

1p




a


3p11 10

yi = 1p


b


2i 1p

4p1

12 22id i= + rArr d = 3i1

2p



156

c

22 29

2iy i= + 2

22 072 mm19yi = = 1p


1 2Dil

λ= 2

2 2Dil

λ=

1 22

1

ii

λλ = 1p


d


iin

= 2p


1 1

1

-025 i ii

= 1p


n = 1p



157

TESTUL 9

A MECANICAtilde


5

21 1 2 2

12 1 2

2

2 2m 1156s

m v m v

mv v vm

=

= =3



a 3p 3p

b

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

4p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

c2 2

08 NT m g m aT

= minus=

2p3p2 2

08 NT m g m aT

= minus= 1p


158

d

1 0 1 1 0

1 1 0

1 1 0

( ) 0

( )g 0 ( )


+ + + + + =

minus + == +

1p

5p


1p

2 2

2

2

0 0

m G TOy G Tm g T

+ =minus =

=

1p

2 1 0

2 10

( )m g m m gm mm


=micro

1p

m0 = 03 kg 1p



aEA = m1gh 2p

3pEA = 1 J 1p

b

ndashf AB

f AB

B A F

B A F

E E L

E E L

=

= +2p

4p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p

c


4p

( )( )

1 1 1 2 1

1 1 1 2

1 1

1 2

m v m m v

m v m m vm vv

m m

= +

= +

=+

1p

unde 1m25s

v = 1p

01 255 m08503 s

v sdot= = 1p

d


( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

2p

4p( ) 22

1 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

1p

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

= 1p



159



VU VC T∆ = ∆ b 3


3pRT

microρ =

[ ] 3

kgmSi

ρ = c

3

41 1

1 1

33 22

L p VLU p VU

=

∆ = rArr =∆

a

3

5 36 gA A

m N Nm mN N


a 3



a15 mol

mν =

microν =

2p3p

15 mol

mν =

microν = 1p

b

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

2p

4p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

1p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ = 1p

c

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

3p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot 1p

d

2 2

3 3

mp V RT

m mp V RT

=micro

minus ∆=

micro

1p

5p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot 1p



160


a

L1231=Aria1231 1p

3p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot

1p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot 1p

b

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

5p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot05p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot 05p

c

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p

3p

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot 1p

d

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= =

1p

4p

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= = 1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η =

1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η = 1p



161




d 3

2 2W RI t= ∆ b 3

3 [ ] 2 1SI

V1 11 N m A sA


4

( )

( )

Aria trapez2

200 300 2500 500 mC

2

b B hQ

Q Q

+ sdot= =

+ sdot= = rArr = a

3

5

( )0

2

0

0

1

288

2110 C

R R t

UR RP

R Rt tR

= + α

= rArr = Ω

minus= rArr = deg

α


3



a


3p

1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω

1p1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω 1p

b

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

4p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω 1p

c

Potrivit legii Ohm

22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

2p

4p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

1p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

= 1p


162

d

1p


( ) ( )2 4

2 1 2 4 3 4

2 4

I I II R R I R R

I I

= +

+ = +

rArr =

1p( ) ( )

2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr =

1p

( ) ( )2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr = 1p



a

1 2

1 2

1 2

1 1 1

07 A

E Er rI

RR r r

I

+=

+ +

=

3p

4p

1

2

2E EE E

==

1

2

2r rr r

==

1p

bU = R I 3p

4pU = 7 V 1p

cP = RI2 3p

4pP = 49 W 1p

d88

e

RR r

η =+

η =

2p3p

88e

RR r

η =+

η = 1p



163

DOPTICAtilde

Subiectul I Punctaj

12

2 smv eU= rArr

2

2SI

2 ms

S

e

eUm

=

d 3

2 2

1

sinsin

i nr n

=

a 3


4 0 00

275 nmextext

hc hcLL


b 3




a

1

1 01 m10

fC

f

=

= =

2p

3p

1

1 01 m10

fC

f

=

= = 1p

b

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

4p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr = 1p

c

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

4p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


d

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

2p

4p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

1p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus1p



164


a

Din grafic15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot

1p2p15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot 1p

b

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

5p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot1p

c


4p

rArr

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot1p

d

( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p

4p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot= 1p



165

TESTUL 10

A MECANICAtilde




a



2

7 msdvt



1

2 msvv a t at

= rArr = =

1p

4p


2 11 1 1 1

1

2 9m2vv a d da

= rArr = =


1 3 3 21 3 3

1 3 31 3 3

22

0

final

final

final

v v a dv a d

v v a tv a t

v

= + = minus= + rArr

= minus= 21

33

21

33

15ms

12m2

vat

vda


1p



b


4 1 3 41 3 4

4 1 4 113 3 3

3


t


3p4p


c



1p


1p


1p



166

d


2 20G N N G+ = rArr = 1p


2p


1p



a


2

2 20

45 m2

2 2

gth

v v g h v g h

∆ = =

= + ∆ rArr = ∆

2p

4p2

2cmvE mg h= = ∆

1p


b


( )c

p


∆ ∆= =

minus ∆ minus ∆

2p

3p

rezultat final 016c

p

EE

= 1p

c


2

22

c

c

L E Fd

mvE

v gh

= ∆ =

∆ =

=

2p

4p

2

2 2mv mghF

d= = 1p


d


2

( 1)

2

f i

i f i

f


mvE fmgh


2p

4p

( 1)( 1) ( 1)



1p




167





a


mpV RTm= ν


micro=

2p3p



A

N N pVpV RT NN RT

= rArr = 2p3p


c


1

p VmRT

micro= respectiv 2

22

p VmRT

micro= 2p


2 12 1

V p pm m mR T T


1p


d


2p



2 3 2






a


primit primit

QLQ Q

η = = minus 2p



1p


b


1 1

1 T TT T

∆η = minus = 1

TT ∆=

η1p


η1p


2 300 KT = p


168


2 22

m m RTp V RT Vp

= rArr =micro micro

2p3p


d


4p


2 3 2

p p pT TT T p

= rArr = 1p


2

1V Vm pQ C T

p

= minus micro

1p




169





a


= = Ω 1p


232 3

R RRR R

=+ 1p


2 3

R RR R R RR R

= + = ++

1p


b




s

EIR r

= =+

1p



c


2 2 3 3

I I IR I R I

= +=

2p3p


1p

d


1 3R R R= +1p


1 2

093As

e

EIR R r

= =+ +

1p


3 3U R I= 1p




170


a


1 1 1 11

33 3 3 3

3

125 A

05 A

PP U I IU

PP U I IU

= rArr = =

= rArr = =

2p


1p

Din legea lui Ohm R

R

URI

= 1p


b



Iminus

= + = + rArr = 1p




d




consumat

095PP

η = = 1p




171

DOPTICAtilde




a


1 20 cmfC

= =


1 1

30 cmf xxf x

= =+




1 25 cmfC



2 1

f xxf x

=+


1p


2p

b



= 1p


c


1

1346 cmFxxF x

= =+

1p


1

y xy x

β = = 1p


1

xy yx

= 1p


d


1

4xx

β = = minus 1p


2 1

1 1 1F x x

= minus 1p





172


a


1ext s

hc L eU= +λ

(1) 1p

4prespectiv 2

2ext s

hc L eU= +λ

(2) 1p


1 21 2 1 2

1 1 s ss s

e U Uhc e U U h

cminus λ λ


1p


b




se obţine 10

1

sc eUh

ν = minusλ

1p


c



4p

Respectiv 22


Se obţine 1 1

2 2

s

s

v Uv U

= 1p

rezultat final 1

2

073vv

= 1p

d


11

ext shc L eU= +λ

(1) respectiv 33

ext shc L eU= +λ

(2)


1p

3p

3 1

1 1s

hcUe

∆ = minus

λ λ 1p




173

TESTUL 11

A MECANICAtilde


c 2

2 2mkg sN s s=

kg m kg m


= adimensional 3

2 c 3

3


5h = 0cE =2

0

2m v m g hsdot

= sdot sdot 0m2 10s

v g h= sdot sdot =

2

2 05 kgc

o

Emvsdot

= =

3

4 c 3

5 d ( ) 60 kW


= = = 3



a


3p 3p



= =+ +

3p

4p

Rezultă 2

m75s

a = 1p

c


sdot=

sdot ∆


4dS πsdot

=

3p4p


d



4pRezultă 1 2





174


a


ti tfE E= 2

0

2 cf pfm v E Esdot

= +

20


= =

2p

3p


b


4pRezultă 1 1



c




h h= sdot =

1p


2 20


= + sdot sdot


2p

Rezultă m1038 kgs

p = sdot 1p

d



Rezultă 0GL = 1p



175



d 2 3

2 3


mol k

minussdot


sdot

3

2 c p vRc c= +micro

p vRc cminus =micro

3


2 8VV = 12

1

8TT

γminus= 53

p

v

cc

γ = =

2 232 3 3

2 11

8 (2 ) 4 4T T TT

= = = rArr = sdot 3

4 b 3





a



3 311

1

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro 1p

3pSe obţine 3 31

11

025 10 mm R TVp


sdotmicro1p


b


NN

ν = 1p


mν =

micro1p


mNmicro

= 1p


c


Np V R TN

sdot = sdot sdot 1p


V= 1p

Se obţine 1

1

Ap NnR T

sdot=

sdot1p



176

d


1 2

( )m R T m m R TV V



1 1 2

1 2

( )m R T m m R TV V



2p

4punde 1

11

m R TpV

sdot sdot=


1 22

2

( )m m R TpV

minus sdot sdot=


1p


1 2

( )m R T m m R TV V






a



5p



4 1

34 3 3 33 1

4ln ln 2 2 ln 2 4 ln 22


= = = =


2p


b






c


max

1 TT

η = minus 1p

4p1 1

min 1p VT TvR

= =

3 3max 3

p VT TvR

= =

1p

Obţinem 1 1

3 3

1 12 2

p V pVp V p V


Rezultă 75η = 1p

d


3p 3p



177



c SI

V RA


SI

WR RI t


3

2 d 11

lRS

ρsdot= 2

2

lRSρsdot

= 1

2

14

RR

= 3

3 b 34 d 3




a


1p

4p





c


3

13 AABUIR

= = 2 24 AI = 2p




1 2 1 2 1 2( )I R R r r E Esdot + + + = + 3p4p




a


4extPP =

22

16ER I

rsdot =

sdot1p


2

P

EIR r

= +

1p


Rezultă 12 2784 016R = Ω Ω 1p


178

b



r= =

sdot


Rezultă 15 kJW =

c



Rezultă 32 WP = 1p

d


1 1 1 11

8E r IE R I r I RI


1

1

RR r

η =+

2p3p

Rezultă 80η = 1p



179

DOPTICAtilde


a 22


2 c 3


3

i rir

= rArr = = 3

4 c 15

2 1214

1 2

1

5 10 252 10

c

cλ ν sdotν

= = = =λ ν sdot

ν

3

5 c 9

43

600 10 2 6 10 06 mm2 2 10Dil

minusminus

minus


sdot3



a


1 1 1x x f

minus = 1p

4Rezultă 1

21

20 ( 30)20 30

fxxf x

sdot minus= =

+ minus2p


b

Distanţa focală 2

1 1 2

1 1 11nf n R R

= minus minus

unde 1 1n =

( )1 1 11nf R R

= minus minus minus

2p

4p

obţinem ( )2 11 1

2n Rn

f R fminus

= rArr minus = 1p

Rezultă 20 1 152 20

n = + =sdot

1p

c


3p 3p

d


1 2 1

2( 1)

nCf n R

C nR

= = minus

= minus

2p


1 15 1151 112

C nnCn

minus minus= =

minus minus1p

Rezultă 2CC

= 1p



180



8

1 91

3 10600 10

cminus

sdotν = =

λ sdot2p

3p


b


16 152 9

2

3 10 1 10 055 10 Hz54 10 18

cminus


λ sdot 1p



2 1

( )( )

h e UhU

e


∆ = unde 152

2


λ1p


c


cL h h= ν =λ

1p

4pObţinem 17


Dar 17

190 19

0044 101 16 1016 10

eV J Lminus

minusminus

sdot= sdot rArr =

sdot1p


d


11 0 ch h Eν = ν +1p


1p





181

TESTUL 12

A MECANICAtilde




a


1 1

1 1

0

f

F T m aN m gF N


2p


2 2

2 2

0f

f

T F m aN m gF N

minus =

minus == micro

2p


1 2

( ) m125s

F g m mam m


+1p


c


1 1 2

m2s


m1s


4p






2 21 cos180


5p

Rezultă 2 2

1 0482

v vgdminus

micro = = 2p


3pp = 8000 Ns 1p

c


21

112 2

MvW =

2p3p

1 8000JW = 1p

d


4 2 2Mv kx

= 2p

4p

Rezultă 212

kN3204 mMvk

x= = 2p



182





a


1 2


1p

3p1 1 2 21 2p V p Vm m

RT RTmicro micro

= = 1p

1 1 1

2 2 2

35m pm p

micro= =

micro1p

b


1 2


1p

4p

1 2

21 2

2 2

( )

2

m m RTm RTp V

V p

+micro micro

= =micro

1p


12

2 2 1 12 1 21 2

2 1 2 2 2

2 2

1 1( )1 1 1( ) (p p )

2 22

mmp pmp m RT p

p



micro

1p


p = + =


1p

c


1 2

1 2


+micro= =

+micro micro

2p


12

2

12

2 1 2

( 1)

1 1( )

mmm

mmm

+micro =

sdot +micro micro

1p


1 2

kg12kmol

p pp p


+1p

d

0 2 rmp V RT=micro

1p

4p02r

p VmRTmicro

= 1p

0 0

1 2 1 1 2 2 1 2

2 2 2 0(6)3

rm p pm m p p p p

micro= = = =

micro + micro +2p



183


a

1 (273 27)K 300KT = + =

1p

3p


2 1

2 2 600Kp p T TT T

= = = 1p


3 2

V VT T


3 1

p pV V

=


1p

b



1p

4p


13 3 3 3 1 1 3 1 1 3 11 1( ) ( )2 2





sdot1p

c

123 12 23L L L= +



123 1 1 13 3 300JL RT p V= ν = = 1p

d

1 ced

abs

QQ

η = minus 1p

4p31 3 1( )( )



1 1

1 1 1 1

2 (4 ) 11 773 5 13(2 ) (4 2 )2 2

R T TR RT T T T



1p



184



2

b


=+

+


2

2

( )( )

E Rx R xRx rR rx

+=

+ +


=

Rezultă RrxR r

=minus

3

3

c


=





3

4

c


2EI

R r=

+


22

EI rR=

+


2

22 2122 7

rRI EI R r E

+= sdot =

+

3

5


nal 1

2nn n

E EIR r R r

= =+ +

Rezultă 174

E E=3





1

171AEIr R

= cong+

2p

b




22 ( ) 22ABx R xR x x

R Rminus

= = minus

2p

4p

22 ( ) 22ABx R xR x x

R Rminus

= = minus


2p


185

c


max 2ABRR =

2p

4pRezultă min

1 2

EI Rr R=

+ + 1p


dsc

EIr

= 2p3p

12AscI = 1p



a


R r=

+1p

4p


22( )

REP RIR r

= =+

1p


2 2

2 4

( ) 2 ( )0 0( ) ( )



1p


4mEP

r= De aici

2

14 m

ErP

= = Ω 1p

b


2p3p


c


02

0( ) 4R E Ef

R r r=

+ 2p


2

02 4(1 )

4 m

f fERP f



2p


186

d

1

2

22

2

22

RR r

RrR

η =+

η =+

2p

4p1

2

1223

η =

η =1p

2

1

43

η=

η 1p



187

DOPTICAtilde




a



1p

5p

2 1 2 1

1 1

n n n nx x R

minusminus = 1p

3 2 3 2

2 2

n n n nx x a R

minusminus =

minus 1p


2 1 1 2

n n n n n nx x R R


b


1 2 1

y n xy n x

= 1p


1 3 1

y n xy n x

= 1p



3 1 1

n x yn x y

β = = 1p

c


2 1 3 2

1 2

imnx f n n n n

R R


+

15p

3p


2 1 3 2

1 2

obnx f n n n n

R R

minusrarr infin =

minus minus+ 15p

d


3pRezultă

22( 1)Rf

n=

minus 2 2

1 1

y xy x

β = = 2 1

1 1 1x x f

minus = 2p



188


a




b

Putem scrie 2

kx rD l










2Dx a l b a l bl


c


4pRezultă

2lDi

lλ

= 1p


Rezultă 2icircDi

lλ

= 1p

d


D D fi il f l f

λ λ∆ = minus =

minus minus2p

4p


2 2D g Di gi

l l+ λ λ

∆ = minus = 2p




190

TESTUL 1

A MECANICAtilde




a


4punde 1 2 1 2( ) sin ( )t

HG m m g m m gL



= + sdot sdot minus

1p


b


5p

unde 2 2

1 2 1 2( ) cos ( )nL HG m m g m m g

Lminus




1 2( )fF L

m m g L H

sdotmicro =

+ sdot sdot minus1p


c


4p




Lminus





d


3pDeci 2 2


L+ micro sdot minus

= + = sdot sdot 1p

Rezultă 28 NF = 1p



191


a


4p

Unde 2 2

0 2 2

M MM cM pM



Obţinem 2

max 2Mvhg

= 1p


b



4 52

QQ cQ pQ cQ cQ cQ

m vE E E E E E

sdot= + = + = =

1p

4pDar M QE E= 1p


5Q Mv v= 1p


c



Deci 2

2M

rm vF m g

dsdot

= sdot + 1p


d


2p3p




192





a


3pObţinem 1 11

1

p VR T

sdotν =

sdot1p


b


4pObţinem 2 2 2

22 A

p V NR T N

sdotν = =

sdot1p

Dar 22

A

NN

ν = 1p


c




1 2


ν + ν1p


d


4p



Rezultă 2 2

1 1

05VV

prime ν= =

prime ν1p



193


a



Dar 12341 0U∆ = 1p


b






c




d


4p






194





a


PR R R R R= + + + 1p

3p



b



P S

EIR r

=+

1p


Rezultă 02 AI = 1p

c




Rezultă 4 Vu = 1p

d


S

EIr

= 2p

4pObţinem SC

EIr

= 1p




195


a


EIR r

=+

1p

4pUnde 12 34

12 34BD

R RRR R

sdot=

+ 1p

Iar 12 1 2R R R= + 34 3 4R R R= + 1p

Rezultă 1 AI = 1p

b


5p









EIr

= 2p3p




196

DOPTICAtilde




a



= 1p


b


1 1 1x x f

minus = 1p

3pObţinem 1

21

x fxx f

sdot=

+1p


c


1 1 1x x f

minus =prime prime

1p


1

12

xxprime


1p


β1p


d


1 1 1x x f


1p


1

xxprimeprime


1p

Obţinem 1

fx f


1p




a


Dil

λ sdot= 1p

3pObţinem

2l iD

sdotλ = 1p



197

b


1p




∆α = = 1p


c


D iil

λ sdot= = 1p

4pDeci 1 2

DD = 1p


Rezultă 1 md = 1p

d


Dil

λ sdot= 1p

4pUnde 2

1 cn n

λλ = sdot =

ν1p

Deci 12 2 2

D iil n n

λ sdot= =

sdot1p




198

TESTUL 2

A MECANICAtilde



∆ SIpF Ft 3

2 c 15 mgiT G F= + = 3

3 a 2

0max2 2 10 m 0

2vd h r

g= = = ∆ = 3

4

b 2 2

5 45 4

4 5

452 2

4 s 5 s

at atx x x a

t t


= =22 msarArr =

2 210 9

10 9

9 10

19 m2 2

9 s 10 s

at atx x x x

t t


= =

3

5


3



a


1 1

2

0

0f

f

F T FT F

minus minus =

minus =

2p

4p

1 1 1 2( )F m m g= micro + 1p

2 2 kgm = 1p

b


1p


2 025micro = 1p


199

c


3 1

2

0

0e f

e f

F F FF F

minus minus =

minus =

2p

4p

3 2 1 2( )F m m g= micro + 1p

3 15 NF = 1p

d


4 1 1

2 2

e f

e f

F F F m aF F m a

minus minus =

minus =

1p

4p24 2 1 2

1 2


minus micro += =

+ 1p

eF k l= ∆ 1p

2 2( ) 1333 Nmm a gkl

+ micro= =

∆1p



a

00 50 mspv

m= =

1p

3p0u

vtg

= 1p

5 sut = 1p

b


4p0

32 2pA pA

cA c

E EE E= rArr = 1p

20

3vh

g= 1p

833 mh = 1p

c

0 3c cAE E= 1p

4p0

3vvrArr = 1p

29 msv = 1p


d

20

max 125 m2vh

g= = 1p


0 104 mc

f

Ehmg F

= =+ 1p

max 12hh

= 1p



200



1

2


= ν ∆ = ν ∆

∆rArr = γ

∆

3

2c

11

1 1 ii i f f f i

f

TTV T V V VT


= rArr =

44

if f i

VV = rArr ρ = ρ

3

3


2 2

V pV p

=


2 1 2

V p TV p T

=


2 2

3p Tp T

= =

3

4c

11

1 21 1 2 2

2 1

03V pp V p VV p

γγ γ γ

= rArr = =

2 1

1 2

N n VnV n V

= rArr =

3

5b 2 2

1 1

22 1

1

1 1

12 kJ

Q TQ T

TQ QT

η = minus = minus

rArr = =

3



a

1 2

1 2

m m+micro =

ν + ν 1p

4p

1 21 2

1 2

pV pVm mRT RT

micro micro= = 1p

1 2 2 1

1 2

T TT T


+ 1p

165 gmolmicro = 1p

b

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν2p

4p1 2

2 1

83 5

TTTT T

=+ 1p

2844 KT = 1p

c

pRT

microρ = 2p

3p30699 kgmρ = 1p


201

d

1 21 2

22 fpp pT T T

ν + ν = ν rArr + =

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν

2p

4p1 2

2 1

116 5

TTTT T

=+ 1p

5 22 1

2 1

11 (2 ) 153 10 Nm2(6 5 )f

p T TpT T

+= = sdot

+ 1p



a

2 3T T= 1p

4p

11 1 1

1 1 2 2 2 12

VTV T V T TV

γminus


1p

1 11 300 Kp VT

R=

ν 1p

2 14 1200 KT T= = 1p

b

323 2

2

ln VL RTV

= ν 1p


2 3

V pV p

= 1p

22 1

2

32RTp pV

ν= = 1p

23 2441 JL = 1p

c

31 31 31Q L U= + ∆ 1p

4p

1 3 1 331

( )( )2


83

V V= 1p

31 1 3( )VU C T T∆ = ν minus 1p

31 1314 JQ = minus 1p

d

min

max

1CTT

η = minus 1p

3p1

2

1CTT

η = minus 1p

75Cη = 1p



202





4 a 2

1 2max

1 2

( ) 4 W4( )

+= =

+E EP

r r 3

5 c 00

0 0

90 mA(1 ) (1 )

U U II IR R t t

= = = =+ α + α

3



a


e e

EIR r

=+

1p

4p1 3 1 3e eE E E r r r= + = + 1p

2 31

2 3e

R RR RR R

= ++

1p

1 AI = 1p

b

1 1 1U I R= 1p

3p1

1

e

e

EIR r

=+ 1p

1 154 VU = 1p

c

2V pU I R=

1p

4p

2 1

e

p e

EIR R r

=+ + 1p

2 3

1 1 1 1

p VR R R R= + + 1p

11 VVU = 1p

d


e e

E IR r

=+

1p

4p

1 23 3 3

1 2


r r= + = + = +

+ 1p

1 2

1 2

1 2

8 V1 1 3p

E Er rE

r r

minus= =

+1p

061 AI = 1p



203


a

22 2

2

PRI

=

2p3p

2 10R = Ω 1p

b


5p

1 21 3

1 2e

R RR RR R

= ++

pentru K deschis 1p

1 22

1 2e

R RRR R

=+


1 2er r r= + 1p

2 45r Ω 1p

c

22 int 2P r I= 1p

4p1 2I I I= + 1p

1 1 2 2I R I R= 1p

22 6 AI I= = 2 int 162 WP = 1p

d2

2

e

e e

RR r

η =+ 2p

3p

037η = = 37 1p



204

DOPTICAtilde


b 2

1

ff

β = 3

2 a 2 2 22

1 1 1

1 45 cma

x n x xx n x n

= rArr = rArr = 3

3 c 15 mmaa

iin

= = 3

4


c

1 2

1 1 11

l l

l

f n nnn R R

= =

minus minus

lfrArr rarr infin 3

5a 1 1 2 2

192 61 10 J

ex c ex c

ex

L E L E

L minus

ε = + = +

rArr = sdot

3



a2p

3p


b

2p

4p


2 25 cmf = 1p

c

1 12

2 1 1 1 1



4p1 2 3 3 35 cmd x x x= + rArr =

1p

4 3 2

1 1 1x x f

minus = 1p

4 875 cmx = 1p


205

d

3

1

yy

β = 1p

4p1 2β = β β 1p

21

1

15xx

β = = minus 42

3

25xx

β = = minus

1p

3 1 5625 cmy y= β = 1p



a

W N= ε 1p

4p

hcε =

λ1p

W Pt= 1p


= = sdot 1p

b

ex cL Eε = + 1p

4p

c sE eU= 1p

ex shcL eU= minusλ

1p


c

1s

N eIt

=

2p

3p18


= = sdot 1p

d1p

4p




206

TESTUL 3

A MECANICAtilde




a

2 0yG F N+ + =

1p



b

1 2 0x x fF F F+ + =


4p


1 2

2

cossinm

F Fmg F

+ θmicro =

minus θ 1p


c

1 2x fF F F ma+ + =

1p



F F F mgam

1p

2412 msa 1p

d



vv = 1p

2 cos2matP F= θ 1p

1854 W mP 1p



207


a

2

2PkxE = 2p

3p


b


4p2

2

= =i pkxE E sistem

1 2= + =f c c cE E E E

2p


c

2 2 21 1 2 2


1p


1p

21

1 1 2

( )

=+

kmv xm m m

12

2 1 2( )kmv x

m m m=

+1p


d

fc FE L∆ = 1p

4p

21 1


21

2vd

g=

micro1p




208





a1 0 2

lp p g= + ρ 2p3p

51 15 atm 15∙10 Pa p = 1p

b

1 1 2 2p V p V= 1p

4p

1 0 2

= + ρlp p g 1

2=


2 00( ) 0


1p

10 cm x 1p

c

1 1p V RT= ν 1p

4p1p ShRT

ν = 1p


d

2iU RT= ν 1p


2

5 078 2

= νNU RT2

5 02 2


aer 138 J U 1p



b

1 1 2 2 amestec

1 2

V VV

C CC ν + ν=

ν + ν 2p

4p2

=ViC R 1

3 2

=VC R 252VC R= 1p



209

c

min

max

1 ndashCTT

η = 1p



d

cedat

primit

1 ndashQQ

η =

1p

5p



3ndash1 1 13


= ν = ν 1p

p VC C R= + 1p


η = =V

p

C RC




210





a2

12

Ve A

V

R RR R RR R

= + ++ 2p

3p


b


2 2 3

1 2 3

( )A

V


= + + + = = +

111 16

=I A

25 8

=I A

31

16I A=

2p

4p


069 AAI 6875 V VU 1p

c 1

AA

EI R r R

=+ +

0VU = 2p4p


d

er R = 2p

4p1e AR R R= + 1p



a


=E

2p

4p2

max 4EP

r= 1p


b

2middotP R I= 2p

4pEIR r

=+

1p


c

ext

totala

p

p

RPP R r

η = =+

2p

4p2pRR = 1p



211

d

lRSρ

= 1p

3p2

4dS π

= 1p




212

DOPTICAtilde




a2 1 2

1 1 1 minus =x x f

2 22

1 1

y xy x

β = =

2p



b2 1 1

1 1 1ndash x x f

=

2 2

1 1

y x y x

β = = 2p



c

2 1 1

1 1 1 minus =x x f 2 120 cm=x 1p


2 1 2

1 1 1ndash x x f

=

1p


d 1 2

1 1Cf f

= + 2p3p




a

11

2λ

=Dil

2

2Dil

λ=

2p

4p1

1

cλ = rArr

ν1 2 1

12id

cλ ν

=

1p


b

1 1 2 2k i k i= rArr 1 1 2 2k kλ = λ 2p


1p



213



d


δ =l yD

2p


k λδ = = 1p



ν= = 1p



214

TESTUL 4

A MECANICAtilde


1

1

01 kgpm mv

= rArr =

m m

vF m a m

t∆

= sdot = sdot∆

m 2 2 2 s


2 NF =

3

2

a 0N F G+ + =


cosm gN sdot

=α

3




12 2 1 v v v= minus 12km 216 h


5





a




3p 3p


215

c


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α



20

1 1 307 m 2 sin cos

vd dg


2p

5p



u

d vv tt

+= = rArr ∆ =

∆


1p


( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot



c

d vv tt

+= = rArr ∆ =

∆


1p


1p

d

Notăm cu R



1p

4p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




2p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




1p




b


( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot

3p4p

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot1p


216

c




( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

2p

4p


2p

d


11 1 1 1

1

= 14 m s


minus



1p

4p

( )1 1 1 2

11 1 1 1

1

= 14 m s


minus



( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr =

1p( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr = 1p



217



2 2CO C Og = + 44


5

29 10 Pa = 29 atm

m R Tp V R T pV

p


micro sdot

rArr = sdot

3

2

a

12

1213

13

12 13

cu 22

5 cu 082

V

p pp

RQ C T C C R

Q CQ C T C RQ C

T T T


∆ = ∆ = ∆

3

3

d-1 -1


1 1 1

2 2 2 1 2

2

V S hV S h S h


2 75

p

v

C iC i

+γ = = =

1

12 1

2

VT TV

γminus

= sdot


3

4



3





b


1

12 1

2

VT TV

γminus

= sdot

(1) 1p

5p

p

v

CC

γ =






2p


γ = = 1p



218

c1 1

111

1

1344 g

cmm


3p 3p

d

2 2

222

2

4368 g

cmm


1p

4p

3 3

333

3

08 g

cmm


1p

1 2 3 5792 gm m m m m= + + rArr = 1p

32

kg 29 m

mV

ρ = rArr ρ = 1p



a


micro ν 1p

3p1 2 3 4m m m m m= + + +

1 2 3 4ν = ν + ν + ν + ν1p


molmicro = 1p

b

4p 4p

c


5p


1 2 3 41 2 3 4v v v vv

C C C CC


ν

1p


1 1 11 1

2121 2

22

2

p V R TT TTV Tp R T


sdot = ν sdot sdot

1p

112 1

1 1 15

31 2

- 155 - 155

155 10 J

ced

ced

TQ Q R T

R T p VQ



= minus sdot

1p


219

d


1 1 1 1 123 2

2 1

2ln ln ln 22 2

V T V p VQ R T RV V


1p

3p5

23 035 10 JQ = sdot

( ) 131 1 2 1 1 121 105


1p

531 105 10 JQ = sdot


1p



220



1 1 2 2 3 31 2 3

1 1 2 2 3 31 1 2 2 3 3

1 2 3


S S S S


1

2

2775 n m222 n m


3

2

b

( ) ( )1

1 10 1

1 0 1

1 1

UI UR IR

R R


(1)

Analog ( )2

0 2

1

UIR

=sdot + α sdotθ

(2)


3



4


1 2 3

1 1 1 1 pR R R R




3

5

a2

U N SW tl

sdot sdot= sdot ∆


3



a

1K rarr

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

2p

4p1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

1p

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω 1p

b

2K rarr

2

2

2

154 A

EIR r

I

=sdot +

rArr =

1p

4p2

2

2

154 A

EIR r

I

=sdot +

rArr =1p



221

c2 2

3e eRR Rsdot

= rArr = Ω 3p 3p

d3 3 36 A2

3

EI IR r= rArr =

sdot+ 2p

4p




a


400 = 3

b b b b

bb b

b

P U I IUR RI

= sdot rArr

= rArr Ω2p

4p

bb c

EIR R r

=+ + 1p

485 1616 3cRrArr = Ω = Ω 1p


c

icircnchisK rarr b

t

PP



100 VR bU U= = 1 2R R R+ =


2 b RI I I= +

2 2b RU I R= sdot

2p

4p



η = = 1p

d


4p


1 1 1

e sR R R= +

prime

1p


r= rArr =

sdot


2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =



632 Wbb b

b

UP PR

= rArr =

2p



222

DOPTICAtilde


( )( )1


3

2 a 3

3

a1

2

50 cm1 80 cm

fC

ff=



1 2

1 2

1 22

1

2

16 cm

h hf f

h fhf

h

= rArr

sdot= rArr

=

3

4c

39 Vextrextr s s s

LL e U U Ue


5

a

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3



a 3p 3p

b




i i = rArr =

2p

c

2 5 2sin cos tg = 3 3 5

r r r= rArr = rArr

2p

4p1

1

1

tg i = = 50 cmtg = = 775 cmtg

tg

x hh i H H

r hxrH

rArr rArr=

2p


223

d

2p

4p

0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p



a

19min min

max

= 262 10 J = 163 eVR


λ λ 2p



1p

b


λ2p

4p max

min c

extrextr c

EL im

L E=




4p252 m = 66 10

2 sc

cm v EE v v

msdot sdot




224

TESTUL 5

A MECANICAtilde




a


1p



b






d





a


0


4p2 20

2G gFr v v df f


V = 9 ms 1p

b


2p


1 1 1 2( )m v m m vprime= + 1p

Rezultă 1 1

1 1

3 msm vvm v

prime = =+ 1p


225

c


4p

1 2( )0 m m g Fr Df

+minus = sdot 1p

1 2( )m m gFrf

+= 1p


d

21 2

21 1

( )cs

cl

E m m vxE m v



3p

Deci 103

x vprime= 1p



226





a

ν1 = 2Omicro

m 1p

3p1

2

116

νν

= 1p

2

2Hmicrom

ν = 1p

b


4pp 2

ν= ν1RT1

p 2ν

= ν2RT2

2p

Rezultă 21

12

16TT

νν

= = 1p

c


pV2prime= ν2RT1

1p

4p

1

22

1VV

=νν


1p


V2prime= 2

L x S +

1p


d


4p1 2

2m m m+ =


Obţinem 1 2

1 2

2micro micromicro =

micro + micro1p

Rezultă 1 2

1 2

2micro micromicro =




227


a


4p 4p

b


1p

4p


12

1

lnc T VQ RV



1 1 3

13 1

3

4

1 44

PV RTPV RT

T T TT

= ν= ν

= rArr =

1p


c


p

QQ

η = minus 1p


215η = 1p

d

minCarnot

max

1 TT


3pCarnot

3 754

η = = 1p



228





a


V

ERR r+

1p


2vR (2) 2 2 2 2

v v

v

R ERU IR r

= =+

2p


b


EIR r

=+ 1p


EIR r

=+

2p


c2 1

1 1VrR

EU U

=

minus

1p

4p

2 1

1 1AR E

I I

= minus

2p


d

RV rarr infin 1p

3p1V

VV

V

ER EU ErR rR

= = =+ + 1p



a


4p

1 1 3

2 2 4

S

S

R R RR R R

= += +

1p

1 3 2 4

1 2 1 2 3 4

1 1 1 ( )( )tot

tot S S


+ += + =gt =

+ + + 1p


1 2 3 4

( )( ) 4


+ += + rArr = Ω + + +

1p


3pW = 504 J 1p


229

c


4p1 2 3 4

1 2 3 4

EI R R R RrR R R R

=+ +

+ +2p


d


tot

RR r

η = =+

2p

4p


tot

RR r

η = =+ 2p



230

DOPTICAtilde




a


1 2

11 1( 1)

fn

R R

=

minus minus

2p

4p1

2 0RR

rarr infinlt

1p

R = 15 cm 1p

b

2


4p2

1 2

1 1 1( 1)l

fn

R R

=

minus minus

1p

|R1| = |R2| = 15 cmRezultă nl=137 1p

c

2 2

1 1

1y xy x

β = = = minus 1p

3p

2 1

1 1 1x x f

minus = 1p


d


21

1

xx

β = 2 1 1

1 1 1x x f

minus =

1p

4p


2 1 2

1 1 1x x fminus =

2

21

xx

β =


1 21 1 2

1 1

f xx f ff x

= minus minus+

1p


11 1

1 1

f fff x

f x


+

1p

2

1

ff

β = minus


1p



231



2kk Dx

lλ

=k = 5

2p3p



2lDi

l= 2p


c


1p

5p



2knx eD

lminus

=

xk = 04 m

1p


d



2 2k D k

l lλ λ

= 1p





232

TESTUL 6

A MECANICAtilde




a

2

2f

BG F cB cA



4 msBv = 1p

b

0cE∆ = 1p

3p20J

20JpE mgh

E

∆ = =

∆ =2p

c

2 0

2f

BF cB


4p2 2

2 2B Bv vgd d

gmicro = rArr =

micro1p

4md = 1p

d

( ) 0 0


4p


cos ( )sin


1p

40JL = 1p



233


a

1 1F t psdot ∆ = ∆

1p


1p

12mvF

t=

∆1p

1 2kNF = 1p

b


1p

4p

2

2 2

22

p p mv

F t pp mvFt t

∆ = =

sdot ∆ = ∆

∆= =

∆ ∆

2p

2 1 kNF = 1p

c

H F t= sdot ∆

H F t= sdot ∆1p

3p

1 1

2 2

2 N s1 N s

H F tH F t


2p

d

2p

4p

3 2 2 2N sp p mv∆ = = = sdot

2p



234





a 4p 4p

b

A BU U U∆ = ∆ = ∆ 1p

4p2 1 2 1 2 2 1 1


250 JU∆ = 1p

c


5p


650 JAQ = minus 1p


50 JBQ = minus 1p





a

12 34 0Q Q= =

123 3 2 3 2 3 2 2 2 2 3 2 3 2

5 5 5 5( ) ( ) ( ) ( ) ( )2 2 2 2V


2p

5p1 1 2 2 2 1

4 1 3 2 3 4

1 2

3 4


γ γ γ

γ γ γ



-11

23 4 1 4 1 15 5( ) ( )2 2primit


3p


235

b41 1 4


4p


c

1 cedat

primit

QQ

η = minus 1p

3p4 1 1

1-1

4 1 1

5 ( ) 121 15 ( )2

p p V

p p Vγminus

γ


εminus ε sdot2p

d

pL Q= ηsdot = 1p

3p-1 -1

4 1 1 4 1 1-1

1 5 5(1 ) ( ) ( 1) ( )2 2


ε2p



236





a

11

1 1 2 1 1 2 1 48 16 8 8 8 4 p

p

RR

= + = + = = = Ω 1p

5p

1 4 20 24sR = + = Ω 1p

22

1 1 1 3 1 69 18 18 6 p

p

RR

= + = = = Ω 1p

2 6 6 12sR = + = Ω 1p

1 1 1 3 1 824 12 24 8 AB

AB

RR

= + = = = Ω 1p

b

1 1 3

1 1 05 8 4 VABU I R IR

I R= += sdot =

2p

4p

1 11 1 2 2 2

2

4 1 025 A16 4

I RI R I R IR

= rArr = = = =

1 22 1 A4 1 20 4 20 24 VAB

I I IU

= + == + sdot = + =

2p

c

ABe

AB

UIR

= 2p3p

3AeI = 1p

d

ABAB e e

e

E UE U I R rI

minus= + rArr = 2p

3p

1r = Ω 1p



237


a

1 2

( )

( )


I

= + ++

= + +

3p

5p

2

6024 24

2 20 50 05A

II

I II

= +

minus + ==

2p

b( )e e e



43eR = Ω 1p

c


4p1

1

21

24 2A1236 3A12

b

b

PIUPIU

= = =

= = =2p

d

11

12 62

bURI

= = = Ω

22

12 43

bURI

= = = Ω2p

3p




238

DOPTICAtilde




a1 1 1 1

22 1 1 2 1 1 1 1


+= + hArr = hArr =

+ 2p3p


b



4p 2 1 2 12

2 1 2 2 2 1 2 1


+= + hArr = hArr =



c

2 2

1 2 1 1

sistemx xx x



d 5p 5p



239




b


2p3p


c



λ= 2p


relaţia 2

xD l∆ δ

= care devine ( 1) 45 ( 1) 45 ( 1)2 2 2 2



en

λ=

minus

2p


d


4p

( 1)nouke

nλ

=minus

1p



240

TESTUL 7

A MECANICAtilde


3



EE mgh m

gh= rArr = =


892 ms2

poco po c p p o

EmvE E E E E vm

+ = + rArr = rArr = =

3

4 c 3

5

d

0

0 sin 0 cos

f

x

f y f


+ + + =


cos 20 10 3 027

sin 10f

f

F mg FF NN F


sdot α

3



b


4p


1 1

2 1 2

l ml m m

∆=

∆ + 1p

1

2

25

ll

∆=

∆ 1p

c


4p2 2 1eF m g m g= + 3 3G m g= 1p


d


1m gm g kx k

x= rArr = 2p

3p

100 Nkm

= 1p



241


a


2p

4p


2 1 2 2 1 2

2

0 ( )80 N

N N G N m m gN


b





1p

5p


1 22 1 2

2 1 2 2 1 2


Oy N N G N m m g

minus minus minus =




1p


33 2 3

3 3 3 3

sin


Oy N m g N m g

α minus minus =




2p

Rezultă 3 2 1 1 2 1 2 1 2 3

3 2 1 1 2 1 22

1 2 3

(sin cos ) 2 ( ) ( )(sin cos ) 2 ( ) 342


m m m s


= =+ +

1p

c

1 1 1 1( ) 1626 NT m a g T= + micro = 1p

3p2 1 2 1 1 2 1 2

2

( ) 2 ( )4736



d

1

1 1

22 ( )

R TR m a g

== + micro 2p

3p3252 NR = 1p



242



4



υ ∆ = υ ∆ + 2 1


1 2 1 1 2 2 2 12 1 2 1

3( ) ( )2 2



2 2

p aVp aV

= =

1 11 2 2 1

2 2

p V p V p Vp V

rArr = rArr =

Obţinem 2 2 1 12 1 2 1

3( ) ( )2 2



2 2 2

p V RTp V RT

= υ= υ


3

5



2pV V p= rArr =

adiabat 11 1 2 2 2 2


2

2pp

γminus= 3



a


4p


1 1 1 1

2 22 2 2 2

1 21 2

( )( )


p V Vp V V RTRT

= ν rArr ν =

= ν rArr ν =

++ = ν rArr ν =

1p

Rezultă 51 2 1 11 2 1 1 2 2 2 2

2



b


2 1 1

p Vp V

ν=

ν 2p3p

1

2

5ν=

ν 1p

c


1 2

m m m= +

micro micro micro1p

4p1 2

1 21 2

1 2

2m m

m m m

=micro micro


2p

7gmolmicro = 1p

d


4p


1 2 1 2

1 2 1 2

1 21 2

1 2

2 1

1 2 1 2

( )22

V V V V V V

V V V VVV


C C C CC C




1p

1 22 1

1 2

134

7716

V VV

V

C CC R

mU C T KJ

micro + micro= =

micro + micro

∆ = ∆ =micro

1p



243


a


3p 3p

b



4p

1 1 1 33 1 2 1

3 3 1


= rArr = = rArr = =

1p



3 3 4 4p V p V= 4 1p p= 1 1 1 49 p V p VrArr =

4 19 45lV VrArr = =1p

c


p Vp V RT TR

= υ rArr =υ

1 12 2 2 1 1 2 2 1



1 13 3 3 1 1 3 3 1



4 3 19T T T= =

2p

5p

32vC R= rArr 1 1

1 3 3 1 1 13 8( ) 122v


υ2p

1 3 6000 J = 6 kJU rarr∆ = 1p

dmin

max

1cTT

η = minus

1

3

1cTT


88cη = 1p



244



3


rArr = = Ω


= rArr = Ω


η = = rArr η =+

3

4


182 31 1 1 1 11

2 3

k

ke

k

E E E Er Er r rE

r r r r

sum + += = =

sum + +

1 1 61 1 1 1 11

2 3

e

k

rr

r r r r

= = =sum + +

3

5

d Puterea maximă 2

max 4EP


2 22 2

2

2 6 04 ( )E E R R Rr r


+


3



a


1

1 1 1 22 3e

e

RRR R R

= + rArr = 1p

4p1 18V 2AU I= = 11 1 eU I R= sdot 1p

11 1

1

2 33 2R UU I R

I= sdot rArr = 1p

6R = Ω 1p

b


2

1 1 1 32e

e

RRR R R

= + rArr = = Ω 1p

5p

1 1 1 1

2 2 2 2



1 1 2 2U I r U I rrArr + = + 1p

1 2

2 1

1U UrI I

minus= = Ω

minus E = 10 V 2p

csc

EIr

= 2p3p

10AscI = 1p

d

2

max 4EP

r= 2p

3p

max 25 WPrArr = 1p



245


a

1R şi


lRS

= ρ

22

lRS

= ρ 1p

3p


1 2 2 1 1 2 1 1 2 2l lR R R R R R R l R l

S Sρ ρ


1p



1 2 1 2

1 1 1 ( )e

e

R R RRR R R R R R R

+ sdot= + rArr =

+ + + 3p4p

5eR = Ω 1p

c


3e e

nE EIR nr R r

= = =+ +

2p4p




e

EIr

= = 2p



246

DOPTICAtilde


2

a


2 2 2 2

2 2 2 1 2

1 1 1

62 4V

(155 )155055





ν minus

140 0 118 10 HzLL h


3

3

b



2

sintg tg1 sin

PB rr PB h r hh r

= rArr = =minus


2

sintg tg 3 07m1 sin

x rx PB hPB r


3

4


2r r

v v

D iil i

λ λ= rArr = =

λ


v

ii i fi i f fi


f = 30

3

5


4l

n= =

2

tg tg


rl r h lh

l lr h h ml l

= rArr =

= = =minus

3



a



1 2

1 20 cm1 1 1( 1)( )

Rfnn

R R


1p


2 1 1

1 1 1 ( 20)( 30) 12cm20 30

fxxx x f f x


+ minus minus 2p


1 1 1

( 12) 2 08 cm30

x y x yyx y x


minus


b


1 2

11 1( 1)( )

a

a

a

n Rf n n nn R R

minus= =

minusminus minus2p

3p

80 cmf = minus 1p


247

c


R ffnn

R R

minus= = = = minus


2p


1

2 2 1 1

1 1 1 ( 10)( 30) 15 cm10 10

f xxx x f f x


+ minus minus 1p


2 2

1 1

x yx y

β = = rezultă

2 12

1

( 15) 2 1cm30

x yyx

minus sdot= = =

minus 1p

d


a

a a

f fFF f f f f

sdot= + rArr =

+1p

4p

1 15 cm1 1 2( 1)( 1)( )( )

aa

a

Rfnn

R R


minus1p

( 20) 15 30 cm30 20


minus1p

12

2 1 1

1 1 1 ( 30)( 30) 15 cm30 30

F xxx x F F x


+ minus minus 1p




2Dil

λ= = 2p


i= rArr = = 1p

b


lλ

=


Dxl

λ= =

1p


λ= + sdot 1p


Dx x x xl


c


Dδ = minus =



∆ = minus minus

2p

5p


λ = minus minus



= minus rArr =minus

2p

0 33 45 mm2

Dx xl


dDin relaţia 1

2KK Dx K

lλ

= = rArr 22l x

Dsdot

λ = 2p3p

2 480nmλ = 1p



248

TESTUL 8

A MECANICAtilde




a

2

2cpEm

= 2p3p


b

pFt

∆=

∆ 1p

4pF ma= 1p

pam t∆

=∆

1p


c

amFFF fx

=++ 21 1p





d

2

2atd = 1p

4p2 1d d∆ = minus

21

1 2atd =

22

2 2atd = 1 1 st = 2 2st = 2p




249


a


4p1iE m gh= 2

1

2fm vE = 2p


b


2gth = 1p


2

2atd vt= + 1p


1p


c


1 1 112 2


1 1 1 2i fp p m v m m v= hArr = +

2p

4p( )1

1 2

2m g h dv

m mminus micro

=+

1p


v = 1p

d


1 2

2sistem

m m vE

+=

2

2sistemk lE ∆

=

1 2m ml vk+

∆ =

2p

3p


l∆ =

1p



250





a

pV RT= ν 1p

4p1

2

2

1

2

1

2

1

micromicro

νν

sdot==mm

pp

2p

rezultat final 47

2

1 =pp

1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0

2p V RT= ν

1p

4p0 finalp V RT= ν

1p


d


2p




b1 1L p V p V= ∆ sdot ∆ = 2p


c

min

max

1CTT

η = minus 1p




251

d

efectuat

primit

LQ

η = 1-2 2-3primitQ Q Q= + 2p

5p


p

V

CC

γ = p VC C R= + 1p

RC γ=

γ minus 1V

RC =γ minus

1p

12 1γ minus

η =γ +

1p

rezultat final 2 11

19η = = 1p



252





a

4 52

4 5s

R RR RR R

= ++ 2p

4p3

13 3

se

R RR RR R

= ++ 1p


b

1e

EIR

= 1


3 3

sp

R RRR R

=+

2p


c 1

EIR

=

2p3p


d3

1 3

EIR R

=+ 3p

4p




alR

Sρ

= 2p3p




c

becuri b

sursatilde

P UP U

η = =

3p4p


η = 1p

d

middotx

xRS

ρ= 1p

4p2 x

UIR

= 2p




253

DOPTICAtilde




b

2 1 2

1 1 1x x f

minus =

2 2

1 1

12

y xy x

β = = =

2p

4p012

21

=+ff 1p


c

( )1

1 11nf R

= minus 2p



d


4p2 1 1

1 1 1x x f


2 1 2



1 2

16 cm

x fx xx f

= = minus+

1p






b

2 1r rδ = minus 1p



enδ minus δ

=minus

1p


c 2Dil

λ= 2p

3p



254

d


δ = 1p

4p



δ = δ + 1p


= minus 1p

rezultat final( )12

ed nh

lminus

= 75 mh = micro 1p



255

TESTUL 9

A MECANICAtilde



5 20 205


= rArr = = ms 3

4

d [ ]

( )

21

2

0 0

m1s

2

sC

atx t x v t

C a

=

= + +

rarr

3

5

a ( )

1 1 2 2

1 2

1 2

1 2

1 2

1 2

N12m

s

s

s

s

F k F kF k

F Fk F Fk k

k kkk k

k

= ∆ = ∆

= ∆ + ∆

rArr = =∆ + ∆ +

rArr =+

=

3



a

1p

4p

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

= 1p

b2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+

2p4p

2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+ 2p


256

c


3p( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot =

1p

( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot = 1p

d( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

2p

4p( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

1p

( )1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

= 1p



a 05 JA

A

E mghE

==

2p3p

1p

b

2p

4p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=

1p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=1p

c


( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

4p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

= 1p

d


2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p

4p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

= 1p



257



b [ ] Jmol KSi

C =sdot

3

2 a 2100 kJ

Q m c tQ

= sdot sdot ∆=

3

3d 1 1 2 2

1 2


N NN Nmicro + micro

micro =+

micro = =

3

4

c 2

1

11 1

2

ln

ln

280 J

VL RTVpL p Vp

L

= ν

=

= minus

3



a0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p

0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

b

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

4p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot 1p

c

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

4p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot 1p

d

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

4p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= = 1p



258


a

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

3p

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

b

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

5p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

c( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

=

2p3p( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

= 1p

d

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p

4p

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p



259



d 2 WRtU

= [ ]SI1st = 3

2 b nEIne r

=+

3

3a

2

32 2 323 7221320

AB

ABAB

AB

RR RR R RR RR

R R RRR R

sdot sdot= + = +

+

sdot= =

+

3

4

c ( )0

00

02 1

1

10 grad

R R tR RR R A t

Rminus minus

= + α


rArr α =

3



a

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω

2p3p

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω 1p

b

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

2p3p

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

1p

c

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

5p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus 1p


260

d


( )( )

1 2 23

1 1 1 1 23 23

2 2 2 2 23 23

I I IE I R r I R

E I R r I R

+ =

= + + sdot

= + + sdot

1p

4p

cu 2 323

2 3

23

12

8 A3

R RRR R

I

= = Ω+

rArr =

1p

Din 2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr =

1p2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr = 1p



a


EIr R

=+

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus1p

4p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p

rArr r = 20 Ω 1p

b

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

3p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

= 1p

c

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

2p

4p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

1p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr = 1p

d

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

4p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

şi eR = r 2p



261

DOPTICAtilde



2

2c

mvE = 3



3

4

d 2 2 22 1

1 1 1

x y yx xx y y

β = = rarr =

Din grafic 1 2

2

1 2

1 2

40 cm 10 cm40 cm

20 cm

x yx

x xf fx x

rArr = =

rArr =

rArr = rArr =minus

3

5 b 9

63

550 10 1 275 10 m2 2 10

275igrave m

Di ie

i

minusminus

minus


sdot=

i = 275 μm 3



a


3pcorespunde cu 1 2

05


1p1 2

05


b

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

4p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr = 1p

c

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

4p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = = 1p

d

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

4p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p



262


a 4p 4p

b0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


2p3p0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


c

00

86

0 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot

2p

3p0

08

60 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot1p

d

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

2p

5p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot 1p



263

TESTUL 10

A MECANICAtilde




a



4p


1 12

2

m v m vm vvm

=

=

1p


b


112 2

v gtda

micro= =

2p3p


c


222

2

0375m s2vad

= = 1p

3p


aa gg



d


5p

21


22


1p

( )2 2 2

1 21 2 1 2 1 2( )





264


a






b


3p20

0(1 cos ) 2 (1 cos )2



c


m= = 1p

4p




m M= =

+


ecuaţia vitezei 2

2vd

g=

micro

1p


d



m M m M= minus =

+ + 2p

4p




265





a


pV RT

mpV RTm= ν


1p

4pm pV RT

microρ = =

1p


3128 kg mρ = 1p

b


1 1 11

p V RTmp V RTm


1p

3p1 1

1

p VmRT

micro=

1p


c


2 1 2 1

22

1

p V RTp VRT

= ν

ν =

1p



1




1



d


1p

5p

1 11

1 1 1 2 2

2 2 12

1



1p

1 2 1 1 1 2 2( )p V V RT p V p Vν+ = = + 1p

1 1 2 2

1 2final

p V p VpV V

+=

+ 1p




266


a


3p 3p

b



2 22 1 2 1 2 1 2 12 1

( )( ) ( )( ) ( )2 2 2


1p

3p

2 22 1

2LkV V

=minus

1p


c


21 1 1


R R= ν rArr = =

ν ν

1p

5p2

2 2 22 2 2 2

p V kVp V RT TR R

= ν rArr = =ν ν

1p

2 22 1 2 1( )kT T T V V


ν 1p

2 22 1

3 ( )2VU C T k V V∆ = ν ∆ = minus

1p


d


1 2

1 2

1 2 1 2 1 2

100 J300 J

400 J

LU

Q U L

minus

minus

minus minus minus

=∆ =

= ∆ + =

1p

4p


23 12 3 2 2

2 2

2 3 2 3

0

ln ln 245 J

245 J

UV VL RT kVV V

L Q

minus

minus

minus minus

∆ =

= ν = = minus

= = minus

1p


1p




267





3p 3p

b


1p

4p



4 4s

ee s

r rrr r

= rArr = =1p


015er = Ω 1p

c


1 23 ( )4

E i r I R RI i

= sdot + + = sdot


142A34

neIr R R

= =+ +

1p

4p




= 1p


d2R Sl =ρ

3p4p




268


a


1 2

1 2 1

( )

1

E E I r r RE EI A

r r R

+ = + ++

= =+ +

1p

3p


1

1 1

2 2

b

b

U IRU E IrU E Ir

== minus= minus

1p

rezultat final

1

2

8 V55V25V

b

b

UUU

===

1p

b


1 2

1 2

035Ass

E EIr r R

+= =

+ +

1p


1 2

266pR RR

R R= = Ω

+ respectiv 1 2

1 2

214 App

E EIr r R

+= =

+ +1p


1p


c



1

ext s

PfP

=

1p


d


2ext p pW R I t=

1p


1 2 1 2

057p pext

total p

R I RWW E E r r R

= = =+ + + 1p


total

WW

= 1p



269

DOPTICAtilde




a


2 2

1 1

12

y xy x

β = = = 1p

3p


1 1 1 125Cf x x

= = minus = δ 1p


b


1 1

10 cmf xxf x

= =+


4p




1 40 cmfC



2 1

f xxf x

=+



c


1

025xx

β = = respectiv 22

1

133

xx

β = = 1p




d


= = 1p


1 1 1F x x

= minus 1p


1

FxxF x

=+

1p




270


a


1 0108 mmxik

= =


Did

λ= se obţine 1

1i d x dD k D

λ = =

1p

3p

Din 22 2 2

D di id D

λ= rArr λ = 1p


b


3 1 13 3 Dx id

= = λ

1p


d= = λ 1p




c


1p


1 1 2 2 11


λ 1p


1p


d


1

nλ

λ = respectiv 2

2

nλ

λ =



2 12

k Dxn d

+ λ=

1p



n dλ

= = 1p






271

TESTUL 11

A MECANICAtilde


c 2 2 2 2

2 2 2 cp m v m v E

m msdot sdot

= = =sdot sdot


2 c 2


= + sdot + 2

55 2

g tx sdot=

24

4 2g tx sdot

= 5 4 45 mx x x= minus = 3


αmicro =

α3

4 b 1N100m

Fkl

= =∆

2N200m

k = 1

2

12

kk

= 3

5 b p F t∆ = ∆ F G= m100 kgs




a




2 2 1

1 2 1 2


= =+ +

1p

Rezultă 2

m375s

a = 1p

b




23 3F T T= sdot =

2p


c


4pRezultă


sin cosmM = =



d


V = a Δt2p

3p

Obţinem m5s

v = 1p



272


a


i fp p=

21 1 2 0

1 2

m v m vu

m msdot + sdot

=+

2p

3p

Rezultăm60s

u = 1p

b


4p




Obţinem Bp =m56kgs

1p

c



4pObtinem

21 2

1 2( ) ( )

2B

Bt


Rezultă 1700 JBt

E = 1p

d


2C f

Bc G F


4pDeci 2

2sin cos2C

t Bc t t




2p




273



b 12


2 b 201204 10A

A




4 b 1 2 1 2

1 2

m m m m+= +

micro micro micro 3

1 2


+micro micro

3


3 2 11mp R T m m mV


3



a


Np V R TN


11

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro

1p

4pSe obţine 1

11

AN P NnV R T

sdot= =

sdot2p

Rezultă 25

1 3

molec12 10m

n = sdot 1p

b


22

p VR T

sdotν =

sdot2p

3p


c


11

m R TpV

sdot= sdot

micro 1p

4p


11

pR T

sdotmicroρ =

sdot

22

2

pR T

sdotmicroρ =

sdot1p

Se obţine 1 1 2

2 2 1

p Tp T

ρ sdot=

ρ sdot1p

Rezultă 1

2

113

ρ=

ρ 1p


274

d


2 2 2 2

p V R Tp V R T


SF 1 1 1

2 2 2

p V R T

p V R T



1p



1 1 2 2

1 2

1 2

1 2

p V p VT Tp V V

T T

sdot sdot+

=+

1p




a


1

ln ln 2vL R T R Tv


34 3 4 3 3

1( )2


141 1 1

4

1ln ln2

vL R T R Tv


4p





3p

34 12465 JU∆ = minus 1p

c


4pObţinem

1

4 3 14





d


p

LQ

η = 1p


2p

Obţinem 505η = 1p



275



c 2

2 UW R I t tR


= 3

2 c 3


4 c eSC

e

EIr

= 65er = Ω 24 V

5eE = 4AeSC

e

EIr

= = 0 VU = 3

5b

2 2U UP RR P

= rArr =

2

10 mS UlP

sdot= =

ρ sdot3



a


e e

EIR r

rArr =+

1p


1 2

1 2

1 2

30 V1 1e

E Er rE

r r

+= =

+ 1

2err = = Ω 1p


1 2

4s se

s s

R RRR R

sdot= = Ω

+ 1p

Obţinem 6 AI = 1p

b



Obţinem 1

2

1II

rArr = 1p

c



10 A3

e

e

EIR r

prime = =+

2p






276


a




b




c

Randamentul va fi e

e

RR r

η =+

1p

4p




2p

Rezultă 886η = 1p

d


4p


12

21

8022 3

2

e

RR RR RR

sdot= + =

+

30 A43e

EIR r

= =+

2p




277

DOPTICAtilde




a


1 1 1 1 1 12x x f x x f


2p

4pRezultă 2

1 32

cf x

minus= =

minus1p

Obţinem 2

3 2015 10

c minus= = δsdot

1p

b

2 1

1 1 1x x f

minus = 1p

4pobţinem 2

1 21

1 22

x x xx



c


2 1

1 1 1x x f

minus = 1p



1

x fxx f

primeprime =

prime +1p

Rezultă 275 5 15 cm75 5


1p




D a ii Da

λ sdot= rArr =

λ2p

3p

Rezultă 3 3

9

2 10 1 10 5 m400 10

Dminus minus

minus

sdot sdot sdot= =

sdot1p

b


4pRezultă

axD

δ = 2p

Obţinem 3 3

72 10 16 10 8 10 m4


δ = = sdot 1p


278

c



4pObţinem

9

max 3

400 10 442 10

xminus

minus

sdot sdot= sdot

sdot1p


d


naλprime =

1p

4pDin Dia

λ= rezultă ii

n= 2p

Rezultă 3

3 31 10 3 10 075 10 m4 43

iminus




279

TESTUL 12

A MECANICAtilde




a


00

( )

05

kt M m m akta t

M m m

= + +

= =+ +

2p


2

2 2

m8 02s

Mg ma

a

micro =

= micro =

2p

b


0 1( )Mg ma

kt Mg m m amicro =

minus micro = +

2p

3p


m( 2)s

kt Mga tm m


+ 1p

c


4p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + =

2p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + = 1p

d


4p


( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus = 1p



a


30 5 10 J



Deci 0cE∆ = 1p


280

b




3p4p


c



4p


d


1p


2p




281



5

b

1 11

1

abs ced

abs

Q QQ

minusη = respectiv

2 22

2

abs ced

abs

Q QQ

minusη =


1 2

abs abs

L LQ Q

η = η = iar

2 1abs cedQ Q=


3



a


1p

3p0p VmRT

micro=

1p


b


4p


A

Np V RTN

= 1p


A

N RTpVN+ α

= 1p

Rezultă 0

1 14pp

= + α = 1p

c


kmolmicro = şi


kmolmicro =

1p

4p

012i ii

A i

N m iN

ν = = =micro

1p

1 2

1 2

m m m= +

micro micro micro

1p

0

1 2 1 2

1 2


A

A A

Nm N

m m N NN N

micromicro


micro micro

1p


282

d


4p


= ν ν = 1p


1 2Q Q Q= + 1 1 1 2 2V VQ C T C T= ν ∆ + ν ∆1p

( 5) 450 J2

pV TQT

α + ∆= sdot =

1p



a


2 1

p pT T T nTT T

= = = 1p

3p


3 2

V V T nkTT T

= = 1p


1 4

V V T kTT T

= =

1p

b


1p

4p


= minus minus


= + minus 1p



1p


283

c


Ccald

T T nkT nkT nk


Cη gt η adică 1 2 2 2 25 2 3

nk nk n knk nk n



adevărat

1p

4p



minusη = =



minusη = =

1p




minus minus


5n n DA q e d

nminus

lt gt minus




minus minus minus

1p



krarrinfinrarrinfin

η = lt

1p

d


cedQL

ϕ =

1p

4p


1 5 2 32 2 2 2

cedQ nk nL nk n k

minus minusϕ = = =

η minus minus +


absQL

ε =

1p


1 3 2 512 2 2 2

nk knk n k

+ minusε = minus =

η minus minus +



1p







284



2




ERIR r r x xr

=+ + +


ERIr R r

= =+

3

3 c 3

4




2 2

U UI R R R= =


R UUab I= sdot =


abR

U UIR R


2

4RUP RI

R= =

3

5


0 i D

i D

U U EI U U E

r R

le += minus minus +


1 5 V2

i

e ii

UU U U

le= +

gt


3



a


n

ki

k k k

I I

U E I rU IR

=

=

minus = minus =

sum


k k

E RI Ir r

= minus

2p

4p


1

11

nk

i kn

i k

ErI

Rr

=

=

=+

sum

sum și respectiv 1

1

11

nk

i kn

i k

ERrU IR

Rr

=

=

= =+

sum

sum2p

b


1 1

lim1 11

n nk k

i ik kgol n nR

i ik k

E ERr rU

Rr r

= =

rarrinfin

= =

= =+

sum sum

sum sum 1p

4prespectiv 1

1

lim 011

nk

i kgol nR

i k

ErI

Rr

=

rarrinfin

=

= =+

sum

sum 1p


0 0n n

ksc sc k sc

i ik

ER I I Ur= =



285

c


1

1

1

1

11

nk

i kn

i k echivalent

echivalentn

i k

Er

r EIR rR

r

=

=

=

= =++

sum

sum

sum

2p

4p

adică 1

1

1

nk

i kechivalent n

i k

ErE

r

=

=

=sum

sum 1p

respectiv

1

11echivalent n

i k

r

r=

=

sum

1p

d

echivalentE E= 1p

3pechivalent

rrn

= 1p

EI rRn

=+

1p



a


0 2

A B

A

B A

I I IE Ir RI

RI RI

= + = + = minus

1p

3pRezultă

3 22

3 23

3 2

B

A

EIr R

EIr R

EIr R

= + = +

= +

2p


286

b

( 2 )3 2CD B A

E R xU RI xIr R

minus= minus =


5p

Pentru max

min

03 2

3 2

2

CD

CD

CD

ERx Ur R

ERx R Ur R

RU o x

= = + = = minus +

= =

1p

Graficul

1p


CDdU Edx r R

α = = minus+

1p


tg CDU

R = minus = Ωα


2 mrE Uα

= minus + = 1p

c


2

1 2



1p

3p

Rezultă

2 2

(2 )E x RIRx R x

minus=

+ minus1p

Pentru

0 2A

2A

02

Ex IR

Ex R IR

Rx I


= = 2

max max 20 WP RI= =

1p



=


xEI cu xy nx rRn

= =+

1p

4p


= =+

1p



R= 1p




287

DOPTICAtilde




a


2 1

1 1n nx x R

minusminus = 1p

5p



1 1n nx x R

minusminus =

minus 1p

Se deduce

12

1( 1)Rxx

nR n x=

+ minus 1p


b


1 1

y nxy x

= 15p

4pLa a doua trecere

2 2 1 1

y xy x

= 15p

Dar 1 2y y= deci

2 03 cmy = minus 1p

c



1 1 2x x R

+ = Rezultă 12

1

2 214 cm2

xxx R

= =minus

1p


21

007 cmxyx

minus= = 1p


d


12 1

1

5 cm 7cm2Rxx R xx R


3p


2 x

Ryy yx R


1p




288


a



2 0 2 2 2 2( ) ( ) ( )

nn a l b n r e n e


minus 1p


b


D l∆

= 1p


Rezultă ( )2 2 2 20 2 2 1 1( ) ( ) ( ) ( )



c


4pRezultă

2Dinlλ

= 2p

d



0 1 1n n= = 1p



289

Cuprins






4

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 d 34 c 35 a 3



a



= 1p


b




deci 0fF F =- 2p3p

Rezultă F = 4 N 1p

d



2p

4pObţinem 1

2 fF F Fam m

= =- 1p






b






5

c



Obţinem 2Ahh = 1p


d





Rezultă 2 JGL = 1p



6


Subiectul I Punctaj

1 b 32 b 33 a 34 c 35 d 3




11

mν =

micro2p

3p


b


A

NN

ν = 1p


22

mν =

micro 1p


2

mN N= sdotmicro

1p


c


1 1

m R TpVsdot

= sdotmicro 1p

4pRezultă 5 2



2 2

m R TpVsdot

= sdotmicro

1p


d


4p

unde 1 2m m+ν =

micro 1p


1 2 1 2 2 1

1 2

( )m m m mm m m m



1p




7


a


1 0

311

1

300 K

831 831 10 Pa

= 3 dm

Tp p

R TVp

=


=

1p


32 1

32 1

2 600 K

831 10 Pa

=2 6 dm

T Tp pV V

= =

= = sdot

=

2p


313

33 2

600 K

4155 10 Pa2

=2 12 dm

T Tpp

V V

= =

= = sdot

=

2p

b


12 2 1 V 2 1 17( ) ( ) ( )2


4p


23 2 12




12 23 17 2ln 22


Q Q Q R T 1p


c


12 V 2 1 15( )2



d


4p


23 2 12







8


Subiectul I Punctaj

1 c 32 a 33 d 34 d 35 b 3




3p


b



S

EIR r

=+

2p

Rezultă 1 AI = 1p

c


torul R este 1

1

EIR r

prime =+

2p




1SC

EIr

= 3p4p




9


a


5p


1 1 1

PR R R= + 1p


S PS P

E ER RR r R r

sdot = sdot

+ + 1p



b


S

RR r

η =+

1p



P

RR r

η =+

1p


c


1 11

EP RR r

= sdot

+ 1p

3pDeci 1

11

( ) PE R rR

= + sdot 1p

Rezultă 6 VE = 1p


EIr

= 2p3p




10

DOPTICAtilde

Subiectul I Punctaj

1 b 32 a 33 d 34 a 35 c 3




f= 2p

3p


b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+1p


c


1

xx

β = 1p

5pRezultă 12

β = minus 1p




3p




11



1

cλ =

ν2p

3p


b




2 1

( )

S S


=minus

1p


c



Obţinem 10 1

Se Uh


0 2Se U

hsdot

ν = ν minus 1p


d



4pRezultă 22







12

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1a [ ] J Nm= =SIL 3

2 c 2

2 2 00 2 270 m

2vv v ad d

a= + rArr = = 3


2

10 6

6 ms

v v at v t

a

= + = +

rArr =

854 kgFma

= = 3

4 b 2

2

2

2

C

C

mvE p mv

pEm

= =

rArr =

2

2

2

2

C

C

mvE p mv

pEm

= =

rArr = 3

5 b 2

0max

0

20 m 10 m2

5 JC C p C

vh hg

E E E E

= = rArr =

= + rArr =

3



a


1 1

1

t f

n

G F ma

N G

minus =

= 2p

4p


21 25 msa = 1p

b


2 20 1 02 0v v a l v= + = 1p

3p1 12v a l= 1p

1 10 316 msv = = 1p

c


2 2

2

fF ma

G N

minus =

=

1p



22 01a

gmicro = minus =

1p


13

d


1t f

n

G F FN G

= +

=

2p

4p


5 NF = 1p



a

1 1f fL F d= minus

1p

4p1f

y

F N

N G F

= micro

= minus

1p


1p

1

75 JfL = minus 1p

bcosL Fd= α 2p

3p

255 JL = 1p

c

1C fE L L∆ = + 1p

4p

2

2CmvE =

1p

2 CEvm

=

1p

2 3 34 msv =

1p

d

2C fE L∆ =

2 2f fL F D= minus

1p

4p2f

F N

G N

= micro

=1p

C

f

EDF

=

1p

D = 6 m 1p



14


Subiectul I Punctaj

1b mpV RT=

micro

30 kPamp RT

V= =

micro

3



4c

3127 kgmpRT

microρ = = 3

5

b 1 1 2 23 32 2

U RT U RT= ν = ν

2 1 2 12 2U U T T= rArr =

11

2 200 J3UL R T RT= ν ∆ = ν = =

3



a


1 2


= = 1p

4p


1 2

1 2

V p pmR T T

micro∆ = minus

1p

1 2

1 2 2 1( )mRTTV

p T p T∆

=micro minus

33324 mV =

1p

b

11

1

p VRT

ν =

2p3p

1 026 kmolν = 1p

c

22 2

mp V RT=micro

2p

4p22

2

p VmRT

micro=

1p

2 512 kgm = 1p

d

11

1

pRT

microρ =

3p

4p

31 25 kgmρ = 1p



15


a


3p 3p

b


1 2

3V V V VT T

= rArr = 1p

4p12 1 2 1( )L p V V= minus 1p

121

2 1( )Lp

V V=

minus 1p

5 21 2 10 Nmp = sdot

1p

c


23 31cedQ Q Q= + 1p


1p

131 1 1 1

3

ln ln3VQ RT p VV

= ν = minus

1p


d


4pp VC C R= + 1p

1 1 15 5absQ RT p V= ν = 1p




16


Subiectul I Punctaj

1b 1 Wh = 3600 J 6720 10 JW = sdot

3

2 c 1 1 2

2

E I EIR r R R r

= =+ + +

1 1 2

2

E I EIR r R R r

= =+ + +

1 22

1

2 8R R r RR r+ +

rArr = rArr = Ω+

3

3 b 23ABRR = = Ω 3

4 c V1sc

EIr

= =Ω

3

5c

2

bec 90n

n

URP

= = Ω


3



a

1e pR R R= + 1p

3p2 3

2 3p

R RRR R

=+

1p

44eR = Ω 1p

b

1e

EIR r

=+

1p


88VU V= 1p

c

1 2 3I I I= + 1p

4p2 2 3 3I R I R= 1p

1 23

2 3

I RIR R

=+

1p

3 08I A= 1p


17

d


1 23

1 2e

R RR RR R

= ++

1p

4p125 A

e

EIR r

=+

1p

1 2

1 1 2 2

I I I I R I R

= +=

1p

12 083 A3II = =

1p



a


1 3 3E Ir U I R= + +

2p

4p

13

3

E Ir UIR

minus minus=

1p

3 15 AI = 1p

b


2 2 3 3I R I R= 1p

4p3 3

22

75I RRI

= = Ω

1p

22 2 2P R I= 1p

2 1875 WP = 1p

c

1e pR R R= + 1p

4p

11 30UR

I= = Ω

1p

2 3

2 3p

R RRR R

=+

1p

4875eR = Ω 1p

d

1 1

s

P UP E

η = =

2p3p

05 50η = = 1p



18

DOPTICAtilde

Subiectul I Punctaj

1 c 3

2 a 2

1

sin 3sin 60sin 2

i n r rr n

= rArr = rArr = 3

3 b 1 2 3 3 4C C C C C= + + rArr = δ 3


3

5b 2 2

1 1

2 1

3

3 150 cm

y xy x

x x

β = = = minus

rArr = minus =

2 1

1 1 1 375 cmff x x

= minus rArr =

3



a

2 21

1 1

3y xy x

β = = = minus

1p

4p2 13x x= minus 1p

11

43fx = minus

1p

1 40 cmx = minus 1p

b

32

1

3xx

β = =

1p

4p3 13x x= 1p

1 32

1 3

x xfx x

=minus

1p

2 60 cmf = 1p

c 3p 3p


19

d

1 2

1 1 1F f f

= + 1p

4p

20 cmF = 1p

14

1

FxxF x

=+

1p

43

1

1xx

β = = minus 1p





b

0L h= ν 1p

4p

150 055 10 Hzν = sdot

1p

151

1


λ 1p


c2

2

hcε =

λ 2p

3p


d

2 ex CL Eε = + 2p

4pCS

EUe

=

1p

079 VSU = 1p



20

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 a 34 a 35 d 3



a


0eF G+ =

1p


11

m glk

∆ = 1p


b

fF F= 1p

4pN G= 1p

fF N= micro


∆ =m glk 2 125 cm∆ =l 1p

c

e fF F ma+ =

1p

4peF F= 1p

2

2

minus micro=

F m gam 1p


d

22

2pk lE ∆

=

2p

4p2

Flk

∆ =

1p




21


a

F G ma+ =

1p



b

vat

∆=

∆1p

3p0 = +v v at 0 0v = 1p


c

total∆ =cE L 2p

4p2

2cmvE∆ = 1p


d


4p

2

1 2

= +mvE mgh

2

2 2PmvE =

1p

2( )2∆

=a th

1p


1p



22


Subiectul I Punctaj

1 a 32 d 33 b 34 a 35 c 3



a A

N mN

=micro

2p4p


b


4p0

0

mV

ρ =

mV

ρ = 1p

0

0middotmVV

V m=

∆ρ +1p



3p


d


4p0 p VRT

ν =


2p



a


2 2

1 1

=p Vp V

2p

3p

Rezultat final 2

1

2pp

= 1p


23

b



2 14T T= 1p


U RT∆ = ν rArr 12 10800U J∆ = 1p

c


( )=L A p V 2p

4p1 1 1

2 2p V RTL ν

= =


2p

d



2 14 =T T 3 12T T= 1p





24


Subiectul I Punctaj

1 d 32 b 33 a 34 c 35 c 3



aRS= R1+ R2 2p


b

UAB = R2 I 1p

4ps

EIR r

=+ 2p


c s A

EIR R r

=+ +

3p4p

I primecong 12 A 1p

d 1

EIR r

=+

3p4p




a


4pISC = 10 A 1p

ISC = scEIr

= 1p


b


4pEIR r

=+

1p


c

echivalent

echivalent

RR r

η =+

2p


91η 1p





25


1 a 32 b 33 b 34 d 35 a 3




b

2 1

1 1 1x x f

minus = 1p

4p1fC

= 1p


c

2 2

1 1

β = =x yx y 2p

3p


d

1 1 60 cm= + = minusx d x 1p

4p12

1

x fx x f

=+

1p





3p


bfW h= ν 2p

4p








26

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1c

5

8

km = 3 10 uas = 012

1ua min1 km = 15 10

cc

sdot rArrsdot

3


m

m

km = 200 m 48 h

= 15 s

dvt

d vt

= ∆ rArr =∆

3

3c

25

3

= 75 10 J2 = 208 kW h

11 J = 1 W s = 10 kW h3600

c c

c

m vE EE

minus


sdot sdot

3

4



= 75 Nm m

m v vF F

tsdot +

= rArr∆



5


0 0 02

0 0 2

0

4 1 3828 cm4

F l F lE lS l E S

d FS l l ld E

l l l


∆ = minus

3



a


4p2 = 2 m sxx

RR m a a am






27

d


5p


1p



1p



2p



a



= 11314 mdrArr

2p

5p2p

2 1 2

= 14 s

= 12 h 24 min 34 s

dv tt

t t t t

= rArr ∆∆

∆ = minus rArr1p

b

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr =

2p3p

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr = 1p

c


4p


2p

( )1 20

879 MJr

r

F

F

m g h h L

L


rArr = minus

1p

d

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

2p

3p

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

1p



28


Subiectul I Punctaj


3



3

b 1 1

2 2

1 1 11

22 2 2

5 cu 80 K 52 1673 3 cu 80 K2

V V

V V



ν

ν3

4

d 23 23 23

2323

40 J0

Q U LU

L= ∆ +

rArr ∆ = minus=


123112

1231 12 23 31

0110 J

UU

U U U U∆ =

rArr ∆ =∆ = ∆ + ∆ + ∆


3

5

c 22 72 10

VA

VA

Q C TQ NN NNC T

N


3



a

1 2 m m m= = 1p

3p1 2

2 1

125ν micro= =

ν micro2p

b1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


1p

4p1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


c

3

kg 157 m

mp V R TpR Tm

V


rArr ρ =

3p

4p

3

kg 157 m

mp V R TpR Tm

V


rArr ρ = 1p


29

d5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

4p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot 2p





c

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

4p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

2p

d

5 1 3

pp v

v v

p v

CC C R RC C

C C R


γ minus= +

1p

5p


( )( )

1 2 1 2

1 2

1 2

1 2

1

1v v v v v v


f f



2p

2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = =

1p2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = = 1p



30


Subiectul I Punctaj

1a 2

2

514 nC

U S qI U r tl t ql

S rq


=

3

2

a ( )( )

1 0 1 2 1

1 2 2 12 0 2

3 1

1 - 1

4 10


Kminus minus


α = sdot

10 0

1

9 1

RR Rt


3

3




3

4b

=12800 C

PIP tU q q

q UIt


3

5


= rArr = Ω


= rArr = =


3





1

9 E EI R r R


+ 2p

b



5p


1817 18

E EI R R rR R r I

R


rArr = Ω cong Ω1p




1p

2 05 mml lR S SS R



31

c


1p

4p1

1

105 A 11 s

EIR R r

I t


primeprime = rArr =

2p

1

1

105 A 11 s

EIR R r

I t



dI = 0 1p

3pU = E 1pU = 20 V 1p



a = 2400 mA h = 8640 C q sdot 2p


b1

06 AUI IR

= rArr = 3p 3p

c 14400 s = 4 hq qI t tt I


3p 3p

d


1p

5p


UI IR



2 22

6480 s = 18 hq qI t tt I


1p

2 22

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆

1p2 2

2

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆ 1p



32


1


3

2 c 3

3




3



D iil n n


3


0

2 4

s

s

h c h c e U

h c h c e U



3



a

22 1

12

1

2 4 cm

2 2 cm

xx x

xx

x


2p3p


b

2 1

1 1 1

1

25 4 cm

x x f

Cf

Cf

minus = =

rArr = δrArr =

2p4p


c

2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus

2p

4p2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus 1p



33

d


4p

2

1 2 1

1 1

2 1

22

8 cm 1875 1 1

xx x x

x x C

C Cx x


primeminus = +



3p




b

1

1

22

kDxD li

l


sdot= sdot

1p


2

R V Dx

lλ minus λ sdot

∆ =sdot

2p

076 mmx∆ = 1p

c2

2

2 2 192 mm

2

k x ixDi

l


= sdot 3p 3p

d


4p

1 21 22 2

D Dk kl l


sdot sdot1p


Soluţia 1

2

32

kk

= =

1p


0Cprime lt rArr


34

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj

1 a 32 b 3

3


0 sin cos 0sincos

Gt Ff mg mg

tg


micro = = αα

3



5



= + = + =

3



a


4p 4p

b


şi 2a


1p


2 2

T m g m aT m g m a

minus sdot = sdot



1 2



+ 1p

c T = m1(a + g) = 2222 N 3p 3p

d


1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p

4p

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p



35


a

2p

3p

cos5640 J

L F dL

= sdot α=

1p

b

NFfL Ff d

Ff= minus sdot

= micro

1p

4pN

FfL Ff dFf

= minus sdot

= micro 1p




4p4504 JfEc = 2p

d564 W

med

med

LPt

P

=∆

=

2p4p

564 W

med

med

LPt

P

=∆

= 2p



36


Subiectul I Punctaj

1 c 3

2 a 33 b 34 a 35 c 3



a3096 kgm

pRT

microρ =

ρ =

2p3p

3096 kgm

pRT

microρ =

ρ = 1p

b

a

a

N m N mNN

sdot= rArr =

micro micro1p

4p



a

a

N p VNp V RT NN RT

= rArr = 1p


RT= = sdot 1p

c

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus=

1p

4p

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus= 1p

1

2

11

pp f

=minus

1p

p2 = p1(1 ndash f ) = 96 ∙ 105 Pa 1p

d


4p


1p

21 O 2

1 2


ν + micro ν

ν + ν2p



37




3p 3p

b


2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= =

2p

4p2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= = 2p

c9kJ

p

p

LO

LO

η = rArr

= =η

2p

4p

9kJ

p

p

LO

LO

η = rArr

= =η

2p

d


4pT2 = T3 1p

ΔU = 0 1p



38


Subiectul I Punctaj

1 a 32 c 33 c 34 a 35 b 3



a


1 1 1

PR R R= + 1p



R = 11 Ω 1p

b




cE = I3r + UAB 2p

3pUAB = 22 V 1p

d


1

UIR

= 1p



2

UR


U1 = U2 = 96 V 1p




3pUb = 120 V 1p

b



Uf = 4 V 1p




W = 4464 kJ 1p



39


1 c 32 a 33 d 34 c 35 c 3





3p

Rezultă cvn

= v = 212 ∙108 ms 1p

b



r = 450 1p090 45rθ = minus = 1p

c




1p

4p1sin sin2

n l n π= 1p

1

2

sin nln

= 1p

l = 450 1p


3p 3p




1Cf

= 2p3p


c

2 1

1 1 1x x f

minus = 2p



1 1

x yx y

β = = 1p

y2= y1=10 cm 1p


2p3p




40

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj

1 c 32 a 33 c 34 c 35 b 3



a

2p

4p

2p

b





M m M m


= =+ +

1p

21msa = 1p


09 NT = 1p

d



17 NF = 1p



a


4pfc u FL L L= + 2p

2500JcL = 1p


41

b

u

c

LL

η = 2p

3p2000 08 802500

η = = = 1p

c

cossin sinfF f


1p

4p

u GL L mgh= = 1p

f

f

FF u

u

LL ctg L

L ctg


α1p

1 0254

micro = = 1p

d

2 60 40 s3

t∆ = sdot = 1p

4p

u cu c

L LP Pt t

= =∆ ∆

1p

2000 50 W40

2500 625 W40

u

c

P

P

= =

= =2p



42


Subiectul I Punctaj

1 d 32 d 33 b 34 c 35 c 3



a

1 1 1 1 1

2 2 2 2 2

( ) ( 2 )2 2 2

( - ) ( - 2 )2 2 - 2




+sdot


4p5

51

55

2

10 1 5 10 Pa18 9

10 1 5 10 Pa02

p

p

sdot= = sdot

sdot= = sdot

2p

b

F



133(3) NF = 1p

c

1 1 1

2 2 2

( 2 ) ( 2 )





2p

4p

51

52

510 Pa6625= 10 Pa

6

p

p

prime =

prime2p

d

1

2 p ll hpsdot

+ = 1p

4p1

1 12

p lhp

sdot= minus prime

1p

01m = 10 cmh = 1p




43



3p4p



c

212 1 1 1

1


= ν = = sdot =


1p

5p

5 51 11 1 2 2 2 1

2 1



3 523

5 2 10 3 10 (1 27) 1500 17 2550J2



1620 2550 1530 600 JQ = minus + = 1p

d

U Q L∆ = minus 1p

3pL Q= 1p

600JL = 1p



44


Subiectul I Punctaj

1 a 32 a 33 d 34 b 35 d 3



a

1 1 1 1 1AU I R I= rArr =

2 2 2 2 075AU I R I= rArr =2p

4p( )1 1 2 1 1 2

2 2 1 2

( )6V

( )E I R r R R I I

EE I R r I I


2p

b

1 11 1 2 2

2 2

( )( ) ( )

( )E I R r

I R r I R rE I R r

= +rArr + = + = +

2p

4p2 2 1 1 2 1

1 2 1 2

I R I R U UrI I I I

minus minus= =

minus minus 1p

2r = Ω 1p

c

66

15012 10 603 10fir

lRS


sdot2p

4p6 075A 075 60 45C8fir

EI q I tR r

= = = = sdot ∆ = sdot =+ 2p

d0

00 0

1(1 ) 1 1

fir

fir firfir



α2p

3p

40 Ct = 1p



a

ss

s

EIR r

=+ 1p

4p10sE E= 1p

10sr r= 1p

4 133 A3sI = = 1p


45

b

pp

p

EI

R r=

+ 1p

4p10prr = 1p

10

110p

ErE E

r

= = 1p

05ApI = 1p

c


4p2 pU I R= 1p

1 1 VU

1p

2 35VU = 1p

d1

s

RR r

η =+

2p

RR r

η =+

1p

3p

17 259227

η = 235 972236

η =

2p



46


1 d 32 c 33 c 34 c 35 a 3



a 4p 4p

b

sin sini n r= 1p

4psin sin n r i= 1p



c


4p

sin sin

3sin 12sin

23

i n r

irn

=

= = =1p

30r = deg 1p



d cosADAB

r= 2p

3p

115cmAB 1p



47


a


1

2

1

2

s

s

ch L eU

ch L eU

= +λ

= +λ

2p

5p


2

2 1

1 1 2

1 2


( )( )

s s

s s

hc e U U

e U Uh

c

minus = minusλ λ

minus λ λ=

λ minus λ

2p


b1

1s

cL h eU= minusλ 2p


c0 0

Lh Lh

ν = hArr ν = 2p3p

150 0924 10 Hzν = sdot 1p

d

2max

2smveU = 1p

4pmax2 seUv

m= 1p





48

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj


0 m 8 N4 m 0 N

x Fx F

= rArr == rArr =


(6 2) 2 8 J

2L + sdot

= =

3


22 2i f

kx mv kxE E vm

= rArr = rArr = = 3

3

a Viteza medie mdvt

= (1)





= şi 2 2dtv



1 22 2

mm m


+= rArr = rArr = =

++km h


3

4 a 35 d 3



a


1km20h

v =

2km40h

v =

1130min h2

t = =

2115min h4

t = = 2p4p


b

Viteza medie mdvt

= 2p


t = = 1p




d


21

2cmvE =

2

22

2cmvE =

Rezultă 2

1

222

1

c

c

E vE v

=2p

3p

2

1

4c

c

EE

= 1p



49


a




hl =α

iar 1

0NL =


2p

ECB = 32 J 1p

bLFf

= LFf1 + LFf2



c


1p

4p2




= = 1p

d



CDEE = 1 1

15





50


Subiectul I Punctaj

1 b 32 c 33 b 3





3

5 c 3



a



1 2

12p p p pT T

= rArr = 1p

2p = 5

2

N12 10m

sdot 1p


max1 max 1

p p p TTT T p

= rArr = 2p4p

max 12 600KT T= = 2p

c


1 00 1

91p Tp T

ρρ = = ρ 2p

3p

31 0819kgmρ = 1p

d


mp V RT=micro

şi 21 2



12 112





a


3p

2p


51

b


VVrArr = 1p


2 3

V VT T



2 12 2888KT T= = 1p


3

1 2p

vv

C RC RC


4p

1 3 1 1 1 1 13 3 3(2 ) 1800 J2 2 2


d

2

1

| |1 QQ

η = minus 1p

4p

1 2 3 3 2( )5

1 2p

Q Q Cp T TRC R

rarr= = υ minusγ

= =γ minus

Rezultă 1 1 1 15 252

Q RT p V= υ =

1p

12 1 2 3 1 1 3 1 1 1 1 1 1 1

2

3| | | | ln ( ) ln 2 (ln 2 15) 21932v



122η = 1p



52


Subiectul I Punctaj

1 c 32 c 33 d 34 d 3




a



1 21

PRI

= 2p

R1 = 5 Ω 1p

b



EIR r

= =+

1p



e

RR r

η =+

2p3p

80η = 1p

d


4pPuterea maximă

2

max 125 W4 EPr

= = 2p



a


5p


2

V

p

V

RRR RR

= = Ω+

32e pRR R= + = Ω 2p


EIR r

= =+

1p



3pP = 75 W 1p


3pPE = 100 W 1p


e

RR r

η =+ 2p

4pη = 75 2p



53



2max

max max2

2S

C S SmV eUE eU eU V

m= rArr = rArr = 3

2 b 33 a 34 c 3

5


1

2

R RR R

= minus= +

1

1 2

1 1 1 1 2( 1)( 1)( ) ( 1)( ) 5mnC n nR R R R R







3pDin 2

2 1 21

4 20cmy y y yy


c


4pCum 2

2 1 11

4 4 5x x x d xx


1

2 1

16cm54 64cm

dx

x x

= minus = minus

= minus =1p

d


2 1 1 2

1 1 1 ( 16) 64 128 cm16 64

x xfx x f x x


minus minus minus2p


1 1 2( 1)( 1) ( )

Rfnn

R R


2p

2( 1) 128 cmR n f= minus = 1p



54


a


3p 3p

b

302



sinsin irn

rArr =

3sin8

r = 1p

c


5p

tg hOI

α = rArr 1038mtg 3

h hOI = = =α 1p


= rArr2

sintg 0809m1 sin

rAB H r Hr

= sdot = =minus

2p


d


4p



53

i n r i n i

i i i n

i

= rArr = minus


rArr =

2p



55

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj

1 a 32 d 33 a 34 b 35 a 3



aG1 = m1g 1p


b




1 2

g m ma

m mminus

=+

1p


c


4p1 2

1 2

2m m gTm m

=+

2p


d

eF k l= sdot ∆ 2p

4p2Tlk

∆ = 1p




a





fFL = minus 1p

b

totalcE L∆ = 1p

4p

2

2cmvE∆ =




1p


56

c

util

consumat

LL



11 ctg

η =+ micro α

ctg xh

α = 1p


d

c totalE L∆ = 1p

4p

2

2cmvE∆ = minus

total fFL L= 1p

2

2mv mgd= micro

2

2vd

g=

micro 1p




57


Subiectul I Punctaj

1 d 32 d 33 d 34 a 35 c 3



a

1 1 1p V RT= ν 1p

3p

11

1

RTVp

ν=

1p


b


1 2

V VT T

= 2p

4p1 22

1

TVTV

=

1p


c


4p11

mV

ρ =

22

mV

ρ = 2p


d

( )12 1 2 1L p V R T T= ∆ = υ minus 1p


52VC R= 1p





58


a

11

NnV

= 1p

3p1 1 1p V RT= ν A

NN

υ = 1p

rezultat final 11

1

Ap NnRT


b

123 3 1U U U∆ = minus 1p

4p1 1VU C T= ν

3 3VU C T= ν 1p

( )123 3 1VU C T T∆ = ν minus 1p


c

43-4 3

3

ln VL RTV

= ν 1p


33-4 3

4

ln pL RTp

= ν 34 3 ln 2L RT= ν 1p


1p

d

cedat 4-1Q Q= 1p


1p

4p VC C R R= + = 1p




59


Subiectul I Punctaj

1 d 32 a 33 d 34 a 35 d 3



a

0

0p

R RRR R

=+

2p

4pe pR R R= + 1p


b


EIR r

=+

1p


( )0

A

E I r RI

Rminus +

= 1p


c

V eU R I= 1p

4pu r I= sdot 1p


d

A

EIr

= 1p

3p 0VU = 1p




a

21 1 1P R I= sdot


4p1

1

EIR r

=+

22

EIR r

=+

1p




60

b scEIr

=

2p3p


c

ext s

totala s

P RP R r

η = =+

2p

4p1 2sR R R= + 1p


η = = 1p

d rEP4

2

max =

2p




61


1 a 32 d 33 c 34 b 35 b 3



a1

1

1Cf

=

11

1fC

=

2p3p




c

112

111fxx

=minus

1p

4p2 2

1 1

y xy x

β = =

1p


d

1 2

1 1 1

sf f f= + 1p

4p2 1

1 1 1

sx x fminus = 1p







c0extL h= ν 2p

4prezultat final 14


dc sE e U= sdot 2p

4p




62

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj

1 a 0

0

FllE s

∆ =sdot

3

2 c 2

m4 ms 22 s s

va at

∆= rArr = =


3

3 b


= sdot =

3

4d

355 10 2750 N20

PP F v Fv

F

= sdot rArr =

sdot= =

3


cos ( sin )cos sin


α + micro α 3



a


3px24 = 3 ∙ (4 minus 2) 1p

x24 = 6 m 1p

b

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

=

2p

4p

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

= 2p

c

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

5p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1ptotal 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


d

total

total

9 m2254 s

m

m

xvA

v

=

= =

2p

3ptotal

total

9 m2254 s

m

m

xvA

v

=

= = 1p



63


a


3p2

0 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

=

1p20 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

= 1p

b


0

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus = 1p

4p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

c 056B

B

p

p

E mgh

E J

=

=

2p4p

056B

B

p

p

E mgh

E J

=

= 2p

d


4p024 J

f AB

f AB

pF B A

pF

L E E

L

= minus

= minus

2p

024 Jf AB

f AB

pF B A

pF

L E E

L

= minus

= minus 1p



64


Subiectul I Punctaj


2 d [ ]SI

J1mol K

C =sdot

3

3c 2

1

ln

1929 J

VL RTV

L

= ν

= minus

3


5 a A

NN

ν = 3



a

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

2p

4p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

1p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot 1p

b0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

c

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

2p

4p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

1p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr = 1p

d

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

4p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

= 1p



65


a

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

4p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

= 1p

b

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

2p

4p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

1p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr = 1p

c

L12341 = Aria12341 1p

4p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot 1p

d( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot

2p3p( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot 1p



66


Subiectul I Punctaj


2 a [ρ]SI = 1 Ω m 3

3 c q = I Δt 3

4 b 0 0 00

112 3 6eR R RR R= + + = 3

5 d 6 2003

UR RI

= rArr = = Ω 3



a2 3

12 3

11

e

e

R RR RR R

R

= ++

= Ω

2p3p

2 31

2 3

11

e

e

R RR RR R

R

= ++

= Ω 1p

b


4p


2

3

RR

1p

rArr 22

3

1 R

I RI =

+

1p

rArr 32

2 3

IRIR R

=+ rArr I = 04 A 1p

c

e

EIr R

= rArr+ 2p

4pe

Er RI

= minus 1p


d

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

2p

4p

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

1p2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =1p



67


a

Din legea Joule

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

3p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr = 1p

b

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1p

4p

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


c

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

2p

4p

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

1p2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω 1p

d

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

2p

4p

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr = 2p



68


1 c 1

SI

m ms s

cminus

= = ν sdot 3


3 c Negativă 3


3

5 a sin 2 141sin

in nr

= rArr = = 3



a 3p 3p

b

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

2p

4p

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p

c

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


2p

4p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

d

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

4p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1psistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr = 1p



69


a

00

814

0 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot

2p

3p0

08

140 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot1p

b 8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

2p

3p8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

1p

c

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

4p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ



d


34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

2p

5p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr = 1p



70

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj

1 a 32 a 33 b 34 c 35 d 3



a


2p4p


b


1 1 1 1

1

(2) ( ) ( )

G T m a G T m a


+ = rArr minus + =

+ + = + rArr + minus = +

1p

4p



1 2 1

( )2


minus += =

+ + +

1p


2 1 1 21 2

1 ( ) ( ) ( )2



+ +rArr 1 1

1

2 ( )2

m m m gTm m

+=

+ 1p

rezultat final 2

m02s

a = 25NT = 1p


1

4 ( )22

m m m gF Tm m

+= =

+ 2p

3p


d




2p

4p1

1

22

m mgNm m

=+

1p


1p


71



a




2p

4p2 2

90 m2 2

at F th gm

= = minus =

1p


1p

b


22

2

45 m2

2

FF g tv g t mm h hg

v gh


2p

4p



c



3pmax2v gh= 1p


d



4p2 maxL mgh= 1p





72


Subiectul I Punctaj

1 c 32 a 33 d 34 c 35 b 3



a


11

1

AA

A

pV RTNpV RTm NN

NN RTVpN


=

3p4p


b


11 A A

m N NmN N

microν = = rArr =

micro

1p2p


c


1A

NN

ν = 2 12 1

2 2A A

N NN N


3 3A A

N NN N

ν = = = ν 3p


1 2 3

2 36

amestecamestec

total


ν ν + ν + ν 1p


d



pν

= 1p


ν + ∆ + ∆= = 1p




73


a


3p 3p

b


4p


1

p Vp V RTRT

= ν rArr ν = 1p


2 11 2

3p p T TT T

= rArr = 1p


c


5p


2 22 2 3 3 3 2

3

32

p Vp V p V V Vp

= rArr = =

1p


2 3 2 1 12

3ln 3 ln 4800 J2



3

32

p TT Tp



1p


1p


1 2 2 3primit

L LQ Q Qminus minus

η = =+

1p3p




74


Subiectul I Punctaj

1 a 32 b 33 b 34 c 35 d 3




sc

E EI rr I

= rArr = 2p3p


b


4p


2 3 4

( )R R RRR R R

+=

+ + 1p



2 3 4

( )e

R R RR R RR R R

+= + +

+ +1p


c


EIR r

=+


eU IR= 1p

4p


1p

2 2 3 4 ( )U I R I R R= = +

1 2 5 1 5 2( ) U U U U I R R I R= + + = + + 1p

1 5

2

( ) eI R R RIR


d






75


a




2p

4p


W WIUt Ut

= = şi 22

23

W WIUt Ut

= =


2p

b


18URI

= Ω 22

9URI

= Ω 2p


1 2p

R RRR R

=+

1p



1 1 1WP R It

= = 2 21 2 2

WP R It

= = 2p4p



p

p

RR r

η =+ fie cu putil

consumat

R EPP EI

η = = 2p3p




76


1 d 32 c 33 a 34 d 35 a 3



a


1 1 1

0 4 4y x x x xy x x



2p4p


b



1 1 1 5Cf x x

= = minus = δ 1p

3p



c


= = 1p


2 1

1 1 1F x x

= minus 1p


1

FxxF x

=+

1p


d


FC

= = 1p


1

25 cm

F xXF x

= =+

1p





a



ν = = sdot 1p


1p


77

b



2extmvLε = + 1p


mε minus


c



LUe

ε minus= 1p


d



5p


1 4λ

λ = (1) 1p


exthcL h= ν =λ


hcL

λ = (2)


hcL

λ =

1p

1 1 11






78

TESTUL 11

A MECANICAtilde

Subiectul I Punctaj

1 b 2 2

2 22

m1kg kg m s2 sSI


2 b 3





m2 sin 10s



sdot sdot= = =

∆ ∆ 3



a


1p

4p



1 2 1 2



+ +

2p




c


1 2 0T Gminus + = 12 fG F= 2p


=micro

1p


d


1p

4p





79




Rezultă 42500 JAt

E = 1p

b


4p

2 2

2 2 f

B AG F


1p

Obţinem 2

2f

BF



c



sinhd =

α1p



1p


d




2p

4pObţinem 1 2

AtE

h sm g

=sdot sdot

1p




80


Subiectul I Punctaj

1d 1 1J kg K

SI

Q Jc cm T kg K


3


R Tp ρsdot sdot

=micro 3

kg8m

pR T

sdotmicroρ = =

sdot3



5 b 3



a


A

m NN

=micro

2p

4pRezultă 0A

mNmicro

= 1p



1p VT

Rsdot

=ν sdot

2p3p


c


4p


2

pR T

sdotmicroρ =

sdot

33

3

pR T

sdotmicroρ =

sdot 1p


3 1 1

2T TT T

ρ sdot= =

ρ1p

Rezultă 2

3

2ρ=

ρ 1p

d


Np V R TN

sdot = sdot sdot 1p

4p

Obţinem 1 12 4A

Np V R TN



1

24

Ap NnR T

sdot sdot=

sdot sdot1p

Rezultă 263

mol036 10m

n = sdot 1p



81


a


4pRezultă 3 11

22 R TV Vp



b


12 23tL L L= + 12 0L = 3

23 22

ln VL R TV






d







82


Subiectul I Punctaj


2 1J A sSI



sdot ∆= = = = Ω

∆

3

3 b 2 2

3 3eR R RR

Rsdot sdot sdot

= =sdot

3

2 3e

E EIR r R r

sdot= =

+ sdot + sdot3

4 d 3


WIR t

= =sdot ∆

50 mAI = 3



a





e e

EIR r

=+

= 5A 1p


b


2 32RI R I= sdot 1 2 3I I I= +2p

4p




1lR

Sρsdot

= 2

4dS πsdot

= 2

1 11 4


ρ sdotρ2p



4pJustificare

1

2 666Ate

EI I cresteR r

sdot= = rArr

+2p



83


a


4pRezultă 2bb

PRI

= 1p


b


1p

4pRezultă b

Rb

PU U sI

= minus 1p

RU R I= sdot

R

b

URI

= 1p


c




d

2R bP R I= sdot 2p

3p




84



= rArr = rArr =λ

3

2 d 3

3 c 22 1 1 1

1


2

i n in n n nr n r

= rArr = = sdot = 3

4 c 1 1 1 02 m = 20 cm5

C ff C

= rArr = = = 3

5 b 3




1 125 10

Cf mminus= =

sdot2p

3p

Rezultă 4C = δ 1p

b


2 1 2 1 1

1 1 1 1 1 1 fxxx x f x f x f x



4pobţinem 2

2 22

2fxx x ff xminus

= rArr =minus

1p



d


4pRezultă 1 2

1 1 100 10025 20

Cf f

= + = minus 2p




a


2p

4pRezultă 25 8

199

66 10 3 10 199 10450 10 4

minusminus

minus


sdot1p


1p


85


0

hcL h= ν =λ

2p3p


c

0eU L= ε minus 1p


eε minus

= 1p

Obţinem 19

19

(49 44) 1016 10

Uminus

minus

minus sdot=

sdot1P


d


2cmE eU ν

= =

1p


m mε minus

ν = = 2p




86

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj

1 b 32 c 33 b 34 a 35 a 3






12

m3 s

F mg maFa gm

minus micro =

= minus micro =

2p

4p1

12

m3 s

F mg maFa gm

minus micro =

= minus micro = 2p

c


2 2

2

cos 0

sin 0f

f

F FN F mgF N

α minus =

+ α minus == micro

2p4p


mgF micro= =

α + micro α 2p


4pRezultă 3 kgFm

g∆ = =

micro2p




2

2MvMgh = 2p

4p




2

0 cos1802c


4p


2p


2 2


2p

4p


2p



87


Subiectul I Punctaj

1 c 32 c 33 a 34 d 35 b 3




1p3p


micro= = 2p

b

Din pV RT= ν 2p

3p

rezultă 4molpVRT

ν = = 4 moli 1p

c


5p

1 2 2

1 1

2871 1 004(3) 4(3)300

U U TfU Tminus



2

1 2 2 2

11 1 1

1 1 1 4(3)

RTp p p TVk RTp p T

V

νminus


scade cu k 1p

d


4p1 0


micro= = =

micro2p



88


a


1 1 1

8atm 300K 1200 1200 273 927 degC2atm

p T pT T K tp T p

= = = = = minus = 1p

4p


Graficul

2p

b

1 1 1p V RT= ν 1p

3p1 1

1

p VRT

ν =

1P

v cong 012 moli 1p

c

323 2

2

ln VL RTV

= ν

1p



2 1


= = = = 1p

223 2 1

1

V ln 11088 JpL pp

= = 1P

d

1abs ced ced

abs abs abs

Q Q QLQ Q Q


4p2

12 23 2 1 21




025 25η cong = 1p



89


Subiectul I Punctaj

1 b 3

2


2 3E E EI I I

R r R r R r= = =

+ + +




1 23

1 2

3 3 2727A3 4

E I IIR r I I

= = =+ minus

3

3 d 3

4


RR R r

η =+ +



R rη =

+



3

5


2

8 8( ) 9 9 4m

RE EPR r r

= = sdot+


5 34

2

rr rR r

plusmn = =


3



a

12 3

12

1 AEI R Rr RR R

= =+ +

+1p

4p2 3

12

2 VBAR RU I

R R= =

+ 1p

22

2 A3

BAUIR

= = 1p

33

1 A3

BAUIR

= = 1p

b

22 2 2qI q I t

t= = sdot ∆

∆2p

4p2

2 A 30 s 20 C3

q = sdot = 2p

c



scEI

R r=

+ 1p


d

p

EIR r

=+ 1p

4p1 2 3

1 1 1 1 142pp

RR R R R

= + + rArr = Ω 1p




90


a


2LL

U UI RR R R=

+ + 2p


Randamentul este 2

2

1 05 50( ) I 2L

RIR R

η = = = =+

1p

b


2 290400J3U tW I Rt

R= = =



290400 003 kWh36 10

W = congsdot

1p


kWhC = sdot = = 1p

c


xxrS


lRS

= ρ


=


UI xRl

=+

1p

4p


2(1 )

UP xRl

=+

1p



max0 605Ux P WR

= = =


min 20163Ux l P W

R= = cong

1p


1p





91


1 c 32 d 33 c 34 d 35 c 3



a


2 1

n nx x

=



= minus

2p

3p


nminus

δ = = 1p

b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=



c


1 1

x yx y

β = = 1p

5p

Rezultă 22 1

1

xy yx

= 1p

Rezultă 2 11

fy yx f

=+

1p








lλ

= 2p3p



4kk Dx

l+ λ

= 2p4p



92

c



4pSe obţine

2Di

lλ

= 2p


d


i i∆ minus

ε = = 1p


ε = minus 1p

Rezultă DD

δε = minus 1p





94

TESTUL 1

A MECANICAtilde




aDeoarece N = Gn 1p


b


0

v vat t

minus=

minus 2p



c



d





a




f g= sdot =

sdot1p



max 2 2cm v pE

msdot

= = 2p3p


c




d


4obţinem rF

r

Ld

F= minus 1p




95





a



obţinem 0

0

p L SR Tsdot sdot

ν =sdot

1p


b


1 2

m m=

micro micro1p

4Dar 1 2 ν = ν + ν sau 1 2 1 2

1 2

m m m m+= +


Deci 1 2

2micro + micro

micro = 1p


c


0

m mm R TV

p

ρ = =sdot

sdotmicro

1p

3pDeci 0

0

pR T

sdot microρ =

sdot1p


d


4Deci 0max

0

p TTpsdot

= 1p




96


a


31

1 11

2 10 Pa

= 4 dm

9627 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

4


32

2 22

4 10 Pa

= 4 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p


33

3 33

4 10 Pa

= 8 dm

38508 K

p

Vp VT

R

= sdot

sdot= cong

ν sdot

1p


34

4 44

2 10 Pa

= 8 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

b


4

unde 34 V 4 3 4 35( ) ( )2


şi 41 P 1 4 V 1 4 1 47( ) ( ) ( ) ( )2



c




d


primit

LQ

η = 1p


Deci 12341

12341 cedat

LL Q

η =+ 1p




97





a


1 1 1

Pr r r= + 1p

3pObţinem 1 2

1 2P

r rrr r

sdot=

+1p


b


1 2P

E r E rEr r

sdot + sdot=

+1p


P

S P

EIR r

=+

1p


Rezultă 05 AI = 1p

c


2

P

P

EIR r

prime =+

3p4


d


4Deci 2 22

2

E I RIr

primeminus sdot= 1p




a






98

b


4

unde S

S

EIR r

=+

1p



RP




S

RR r

η =+

2p3p

Rezultă 50 η = 1p

d



P

EIR r

prime =+ 1p





99

DOPTICAtilde




a


1 1 1x x f

minus = 2p

4Obţinem 1 21

1 2

x xfx x

sdot=

minus1p



1

1Cf

= 2p3p


c


1 1 1 x x f

minus =prime 1p

4

cu 1 2

1 1 1 f f f

+ = 1p

obţinem 12

1

f fff f

sdot=

minus sau 2 2

22 2

x xfx x

primesdot=

primeminus 1p


d


1 1

y xy xprime prime

β = = 1p

4


1 1

y xy x

β = = 1p

Deci 2 2

2 2

y xy xprime prime

= 1p

Rezultă 2

2

12yyprime

= 1p



100


a


3Obţinem xiN

= 1p


b


Dil

λ sdot= 2p

4Obţinem

2l iD

sdotλ = 1p


c


2Dx

lλ sdot

=1p


42

Dxl

λ sdot= 1p



2Dx x x

lλ sdot

∆ = minus =1p


d


2Di

lλ sdot

= 1p

4unde 1

1 cn n

λλ = sdot =

ν1p

Deci 1 12

D inl i i

λ sdot= =

sdot1p




101

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj


2gt hh t

g∆

∆ = rArr = = 3p

2 b pF F t pt

∆= rArr ∆ = ∆

∆

3p

3


L = 22 J 3p



α 3p



b



2p

3p

2 2

1 1

G T m aT G m a

minus =minus = 1 2

g mam m

∆rArr =

+ 2p

c



+ = = 1p


2 1

08 kg01 kg

m mm m

+ =minus =

2p



2p

4p

1 2( ) 126 kg ms= + = sdotp m m v 2p



102




2p3p


b

GL mg h= ∆

1p


2

2atx = 1p


45 JGL = 1p

c



4p

21 2 2c f

h xE mg F= minus

1p

21 (sin 03)

2cmgxE = α minus

1p

1 12 JcE = 1p

d

22 2

1 12c

cc

p p EEm p E

= rArr =

1p

4p



2 24 JcE =

1p

2

1

2pp

=

1p



103



1 2

1 2

m m+micro =

ν + ν

1 1 2 21 2

1 2

A A

N Nm mN N

N N

micro micro= =

=

1 2

2micro + micro

micro = b

3p


1

1 1 ii i f f f i

f

VTV T V T TV

γminus

γminus γminus

= rArr =

1

1iV i

f

VL C TV



c

3p

41 1 2 2

1 32 2 5

1 1

3 3

p V p V

V VV V

γ γ

γ

=

= rArr = b

3p

52 2

1 1

22

1

1 1

065 260 K

Q TQ T

T TT

η = minus = minus

rArr = rArr =

b3p



a

1 2

1 1 2 2( )

m m m

m p V p VRT

= +micro

= +3p

4p


b


4pRezultă 1 1 2 2

1 2

p V p VpV V

+=

+2p

5 28 10 Nm3

p = sdot

1p

c

mpV RT=micro 2p

4p1 21 2pV pVm m

RT RTmicro micro

= = 1p



104

d

2 22

2

m RTp V

=micro 2p

3p5 2

2 241 10 Nmp = sdot 1p



a


3p 3p

b

3 31

1


11

186 10 NmmRTpV

= = sdotmicro 1p

4p

3 312 2 12 4 10 m

2V V minusρ

ρ = rArr = = sdot

1p

5 21 12

2

93 10 Nmp VpV

= = sdot 1p

2 1

212 1

1

ln 25915 J

T Tm VL RT

V

=

= =micro

1p

c

13 450 K

2TT = = 1p

5p

11

3 323 2

3

224 10 mTV VT

γminusminus

= = sdot

1p

5 233

3

083 10 NmmRTpV

= = sdotmicro 1p


4 44 40 Kp VT

mRmicro

= = 1p

d

min

max

1CTT

η = minus 1p

3p4

1

1CTT

η = minus 1p

95Cη = 1p



105


Subiectul I Punctaj


21

2

1 A

AA

A

EI Rr R

EIR R r

= rArr = Ω+

= =+ + a

3


c3

4 22

225 W4

ER r Ir

EPr

= rArr =

= = c

3

5( 1) 10 ka V a

V

UR R n n RU

= minus = rArr = Ω


3



a

e

UIR

= 3 4 15sR R R= + = Ω 1p

4p1

1

5sp

s

R RRR R

= = Ω+ 1p

2 10e pR R R= + = Ω 1p

1I A= 1p

be

EIR r

=+ 2p

4p

2eEr RI

= minus = Ω 1p

c

2 3

2 3

25pR RR

R R= = Ω

+ 1p

4p

1 10s pR R R= + = Ω 1p

4

4

5se

s

R RR R R

= = Ω+ 1p

12 A7e

EIR r

= =+

857 VeU IR = = 1p


106

d


1p

4p1 2

2 1 4

1 26 A7

s

I I I I R I R

I I

= +=

rArr = =

2 3 4

3 2 4 3

3 46 A

14

I I I I R I R

I I

= +=

rArr = =1p

1 4AI I I = + 1p

9 A7AI = 1p



a


4p1 3 1 1 2 3( )P P r R R R R= rArr = + + 1p

( )21 2 1 1 2 3( )R R R R R R+ = + + 1p


b

2

max 84EP r

r= = Ω 1p

3pPmax = 450 W 1p

1 2

1 2

05R RR R r

+η = =

+ + 1p

c

2 Ann

n

PIU

= = 1p

4p1( ) 6 AnE I R r U I= + + rArr = 1p


4 15n

R

URI

= = Ω 1p

d

P UI= 1p


432 WP = 1p



107

DOPTICAtilde

Subiectul I Punctaj

1

2 22 1

1 1

12

1

2

53

23 3

y x y yy x

fxx ff x

fy

β = = rArr = β

= =+


3

22 5π

∆ϕ = δ = πλ c 3

3 375 nmaan

λλ = = b 3

4 206 nmexex

hc hcL eUL eU

= + rArr λ = =λ +

a 3

5 00


c 3



a


4

2( 1)C nR


( 1)2

CRnrArr minus = 1p

15n = 1p

b

2 1 2 1

1 1 1 1 1 Cx x f x x


3p12

1

15 m1

xxCx

= =+


2

1

5xx

β = = minus 1p

c



4

12( 1)

l

nCn R

= minus 1p

1 12

lnn C R=

+ 1p

163ln = 1p


108

d


4p

13 1 1 3 1

1 1 1 1 1 Cx x f x x


13

1 1

025 m1

xxC x

= = minus+ 1p

3

1

083xx

β = = 1p



a




2kk ix x iminus

= rArr = 1p


b

2Dil

λ= 1p

3p2liD =λ

1p

142 mD = 1p

c


4p

2 ( 1)klx e n kD




Pentru 33

1kx i en

λ= rArr =

minus 1p

42 me = micro 1p

d

1 2r vk kδ = λ = λ 1p

4p2 12k k= 1p

1 21 2k k= = 1p



109

TESTUL 3

A MECANICAtilde




a

1

2

AI dtgIB d

α = = 1p

4p1 0 =d v t

2

2 2atd = 2p

Rezultat final 2 tg

=sdot α

vta

15s=t 1p

b

2 0v v at at= + = 1p

3p1

21 1 135 kJ2

= =cm vE 1p

2

22 2 90 kJ2

= =cm vE 1p

c1 1= ∆ = minus

fF c cL E E 2p3p


d







a

0T F G+ + = 1p

4pcossin

T mgT F

α =α =

1p


α=

αF mg

3 N 09 N2

= congF 2p

b cosmgT =

α2p

3p



110

c


3p=AE mgh

2

2BmvE = 1p



d


4

p mv= 1p

final2mv mv= 1p


= =vv 1p



111





a

pV RT= ν 1p

4pν =

microm m

Vρ =

1p

p RTmicro = ρ 1p


b

1 2 3 1 2 3( ) ( )p V V V RT+ + = ν + ν + ν 1p

4p

1 1 2 2 3 3

1 2 3

p V p V p VpV V V

+ +=

+ + 1p

173pp = 1p


c

1 1 1 ∆ = minusU U U 1 1 1 15 5 2 2

= ν =U RT p V 1 15 2

U RT= ν 1p

4p1 1 1 1

5 2

= ν =pV RT U pV

1 1 1 15 2

= ν =pV RT U pV 1p

1 1 1 1 15 10( )2 3



d

1 1 2 2 3 3

1 2 3


ν + ν + ν p

3p1 2 34 9


micro = 1p




112


a


4p

2 1 1 1 2 13 3= =p V p V p p 2 1 1 1 2 13 3= =p V p V p p 1p

1 1 1 1 12 2 4 4L p V p V RT= = = ν

14LRT

ν =

1p


b


4p

12 1 13 2 32


12 34LU∆ = 1p


c


3p23 1 15 6 152



= sdot =Q L 1p

d

P

LQ

η = 1p



92PQ L= 1p


η = = 1p



113








c


2 2

23

U rU r

= = 1p

5pK icircnchis 1 11

1 1

=+p

r RRr R

2 22

2 2

=+p

r RRr R

11 2p p

U R RI

= = 2p


d

1

1G

p

U IR

= 2p

4p1 11

1 1

12 k= = Ω+p

r RRr R 1p




a

2

b

UPR

= 2p3p


b


5p

2 A= =bPIU

1p



UnE I r

1p


c4 (4 )= + +b b aE I r R R 2p


d

2

max 4

=EP

r cacircnd R r= 2p

4p




114

DOPTICAtilde



b 2 1 1

1 1 1x x f

minus = 2p3p


c


4p2 1 2

1 1 1x x f

minus = 1p


d



1 12

1 1

x fx x f

=+ 1p





2Dil

λ= 2p

3p


b


4p2 2 4d i i i= + = 2p


c


4p

2 π∆∆ϕ =

λr 2

3sdot λ

∆ = =y lr

D

Rezultă 2 3π




II

= 1p


115

d


21

15 cm= =+

x fxx f




1

22 sdot=

minusl xl

x

2p

4p


λ= 1p




116

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1

b8

3

m 3 10 s

1 al 631 10 ua

1 an

c

dc dt

t


∆ ∆ =

3

2 d 3

3

a

m m

vF m a m

t∆

= sdot = sdot∆

m5s

v v∆ = =

= 25 NF

3

4

c

fR N F= +




θ = = micro

3

5


2

22 2

2 42

m v F dv v

m v F d


3



a

1

1

0const2m

d va vt

+= rArr = =

∆ 2p

4p11

1

2 100 m sdv vt


∆ 2p

b

22

2

50 sdv tt

= rArr ∆ =∆ 2p

4p

1 2 110 st t t t∆ = ∆ + ∆ rArr ∆ = 2p


d11 2 727 m sm m

d dv vt

minus+= rArr = sdot

∆ 4p 4p



117



b


0

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

2p

4p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

1p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p 1p

c


1 1 231 msin

hd d= rArr =α

1p


( )

2

2 2

2

2 2

0 sin - cos 2

237 m2 sin cos

ABAB R

m vEc L m g d m g d

vd dg



2p

1 2 868 mD L d d D= + + rArr = 1p

d


4p


2 2 1p

2 = 09 m 1p

pp

Ek xE xk

x


1p2

= 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr =

1p



118



11

1 1 2

2 2 12

2

1 8

A

A

m NN N

m N NN


micro

3

2

d

0 V

Q U LL U C T

Q= ∆ +



= 12465 JL

3

3

c

12 12

12 12

1213 13 13

13

12 13

31 2 32

5 61 3 80 1202 5

V

V

RC C RQ R T

Q C TQQ C T R T KQ

T T


3


3

5


2

1

1C

p

LQ

TT

η =

η = minus


3



119



3p 3p

b51

1 1 1 1 15 10 PaR Tp V R T p pV


c


1p

5p

0const 0 0 1i f

QV L U U U pQ U L

= = rArr = rArr ∆ = rArr == ∆ +

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p 1p

d ( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

2p

4p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

1p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p 1p



a

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

1p

4p


1

2 3 3 23

53 51 1 1 1

1

2 = 8 = 15

5 3

p

v

p V V pVp V V pVC

C

γγ

= rArr = rArr =

γ = =

1p

13 1 3

1

15 15 374 lR TV V Vp



120

b

2p

4p

( )12 2 1

12

37395 J

VQ C T TQ


=2p

c

1231 12 23 31 1pL L L L= + +

12 0 L =1p

4p

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

( )

( )

23 23 3 2

23 1 1 13 315 22 4

vL U C T T

L R T T R T



1p

23 186975 J 1pL =


31 12465 J 1pL = minus

1231 62325 JL =1p

d


2 3 2 3

2 3 2 3

A A A

B B B

Q Q QQ Q Q

= += +

32 2

2

lnAVQ R TV

= ν sdot sdot sdot


32 2 2

2 3

ln ln BB

V VQ R T R TV V





( )2 3 23

ln 0BVR T R T TV


( )2 3 2 2 3 23

1

2 3 2 3

4ln ln 3

42 ln 15 2 0 3

B

B A

VR T R T T R T T TV

R T

Q Q



rArr gt

3p 3p



121


Subiectul I Punctaj

1

b

1

2

21

2

156

lRl LLR

l lR lR L

= ρ sdotsdot

= ρ sdotsdot

= =

3

2

a U R I= sdot

2

lRmS R

m m d Sd ll S d S


sdot sdot

2 mU Id S

= ρ sdot sdotsdot

= 3 VU

3

3


2 WRI t

=sdot ∆

1 R = Ω

3

4



02 AI =

3

5

b2

max 4

p

p

EP

r=

sdot

1 2

1 2

6 12 5p pr rr r

r rsdot

= rArr = Ω = Ω+ 1 2

1 2

26 Vpp

p

E E E Er r r

= + rArr =

max 14083 WP =

3



a

1

3 1 1

075 A

EI THORN IR R r

= =+ + 1p

3p1

1

1

45 C 1p

R

R

qI p

tq

=∆

=

1p1

1

1

45 C 1p

R

R

qI p

tq

=∆

= 1p

b

2

3 2 2

06 A

EI IR R r


4p2 2 1

84 1U E I r pU V p

prime= minus sdot=

1p2 2 1

84 1U E I r pU V p



122

c


( ) ( )( )

1 1 2 2

1 1 2 2

90 19p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1p

4p1 2

1 1 2 2

18 V 19

pp

p

E E E Er R r R r


3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= =

1p3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= = 1p

d



( )3

3

2 1

1 1 1 1

1

2

115 106

R

R

I I I

E I R r I

I AI A

+ =

= sdot + +

==

2p

4p

1 1 2 2 1 1883 1

AB

AB

U I R I R pU V p


1p1 1 2 2 1

1883 1AB

AB

U I R I R pU V p




a


1 2 1 2

085 A

E EI IR R r r

minus= rArr =

+ + +3p 3p

b


3 2 2

2 A 1p

EI IR R r


1p




123

c

[ ] 015 mint isin ( ) 1 1 1

685 V

AB

AB

U E I R rU

= minus sdot +

=1p

6p


1p


( ) ( )( )

1 1 2 2

1 1 2 2

079

p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1 2

1 1 2 2

7 V

pp

p

E E E Er R r R r

= + rArr =+ +

3

25 A 1p

p

p

EI I


+

1p


2p

d 3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


=

2p3p3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


= 1p



124

DOPTICAtilde

Subiectul I Punctaj

1



02

2

sinsin n irnsdot

=

2sin 04r =


3

2


1 0

1 1ln gf n

= minus sdot

2

1 1l

a

n gf n

= minus sdot


( )2

1

14a l

l a

n nff n n

sdot minus= =

minus

2 1 4f f= sdot

3

3 c 3

4

d


1

22

2

2

2

extr

extr

m vL

m vL

sdotε = +

sdotε = +


3

5

c

2

11 2

2k

Dx kl

λ sdot= sdot sdot

sdot 2

2 2k

Dx kl

λ sdot= sdot

sdot

22 1 k k

d x x= minus

( )1 2 2 - 02k Dd

lsdot


3




b


4p



0 01 1

1

sin 2 cmsin

r rr d dd r

= rArr = rArr =

1p



125

c

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot

3p4p

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot 1p

d


4p





2n = = 1p




1

13 1 3 3

2Dx ilλ


sdot 2p

3px3λ1

= 36 mm 1p

b

x4λ2

= 36 mm 1p

4p

x4λ2

=i2 1p


12

2

3 4

450 nm 1p

sdot λλ =

λ =1p

c

3

1

3 1

34

13 3

4 3

4 2

3 360 nm2

DxlDxl

x x

λ

λ

λ λ


sdot =

3p4p


d1 1

14 4 4 6 mm

08 2Dx x

lλ λ


4p 4p



126

TESTUL 5

A MECANICAtilde




a


0N Gn Gt Ff+ + + =

1p


2p

1p

b




1p




c


N T T= + 1p

4p( ) 306BT m a g N= + = 1p

( )( )2 22 1 cos 90N T= + minus α 1p

5911N N= 1p

d


3p( )2

2B B


072m sBv 1p



127


a




b




2B



c


2

cc

mvEc p mv= = 2p


2 2C B

Ffmv mv Lminus = 1p

Rezultă

6 m sCv = 12 m sCp = 36 JCEc = 1p

d


max

2 2Cmv kx

=


4p

max cmx vk

= 1p




128






HemNamicro

= 2p3p


b



c


1p

4p


1p

Obţinem

1 1 1

2 2 2 1 2

1 2 1 2


ν= ν =gt ν = ν =

ν + ν = ν + ν


d





a


1 2

p pT T

= 1p


3 4

p pT T

= 1p


T2 = T4 = 1 3 360KTT = 1p

b

LTOT = L12 + L23 + L34 + L41 1p



129

c

TOT

p

LQ

η = 1p



d

min

max

1 TT

η = minus 1p

3p1

3

1 TT

η = minus 1p

30η = 1p



130





a


1 2 3

1 2 3

1 2 3

36V1 1 1

E E Er r rEp

r r r

+ += =

+ +

1p

3p

1 2 3

1 1 1 1 1pp

rr r r r

= + + =gt = Ω 1p


p

EI

R r= =

+A 1p

b


1 2

1 1 1

2 2 2


= +


2p

5p


11 2 1 2

E r E E RI


=sdot + sdot + sdot

2 1 2 12

1 2 1 2


+ minus=

+ + 2p


c

Deoarece

ABU I R= sdot

1 2I I I= + 1p

4p1 2

1 2

1 2 3

1 1 1AB

E Er rU

r r r

+=

+ +2p

UAB = 40 V 1p

d

Icircn relaţia

1 2

1 2

1 2

1 1 1AB

E Er rU

r r R

+=

+ +1p

3p


vRrarr 1p




131


a


4p



sdot=

sdot


U tminus sdot

=sdot

2p

5AQIU t

= =sdot

1p

b


4pRezultă 2pQRI t

= 1p

Rp = 22 Ω 1p


3pE = 120 V 1p

d


x p

x p

R Rr

R R=

+

2p

4p

px

p

R rR

R r=

minus 1p



132

DOPTICAtilde




a



2 smv e U= sdot 1p

4p11

22

S ex

S ex

ch eU L

ch eU L

= + λ = + λ

2p


b


11

S exch eU L= +λ

2p

4p

11



c


4p0exLh

υ = 1p


d2

2

cW Nh Nh= υ =λ 2p

3p




a1 2

11 11e

m

fnn R R

=

minus minus

2p3p


b


2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2p


minus x1 = 80 cm 1p


133

c


1 22 1

1 1 1d x xx x f

= minus + minus =

1 22 1

1 1 1d x xx x f



11

1y xy x

β = = = minusminus

2p


d


1 2

2

1

2 1

1 1 1

2

1 1 1

F f fxx

x x F

= +

minusβ = =

minus

minus =

2p4p



134

TESTUL 6

A MECANICAtilde




atg 1

3tg ϕ = micro = 2p

3p

30ϕ = 1p

b

t fG Fa

mminus

= 1p

4psin cosmg mga

mα minus micro α

= 1p


2310 =577ms3

a = 1p

c



1p

4pt fG F F= + 1p


310 3 10 100 N3

F = sdot = 1p

d



1p

4p

t fG F F+ = 1p

sintG mg= α 1p

3 1 1 100 3 100 3 200 N2 23




135


a

sin 5

1000m54 =15ms3600s

hpl

v

= = α =

= sdot1p


cosUcirc α


rArr sintg cos


α1p



4 352 sin 2 10 15 15 10 W=15kW100


b


4psincos


α1p

7500W=75kWP 1p


2p3p



3p25kJpE E∆ = ∆ = 1p



136





a

1

1 1 2 1 21

22 1 2 2 1

2

2 540 2 54 4 083 450 3 45 5

pVV T V TRT

pV T V V TRT


ν2p

3p

1

2

4 085

ν= =

ν 1p

b

1

1 1

22 2

p VVRT

p V VRT

ν= =

ν 2p

3p

1

2

08VV

= 1p

c

1 1 11 2 1 1 2 2

2 2 2

( ) ( )pV RT

p V V R T TpV RT

= ν rArr + = ν + ν= ν

1p

5p

1 1

1 2 1 22 2

( ) ( )

p V RTp V V RT

p V RT

= ν rArr + = ν + ν= ν

1p

1 1 2 2 2 2

1 2 2

5 08 450 540 ( ) 3 18

p T Tp T T


ν + ν sdot ν 2p


d

1 36lV =

12

2

2 3 36 3 18 54l3 2

V VV


1p

4p 1 2 1 2

1

1 22

9

08 08

V V V V lV V VV

+ = + =

= rArr =2p




137


a 4p 4p

b

212 1

1

41 1 4 1 4

ln 0

5( - ) ( ) 02V

VQ RTV

RQ C T T T T

= ν gt


4p

12 41 229356JprimitQ Q Q= + = 2p

c

23 4 1 4 1

134 4

2

5( ) ( ) 02

ln 0

VRQ C T T T T

VQ RTV


= ν = lt2p

4p


d1 cedat

primit

QQ

η = minus 2p3p

3043η = 1p



138





a

2AP R I= sdot 2p

3p25 WP = 1p

b

0

0 0

0

A

A

A

I I IxI R I Rl

l xE I R I Rl

= + = minus

= +

3p5p

05 mx = 2p

c

ee

E EI RR I

= rArr = 1p

4p0

00

05A

A

A

I I IR lI Ix R

= +sdot

= =sdot

2p

151 151eI A R= = = Ω 1p


3p2250Wh 225 kWhW = = 1p



139


a


3p200 4A50

totaltotal

PP U I IU


b



5p



1p

21

1 1 1 2

22

2 2 2 2

40 102 4

2

60 152 4

2

I PP R RI

I PP R RI

= rArr = = = Ω

= rArr = = = Ω

2p

10VABU = 1p



d

1

1 01 1 3

0

191 1800 C

5 10

RR Rt tR minus


α sdot

2p

4p

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot= t2 =

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot

2p



140

DOPTICAtilde




a

1 2


n ff R R


2p

4p

1 2


n ff n R R


2p

b

2 1

1 1 1

aerx x f= + 1p

4p2

24 cm 48cm5


2

1

xx

β = 1p

04β = 1p

c

2 1

1 1 1

apax x f= + 1p

4p

2

96 cm 192cm5


21

xx

β = 1p

04β = 1p

d2p

3p




141


a

412 10 m 12mm2Dil


3p


b

λapăapatilde nλ

λ = 1p

3piapă

12 3 09mm4apatilde

iin

= = sdot = 1p


c



2p




2p

10nouk = 1p

d






142

TESTUL 7

A MECANICAtilde


3


md dvt t t

= =+

Cum 112 6

d dtv v

= = şi 222 2

d dtv v

= =

6 154

6 2

md vv vd d

v v

rArr = = =+

a

3

4


∆= = minus

∆






Icircnlocuind 35

rArr micro =c

3

5


11 2

2mvmgh v gh= rArr =




∆ b

3



a


3p 3p

b


4p



Randamentul este 2

u

c

L mghL Fl

η = = 1p


l= =

η1p


143

c


0fG T N F+ + + =


2p


lα =

Rezultă 2fhF F mgl

= minus

2p

600 NfF = 1p

dcos

cosf

f f

FF N F mg


α 2p3p

065rArr micro = 1p



a


cE L∆ = 1p

4p



1p

Ff = μN = μmg

Rezultă 2 2

200 2 21 ms 458ms


2p

b



1p


1

2 2mv mv mgR= + 1p


1 11 ms 331msv = = 1p

c



max max2 2mv vmgh h

g= rArr =

2p3p

max 105msh = 1p

d


max 081 m2vh fh fg

= = = 1p



222 2


2p

2 204 v m s= 1p



144




1 2 2 1 1 21 2 2 3 2 3 2

2 3

( )( ) 150 J ( ) 200J 0752

p p V V LL L p V VL

rarrrarr rarr

rarr


c

3

5


max

1CTT

η = minus


1 max 1

1L T LQ T Q

η = rArr minus =


3 150 K8

TT = =d

3



a


1 1

0 1

V VT T

= şi

2 2

0 2

V VT T

= 1p

4p

dar 1 2V V V+ = şi 2 1 12


23VV = iar

1 2 2VV V= = 1p


32TT = şi 0

234TT = 1p

Rezultă 1 4095KT = şi 2 20475KT = 1p

b


4p

1 1 ( )

2 2l lV S V x S= = +


2 2 ( )2 2l lV S V x S= = minus 1p

Obţinem 50 00 1 1 2

500 2 2 0 2

2 N( ) 066 102 2 2 3 m

N( ) 2 2 102 2 2 m





1p

51 2

52 2

N066 10m

N2 10m

p

p

= sdot

= sdot1p


536 NF = 1p

d


mp V RT=micro

şi 2 2 2 20 2 2

1 1 1

m V m Tp V RTV m T

= rArr =micro 2p

5p

Dar 0 01 2

3 32 4T TT T= = şi 1 2

23 3V VV V= =

1p

2 2

1 1

2m Vm V

rArr = 1p

2

1

2mm

rArr = 1p



145


a


1p

3p


2p

b


1 2 1 2


= rArr = rArr = 1p


2 3

p pT T

=


31 13 3

p p ppT T

rArr = rArr =

2p

5 23 10 Nmp = 1p

c


4p1 1

32 32



1 2 2700JU rarr∆ = 1p

d

1

LQ

η =



1p



44 1 4 1 3

p p p p TTT T T T

= hArr = rArr =


2p


1 1

4 2 02218 9

p Vp V

η = = = 22rArr η = 1p



146



41 2sR R= 1 1 1 2

2 3pp

RRR R R

= + rArr = 22 53 3sR RR R= + =

1 1 23AB

RR R

= +


b

3



a

1 2R R R R= + = 1p

4p

22 1

1 1 1 3 22 3

RRR R R R

= + = rArr = 1p

3 22 53 3R RR R R R= + = + = 1p

3

1 1 1 3 1 85 5

5 6258

AB

AB

R R R R R RRR

= + = + =

= = Ω1p

b


UU IR IR

= rArr = = 2p


( ) 264ABAB

EI E I R r VR r

= rArr = + =+

2p

c




AB

EIR r

= =+

1p





d

2

max 4EP

r= 2p

3p

max 8712WP = 1p



147


a


2 4 1

3 4 5

00


minus minus =

2p


1 3 1 1 3 1 2 2

2 3 2 4 4 2 2

2 3 2 5 5



3p



4p19MNU V= minus 1p

c

24 4 4P I R= 2p

3p4 8P W= 1p

d2

1 1 1W I R t= 2p3p

1 10800 J 108 kJW = = 1p



148

DOPTICAtilde


680nma a aa

nnλ

λ = rArr λ = λ =

d 3

2 c 3

3



2MO PO AOMOAO PO

= rArr =

2ON PO O BONO B PO

= rArr =


=09 m2 2


b

3

4


k Dxl

λ=


2

1 1 mm222 2 mm2

Dk xl

Dk xl

λ= rArr = =

λ= rArr = =

c3

5


2 1 1 2

1 1 1 x xfx x f x x

minus = rArr =minus


1

x x xx

β = rArr = β 1 1

11 1

x xf βrArr = =

minus β minusβ


20 10 cm1 1

f minusrArr = =

minus minusb

3



a



1 2

1 1 1 1( 1)( ) n nC n RR R R C


R = 10 cm 1p

b

Din 1 1 20C f cmf C

= rArr = = cm 1p


2 1 1

1 1 1 fxxx x f f x

minus = rArr =+

2p


x minus= =

minus2p


149

c1 2d x x= minus +

2p3p


d


1 1

y xy x

β = = 1p

3p22 1

1

xy yx

= 1p

2 2 cmy = minus




a


1 11 1

2 12 2

( )(1 )

c ch L eU h L eU

c ch L eU h L eU f


2p

5p

11

1

2

08

ch LeU

c eUh L

minusλ

=minus

λ

1 2

1 2

0802


λ λ2p



LL hh


1410 HzL = 1p


1

ch hε = ν =λ

2p3p

191 011 10 J 006eVminusε = sdot = 1p

d


11

22

2

2

2

e

e

m v eU

m v eU

=

=

2p

4p1 1 1

2 2 108v U Uv U U

= = 1p

1

2

54

vv

rArr = 1p



150

TESTUL 8

A MECANICAtilde




a


4p2

1 2atx = 2 0x v t= 2p


= = 1p

b0v v at= + 2p

3p16msv = 1p

c


3prezultat final

-=fracircnare

fracircnare

va


d

1 2= +d d d 2p

5p1 0=d v t 1 1 2= = = 64d x x m 64 m 2p


2 2 fracircnare

vdaminus

= 2 = 32d m




aE mgh= 2p


b

c totalE L∆ = 1p

4p

2

2cmvE∆ = = +

ftotal G FL L L 1p



1p



151

c

c totalE L∆ =

1p

4p 0cE∆ = GL mgh=

f


rezultat final = -f

totalFL mgh = -4500

f

totalFL J 1p

d

= +f f f


4p2f


F oprire

F FL d

+= minus 1p







152





a

pV RT= ν 1p

4p1 1 1 2

2 2 2 1

p mp m

ν micro= = sdot

ν micro2p

rezultat final 1

2

74

pp

= 1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0


4pVf = 2 V 1p


d




4p 4p



a1 1

1 1 11

p Vp V RTRT

= ν rArr ν = 2p3p


b


4p

1 2

1 2 12 1

2 1

34

V

p pV ctT T U C T

T T


=2p




153

c

min

max

1 TT

η = minus 1p

4pmin 1=T T 1p

max 2 1= = 4T T T 1p


η = = 1p

d

P

LQ

η = 1p

4p


1


1-2 2-3 3-1L L L L= + + 1-2 0L = 32-3 2

1

ln VL RTV

= ν 3-1 1 3( )L R T T= ν minus 1p


minusη =

+ 25η 1p



154





a


( )1 1 1 3E I R R= + de unde 1 3AI =2p


b



2422

131 )()( 1p


c


2p4p


d


2p3p




a



bP1 + P2 = Rp I

2 2p3p


c

echivalent

echivalent

RR r

η =+ 2p



d


4p


EIR r

=+

1p




155

DOPTICAtilde




b

2 1

1 1 1x x f

minus = 1p

4p2

1

xx

β = 1p


c

21 CCCsistem += 1p


2 1

1 1sistemC

x xminus = 1p


d

1 2= -x d x 2p

4p

1 22

1 2

x fxx f

=+

1p




a


3p11 10

yi = 1p


b


2i 1p

4p1

12 22id i= + rArr d = 3i1

2p



156

c

22 29

2iy i= + 2

22 072 mm19yi = = 1p


1 2Dil

λ= 2

2 2Dil

λ=

1 22

1

ii

λλ = 1p


d


iin

= 2p


1 1

1

-025 i ii

= 1p


n = 1p



157

TESTUL 9

A MECANICAtilde


5

21 1 2 2

12 1 2

2

2 2m 1156s

m v m v

mv v vm

=

= =3



a 3p 3p

b

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

4p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

c2 2

08 NT m g m aT

= minus=

2p3p2 2

08 NT m g m aT

= minus= 1p


158

d

1 0 1 1 0

1 1 0

1 1 0

( ) 0

( )g 0 ( )


+ + + + + =

minus + == +

1p

5p


1p

2 2

2

2

0 0

m G TOy G Tm g T

+ =minus =

=

1p

2 1 0

2 10

( )m g m m gm mm


=micro

1p

m0 = 03 kg 1p



aEA = m1gh 2p

3pEA = 1 J 1p

b

ndashf AB

f AB

B A F

B A F

E E L

E E L

=

= +2p

4p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p

c


4p

( )( )

1 1 1 2 1

1 1 1 2

1 1

1 2

m v m m v

m v m m vm vv

m m

= +

= +

=+

1p

unde 1m25s

v = 1p

01 255 m08503 s

v sdot= = 1p

d


( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

2p

4p( ) 22

1 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

1p

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

= 1p



159



VU VC T∆ = ∆ b 3


3pRT

microρ =

[ ] 3

kgmSi

ρ = c

3

41 1

1 1

33 22

L p VLU p VU

=

∆ = rArr =∆

a

3

5 36 gA A

m N Nm mN N


a 3



a15 mol

mν =

microν =

2p3p

15 mol

mν =

microν = 1p

b

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

2p

4p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

1p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ = 1p

c

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

3p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot 1p

d

2 2

3 3

mp V RT

m mp V RT

=micro

minus ∆=

micro

1p

5p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot 1p



160


a

L1231=Aria1231 1p

3p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot

1p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot 1p

b

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

5p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot05p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot 05p

c

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p

3p

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot 1p

d

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= =

1p

4p

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= = 1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η =

1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η = 1p



161




d 3

2 2W RI t= ∆ b 3

3 [ ] 2 1SI

V1 11 N m A sA


4

( )

( )

Aria trapez2

200 300 2500 500 mC

2

b B hQ

Q Q

+ sdot= =

+ sdot= = rArr = a

3

5

( )0

2

0

0

1

288

2110 C

R R t

UR RP

R Rt tR

= + α

= rArr = Ω

minus= rArr = deg

α


3



a


3p

1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω

1p1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω 1p

b

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

4p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω 1p

c

Potrivit legii Ohm

22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

2p

4p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

1p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

= 1p


162

d

1p


( ) ( )2 4

2 1 2 4 3 4

2 4

I I II R R I R R

I I

= +

+ = +

rArr =

1p( ) ( )

2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr =

1p

( ) ( )2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr = 1p



a

1 2

1 2

1 2

1 1 1

07 A

E Er rI

RR r r

I

+=

+ +

=

3p

4p

1

2

2E EE E

==

1

2

2r rr r

==

1p

bU = R I 3p

4pU = 7 V 1p

cP = RI2 3p

4pP = 49 W 1p

d88

e

RR r

η =+

η =

2p3p

88e

RR r

η =+

η = 1p



163

DOPTICAtilde

Subiectul I Punctaj

12

2 smv eU= rArr

2

2SI

2 ms

S

e

eUm

=

d 3

2 2

1

sinsin

i nr n

=

a 3


4 0 00

275 nmextext

hc hcLL


b 3




a

1

1 01 m10

fC

f

=

= =

2p

3p

1

1 01 m10

fC

f

=

= = 1p

b

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

4p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr = 1p

c

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

4p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


d

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

2p

4p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

1p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus1p



164


a

Din grafic15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot

1p2p15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot 1p

b

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

5p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot1p

c


4p

rArr

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot1p

d

( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p

4p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot= 1p



165

TESTUL 10

A MECANICAtilde




a



2

7 msdvt



1

2 msvv a t at

= rArr = =

1p

4p


2 11 1 1 1

1

2 9m2vv a d da

= rArr = =


1 3 3 21 3 3

1 3 31 3 3

22

0

final

final

final

v v a dv a d

v v a tv a t

v

= + = minus= + rArr

= minus= 21

33

21

33

15ms

12m2

vat

vda


1p



b


4 1 3 41 3 4

4 1 4 113 3 3

3


t


3p4p


c



1p


1p


1p



166

d


2 20G N N G+ = rArr = 1p


2p


1p



a


2

2 20

45 m2

2 2

gth

v v g h v g h

∆ = =

= + ∆ rArr = ∆

2p

4p2

2cmvE mg h= = ∆

1p


b


( )c

p


∆ ∆= =

minus ∆ minus ∆

2p

3p

rezultat final 016c

p

EE

= 1p

c


2

22

c

c

L E Fd

mvE

v gh

= ∆ =

∆ =

=

2p

4p

2

2 2mv mghF

d= = 1p


d


2

( 1)

2

f i

i f i

f


mvE fmgh


2p

4p

( 1)( 1) ( 1)



1p




167





a


mpV RTm= ν


micro=

2p3p



A

N N pVpV RT NN RT

= rArr = 2p3p


c


1

p VmRT

micro= respectiv 2

22

p VmRT

micro= 2p


2 12 1

V p pm m mR T T


1p


d


2p



2 3 2






a


primit primit

QLQ Q

η = = minus 2p



1p


b


1 1

1 T TT T

∆η = minus = 1

TT ∆=

η1p


η1p


2 300 KT = p


168


2 22

m m RTp V RT Vp

= rArr =micro micro

2p3p


d


4p


2 3 2

p p pT TT T p

= rArr = 1p


2

1V Vm pQ C T

p

= minus micro

1p




169





a


= = Ω 1p


232 3

R RRR R

=+ 1p


2 3

R RR R R RR R

= + = ++

1p


b




s

EIR r

= =+

1p



c


2 2 3 3

I I IR I R I

= +=

2p3p


1p

d


1 3R R R= +1p


1 2

093As

e

EIR R r

= =+ +

1p


3 3U R I= 1p




170


a


1 1 1 11

33 3 3 3

3

125 A

05 A

PP U I IU

PP U I IU

= rArr = =

= rArr = =

2p


1p

Din legea lui Ohm R

R

URI

= 1p


b



Iminus

= + = + rArr = 1p




d




consumat

095PP

η = = 1p




171

DOPTICAtilde




a


1 20 cmfC

= =


1 1

30 cmf xxf x

= =+




1 25 cmfC



2 1

f xxf x

=+


1p


2p

b



= 1p


c


1

1346 cmFxxF x

= =+

1p


1

y xy x

β = = 1p


1

xy yx

= 1p


d


1

4xx

β = = minus 1p


2 1

1 1 1F x x

= minus 1p





172


a


1ext s

hc L eU= +λ

(1) 1p

4prespectiv 2

2ext s

hc L eU= +λ

(2) 1p


1 21 2 1 2

1 1 s ss s

e U Uhc e U U h

cminus λ λ


1p


b




se obţine 10

1

sc eUh

ν = minusλ

1p


c



4p

Respectiv 22


Se obţine 1 1

2 2

s

s

v Uv U

= 1p

rezultat final 1

2

073vv

= 1p

d


11

ext shc L eU= +λ

(1) respectiv 33

ext shc L eU= +λ

(2)


1p

3p

3 1

1 1s

hcUe

∆ = minus

λ λ 1p




173

TESTUL 11

A MECANICAtilde


c 2

2 2mkg sN s s=

kg m kg m


= adimensional 3

2 c 3

3


5h = 0cE =2

0

2m v m g hsdot

= sdot sdot 0m2 10s

v g h= sdot sdot =

2

2 05 kgc

o

Emvsdot

= =

3

4 c 3

5 d ( ) 60 kW


= = = 3



a


3p 3p



= =+ +

3p

4p

Rezultă 2

m75s

a = 1p

c


sdot=

sdot ∆


4dS πsdot

=

3p4p


d



4pRezultă 1 2





174


a


ti tfE E= 2

0

2 cf pfm v E Esdot

= +

20


= =

2p

3p


b


4pRezultă 1 1



c




h h= sdot =

1p


2 20


= + sdot sdot


2p

Rezultă m1038 kgs

p = sdot 1p

d



Rezultă 0GL = 1p



175



d 2 3

2 3


mol k

minussdot


sdot

3

2 c p vRc c= +micro

p vRc cminus =micro

3


2 8VV = 12

1

8TT

γminus= 53

p

v

cc

γ = =

2 232 3 3

2 11

8 (2 ) 4 4T T TT

= = = rArr = sdot 3

4 b 3





a



3 311

1

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro 1p

3pSe obţine 3 31

11

025 10 mm R TVp


sdotmicro1p


b


NN

ν = 1p


mν =

micro1p


mNmicro

= 1p


c


Np V R TN

sdot = sdot sdot 1p


V= 1p

Se obţine 1

1

Ap NnR T

sdot=

sdot1p



176

d


1 2

( )m R T m m R TV V



1 1 2

1 2

( )m R T m m R TV V



2p

4punde 1

11

m R TpV

sdot sdot=


1 22

2

( )m m R TpV

minus sdot sdot=


1p


1 2

( )m R T m m R TV V






a



5p



4 1

34 3 3 33 1

4ln ln 2 2 ln 2 4 ln 22


= = = =


2p


b






c


max

1 TT

η = minus 1p

4p1 1

min 1p VT TvR

= =

3 3max 3

p VT TvR

= =

1p

Obţinem 1 1

3 3

1 12 2

p V pVp V p V


Rezultă 75η = 1p

d


3p 3p



177



c SI

V RA


SI

WR RI t


3

2 d 11

lRS

ρsdot= 2

2

lRSρsdot

= 1

2

14

RR

= 3

3 b 34 d 3




a


1p

4p





c


3

13 AABUIR

= = 2 24 AI = 2p




1 2 1 2 1 2( )I R R r r E Esdot + + + = + 3p4p




a


4extPP =

22

16ER I

rsdot =

sdot1p


2

P

EIR r

= +

1p


Rezultă 12 2784 016R = Ω Ω 1p


178

b



r= =

sdot


Rezultă 15 kJW =

c



Rezultă 32 WP = 1p

d


1 1 1 11

8E r IE R I r I RI


1

1

RR r

η =+

2p3p

Rezultă 80η = 1p



179

DOPTICAtilde


a 22


2 c 3


3

i rir

= rArr = = 3

4 c 15

2 1214

1 2

1

5 10 252 10

c

cλ ν sdotν

= = = =λ ν sdot

ν

3

5 c 9

43

600 10 2 6 10 06 mm2 2 10Dil

minusminus

minus


sdot3



a


1 1 1x x f

minus = 1p

4Rezultă 1

21

20 ( 30)20 30

fxxf x

sdot minus= =

+ minus2p


b

Distanţa focală 2

1 1 2

1 1 11nf n R R

= minus minus

unde 1 1n =

( )1 1 11nf R R

= minus minus minus

2p

4p

obţinem ( )2 11 1

2n Rn

f R fminus

= rArr minus = 1p

Rezultă 20 1 152 20

n = + =sdot

1p

c


3p 3p

d


1 2 1

2( 1)

nCf n R

C nR

= = minus

= minus

2p


1 15 1151 112

C nnCn

minus minus= =

minus minus1p

Rezultă 2CC

= 1p



180



8

1 91

3 10600 10

cminus

sdotν = =

λ sdot2p

3p


b


16 152 9

2

3 10 1 10 055 10 Hz54 10 18

cminus


λ sdot 1p



2 1

( )( )

h e UhU

e


∆ = unde 152

2


λ1p


c


cL h h= ν =λ

1p

4pObţinem 17


Dar 17

190 19

0044 101 16 1016 10

eV J Lminus

minusminus

sdot= sdot rArr =

sdot1p


d


11 0 ch h Eν = ν +1p


1p





181

TESTUL 12

A MECANICAtilde




a


1 1

1 1

0

f

F T m aN m gF N


2p


2 2

2 2

0f

f

T F m aN m gF N

minus =

minus == micro

2p


1 2

( ) m125s

F g m mam m


+1p


c


1 1 2

m2s


m1s


4p






2 21 cos180


5p

Rezultă 2 2

1 0482

v vgdminus

micro = = 2p


3pp = 8000 Ns 1p

c


21

112 2

MvW =

2p3p

1 8000JW = 1p

d


4 2 2Mv kx

= 2p

4p

Rezultă 212

kN3204 mMvk

x= = 2p



182





a


1 2


1p

3p1 1 2 21 2p V p Vm m

RT RTmicro micro

= = 1p

1 1 1

2 2 2

35m pm p

micro= =

micro1p

b


1 2


1p

4p

1 2

21 2

2 2

( )

2

m m RTm RTp V

V p

+micro micro

= =micro

1p


12

2 2 1 12 1 21 2

2 1 2 2 2

2 2

1 1( )1 1 1( ) (p p )

2 22

mmp pmp m RT p

p



micro

1p


p = + =


1p

c


1 2

1 2


+micro= =

+micro micro

2p


12

2

12

2 1 2

( 1)

1 1( )

mmm

mmm

+micro =

sdot +micro micro

1p


1 2

kg12kmol

p pp p


+1p

d

0 2 rmp V RT=micro

1p

4p02r

p VmRTmicro

= 1p

0 0

1 2 1 1 2 2 1 2

2 2 2 0(6)3

rm p pm m p p p p

micro= = = =

micro + micro +2p



183


a

1 (273 27)K 300KT = + =

1p

3p


2 1

2 2 600Kp p T TT T

= = = 1p


3 2

V VT T


3 1

p pV V

=


1p

b



1p

4p


13 3 3 3 1 1 3 1 1 3 11 1( ) ( )2 2





sdot1p

c

123 12 23L L L= +



123 1 1 13 3 300JL RT p V= ν = = 1p

d

1 ced

abs

QQ

η = minus 1p

4p31 3 1( )( )



1 1

1 1 1 1

2 (4 ) 11 773 5 13(2 ) (4 2 )2 2

R T TR RT T T T



1p



184



2

b


=+

+


2

2

( )( )

E Rx R xRx rR rx

+=

+ +


=

Rezultă RrxR r

=minus

3

3

c


=





3

4

c


2EI

R r=

+


22

EI rR=

+


2

22 2122 7

rRI EI R r E

+= sdot =

+

3

5


nal 1

2nn n

E EIR r R r

= =+ +

Rezultă 174

E E=3





1

171AEIr R

= cong+

2p

b




22 ( ) 22ABx R xR x x

R Rminus

= = minus

2p

4p

22 ( ) 22ABx R xR x x

R Rminus

= = minus


2p


185

c


max 2ABRR =

2p

4pRezultă min

1 2

EI Rr R=

+ + 1p


dsc

EIr

= 2p3p

12AscI = 1p



a


R r=

+1p

4p


22( )

REP RIR r

= =+

1p


2 2

2 4

( ) 2 ( )0 0( ) ( )



1p


4mEP

r= De aici

2

14 m

ErP

= = Ω 1p

b


2p3p


c


02

0( ) 4R E Ef

R r r=

+ 2p


2

02 4(1 )

4 m

f fERP f



2p


186

d

1

2

22

2

22

RR r

RrR

η =+

η =+

2p

4p1

2

1223

η =

η =1p

2

1

43

η=

η 1p



187

DOPTICAtilde




a



1p

5p

2 1 2 1

1 1

n n n nx x R

minusminus = 1p

3 2 3 2

2 2

n n n nx x a R

minusminus =

minus 1p


2 1 1 2

n n n n n nx x R R


b


1 2 1

y n xy n x

= 1p


1 3 1

y n xy n x

= 1p



3 1 1

n x yn x y

β = = 1p

c


2 1 3 2

1 2

imnx f n n n n

R R


+

15p

3p


2 1 3 2

1 2

obnx f n n n n

R R

minusrarr infin =

minus minus+ 15p

d


3pRezultă

22( 1)Rf

n=

minus 2 2

1 1

y xy x

β = = 2 1

1 1 1x x f

minus = 2p



188


a




b

Putem scrie 2

kx rD l










2Dx a l b a l bl


c


4pRezultă

2lDi

lλ

= 1p


Rezultă 2icircDi

lλ

= 1p

d


D D fi il f l f

λ λ∆ = minus =

minus minus2p

4p


2 2D g Di gi

l l+ λ λ

∆ = minus = 2p




190

TESTUL 1

A MECANICAtilde




a


4punde 1 2 1 2( ) sin ( )t

HG m m g m m gL



= + sdot sdot minus

1p


b


5p

unde 2 2

1 2 1 2( ) cos ( )nL HG m m g m m g

Lminus




1 2( )fF L

m m g L H

sdotmicro =

+ sdot sdot minus1p


c


4p




Lminus





d


3pDeci 2 2


L+ micro sdot minus

= + = sdot sdot 1p

Rezultă 28 NF = 1p



191


a


4p

Unde 2 2

0 2 2

M MM cM pM



Obţinem 2

max 2Mvhg

= 1p


b



4 52

QQ cQ pQ cQ cQ cQ

m vE E E E E E

sdot= + = + = =

1p

4pDar M QE E= 1p


5Q Mv v= 1p


c



Deci 2

2M

rm vF m g

dsdot

= sdot + 1p


d


2p3p




192





a


3pObţinem 1 11

1

p VR T

sdotν =

sdot1p


b


4pObţinem 2 2 2

22 A

p V NR T N

sdotν = =

sdot1p

Dar 22

A

NN

ν = 1p


c




1 2


ν + ν1p


d


4p



Rezultă 2 2

1 1

05VV

prime ν= =

prime ν1p



193


a



Dar 12341 0U∆ = 1p


b






c




d


4p






194





a


PR R R R R= + + + 1p

3p



b



P S

EIR r

=+

1p


Rezultă 02 AI = 1p

c




Rezultă 4 Vu = 1p

d


S

EIr

= 2p

4pObţinem SC

EIr

= 1p




195


a


EIR r

=+

1p

4pUnde 12 34

12 34BD

R RRR R

sdot=

+ 1p

Iar 12 1 2R R R= + 34 3 4R R R= + 1p

Rezultă 1 AI = 1p

b


5p









EIr

= 2p3p




196

DOPTICAtilde




a



= 1p


b


1 1 1x x f

minus = 1p

3pObţinem 1

21

x fxx f

sdot=

+1p


c


1 1 1x x f

minus =prime prime

1p


1

12

xxprime


1p


β1p


d


1 1 1x x f


1p


1

xxprimeprime


1p

Obţinem 1

fx f


1p




a


Dil

λ sdot= 1p

3pObţinem

2l iD

sdotλ = 1p



197

b


1p




∆α = = 1p


c


D iil

λ sdot= = 1p

4pDeci 1 2

DD = 1p


Rezultă 1 md = 1p

d


Dil

λ sdot= 1p

4pUnde 2

1 cn n

λλ = sdot =

ν1p

Deci 12 2 2

D iil n n

λ sdot= =

sdot1p




198

TESTUL 2

A MECANICAtilde



∆ SIpF Ft 3

2 c 15 mgiT G F= + = 3

3 a 2

0max2 2 10 m 0

2vd h r

g= = = ∆ = 3

4

b 2 2

5 45 4

4 5

452 2

4 s 5 s

at atx x x a

t t


= =22 msarArr =

2 210 9

10 9

9 10

19 m2 2

9 s 10 s

at atx x x x

t t


= =

3

5


3



a


1 1

2

0

0f

f

F T FT F

minus minus =

minus =

2p

4p

1 1 1 2( )F m m g= micro + 1p

2 2 kgm = 1p

b


1p


2 025micro = 1p


199

c


3 1

2

0

0e f

e f

F F FF F

minus minus =

minus =

2p

4p

3 2 1 2( )F m m g= micro + 1p

3 15 NF = 1p

d


4 1 1

2 2

e f

e f

F F F m aF F m a

minus minus =

minus =

1p

4p24 2 1 2

1 2


minus micro += =

+ 1p

eF k l= ∆ 1p

2 2( ) 1333 Nmm a gkl

+ micro= =

∆1p



a

00 50 mspv

m= =

1p

3p0u

vtg

= 1p

5 sut = 1p

b


4p0

32 2pA pA

cA c

E EE E= rArr = 1p

20

3vh

g= 1p

833 mh = 1p

c

0 3c cAE E= 1p

4p0

3vvrArr = 1p

29 msv = 1p


d

20

max 125 m2vh

g= = 1p


0 104 mc

f

Ehmg F

= =+ 1p

max 12hh

= 1p



200



1

2


= ν ∆ = ν ∆

∆rArr = γ

∆

3

2c

11

1 1 ii i f f f i

f

TTV T V V VT


= rArr =

44

if f i

VV = rArr ρ = ρ

3

3


2 2

V pV p

=


2 1 2

V p TV p T

=


2 2

3p Tp T

= =

3

4c

11

1 21 1 2 2

2 1

03V pp V p VV p

γγ γ γ

= rArr = =

2 1

1 2

N n VnV n V

= rArr =

3

5b 2 2

1 1

22 1

1

1 1

12 kJ

Q TQ T

TQ QT

η = minus = minus

rArr = =

3



a

1 2

1 2

m m+micro =

ν + ν 1p

4p

1 21 2

1 2

pV pVm mRT RT

micro micro= = 1p

1 2 2 1

1 2

T TT T


+ 1p

165 gmolmicro = 1p

b

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν2p

4p1 2

2 1

83 5

TTTT T

=+ 1p

2844 KT = 1p

c

pRT

microρ = 2p

3p30699 kgmρ = 1p


201

d

1 21 2

22 fpp pT T T

ν + ν = ν rArr + =

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν

2p

4p1 2

2 1

116 5

TTTT T

=+ 1p

5 22 1

2 1

11 (2 ) 153 10 Nm2(6 5 )f

p T TpT T

+= = sdot

+ 1p



a

2 3T T= 1p

4p

11 1 1

1 1 2 2 2 12

VTV T V T TV

γminus


1p

1 11 300 Kp VT

R=

ν 1p

2 14 1200 KT T= = 1p

b

323 2

2

ln VL RTV

= ν 1p


2 3

V pV p

= 1p

22 1

2

32RTp pV

ν= = 1p

23 2441 JL = 1p

c

31 31 31Q L U= + ∆ 1p

4p

1 3 1 331

( )( )2


83

V V= 1p

31 1 3( )VU C T T∆ = ν minus 1p

31 1314 JQ = minus 1p

d

min

max

1CTT

η = minus 1p

3p1

2

1CTT

η = minus 1p

75Cη = 1p



202





4 a 2

1 2max

1 2

( ) 4 W4( )

+= =

+E EP

r r 3

5 c 00

0 0

90 mA(1 ) (1 )

U U II IR R t t

= = = =+ α + α

3



a


e e

EIR r

=+

1p

4p1 3 1 3e eE E E r r r= + = + 1p

2 31

2 3e

R RR RR R

= ++

1p

1 AI = 1p

b

1 1 1U I R= 1p

3p1

1

e

e

EIR r

=+ 1p

1 154 VU = 1p

c

2V pU I R=

1p

4p

2 1

e

p e

EIR R r

=+ + 1p

2 3

1 1 1 1

p VR R R R= + + 1p

11 VVU = 1p

d


e e

E IR r

=+

1p

4p

1 23 3 3

1 2


r r= + = + = +

+ 1p

1 2

1 2

1 2

8 V1 1 3p

E Er rE

r r

minus= =

+1p

061 AI = 1p



203


a

22 2

2

PRI

=

2p3p

2 10R = Ω 1p

b


5p

1 21 3

1 2e

R RR RR R

= ++

pentru K deschis 1p

1 22

1 2e

R RRR R

=+


1 2er r r= + 1p

2 45r Ω 1p

c

22 int 2P r I= 1p

4p1 2I I I= + 1p

1 1 2 2I R I R= 1p

22 6 AI I= = 2 int 162 WP = 1p

d2

2

e

e e

RR r

η =+ 2p

3p

037η = = 37 1p



204

DOPTICAtilde


b 2

1

ff

β = 3

2 a 2 2 22

1 1 1

1 45 cma

x n x xx n x n

= rArr = rArr = 3

3 c 15 mmaa

iin

= = 3

4


c

1 2

1 1 11

l l

l

f n nnn R R

= =

minus minus

lfrArr rarr infin 3

5a 1 1 2 2

192 61 10 J

ex c ex c

ex

L E L E

L minus

ε = + = +

rArr = sdot

3



a2p

3p


b

2p

4p


2 25 cmf = 1p

c

1 12

2 1 1 1 1



4p1 2 3 3 35 cmd x x x= + rArr =

1p

4 3 2

1 1 1x x f

minus = 1p

4 875 cmx = 1p


205

d

3

1

yy

β = 1p

4p1 2β = β β 1p

21

1

15xx

β = = minus 42

3

25xx

β = = minus

1p

3 1 5625 cmy y= β = 1p



a

W N= ε 1p

4p

hcε =

λ1p

W Pt= 1p


= = sdot 1p

b

ex cL Eε = + 1p

4p

c sE eU= 1p

ex shcL eU= minusλ

1p


c

1s

N eIt

=

2p

3p18


= = sdot 1p

d1p

4p




206

TESTUL 3

A MECANICAtilde




a

2 0yG F N+ + =

1p



b

1 2 0x x fF F F+ + =


4p


1 2

2

cossinm

F Fmg F

+ θmicro =

minus θ 1p


c

1 2x fF F F ma+ + =

1p



F F F mgam

1p

2412 msa 1p

d



vv = 1p

2 cos2matP F= θ 1p

1854 W mP 1p



207


a

2

2PkxE = 2p

3p


b


4p2

2

= =i pkxE E sistem

1 2= + =f c c cE E E E

2p


c

2 2 21 1 2 2


1p


1p

21

1 1 2

( )

=+

kmv xm m m

12

2 1 2( )kmv x

m m m=

+1p


d

fc FE L∆ = 1p

4p

21 1


21

2vd

g=

micro1p




208





a1 0 2

lp p g= + ρ 2p3p

51 15 atm 15∙10 Pa p = 1p

b

1 1 2 2p V p V= 1p

4p

1 0 2

= + ρlp p g 1

2=


2 00( ) 0


1p

10 cm x 1p

c

1 1p V RT= ν 1p

4p1p ShRT

ν = 1p


d

2iU RT= ν 1p


2

5 078 2

= νNU RT2

5 02 2


aer 138 J U 1p



b

1 1 2 2 amestec

1 2

V VV

C CC ν + ν=

ν + ν 2p

4p2

=ViC R 1

3 2

=VC R 252VC R= 1p



209

c

min

max

1 ndashCTT

η = 1p



d

cedat

primit

1 ndashQQ

η =

1p

5p



3ndash1 1 13


= ν = ν 1p

p VC C R= + 1p


η = =V

p

C RC




210





a2

12

Ve A

V

R RR R RR R

= + ++ 2p

3p


b


2 2 3

1 2 3

( )A

V


= + + + = = +

111 16

=I A

25 8

=I A

31

16I A=

2p

4p


069 AAI 6875 V VU 1p

c 1

AA

EI R r R

=+ +

0VU = 2p4p


d

er R = 2p

4p1e AR R R= + 1p



a


=E

2p

4p2

max 4EP

r= 1p


b

2middotP R I= 2p

4pEIR r

=+

1p


c

ext

totala

p

p

RPP R r

η = =+

2p

4p2pRR = 1p



211

d

lRSρ

= 1p

3p2

4dS π

= 1p




212

DOPTICAtilde




a2 1 2

1 1 1 minus =x x f

2 22

1 1

y xy x

β = =

2p



b2 1 1

1 1 1ndash x x f

=

2 2

1 1

y x y x

β = = 2p



c

2 1 1

1 1 1 minus =x x f 2 120 cm=x 1p


2 1 2

1 1 1ndash x x f

=

1p


d 1 2

1 1Cf f

= + 2p3p




a

11

2λ

=Dil

2

2Dil

λ=

2p

4p1

1

cλ = rArr

ν1 2 1

12id

cλ ν

=

1p


b

1 1 2 2k i k i= rArr 1 1 2 2k kλ = λ 2p


1p



213



d


δ =l yD

2p


k λδ = = 1p



ν= = 1p



214

TESTUL 4

A MECANICAtilde


1

1

01 kgpm mv

= rArr =

m m

vF m a m

t∆

= sdot = sdot∆

m 2 2 2 s


2 NF =

3

2

a 0N F G+ + =


cosm gN sdot

=α

3




12 2 1 v v v= minus 12km 216 h


5





a




3p 3p


215

c


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α



20

1 1 307 m 2 sin cos

vd dg


2p

5p



u

d vv tt

+= = rArr ∆ =

∆


1p


( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot



c

d vv tt

+= = rArr ∆ =

∆


1p


1p

d

Notăm cu R



1p

4p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




2p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




1p




b


( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot

3p4p

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot1p


216

c




( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

2p

4p


2p

d


11 1 1 1

1

= 14 m s


minus



1p

4p

( )1 1 1 2

11 1 1 1

1

= 14 m s


minus



( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr =

1p( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr = 1p



217



2 2CO C Og = + 44


5

29 10 Pa = 29 atm

m R Tp V R T pV

p


micro sdot

rArr = sdot

3

2

a

12

1213

13

12 13

cu 22

5 cu 082

V

p pp

RQ C T C C R

Q CQ C T C RQ C

T T T


∆ = ∆ = ∆

3

3

d-1 -1


1 1 1

2 2 2 1 2

2

V S hV S h S h


2 75

p

v

C iC i

+γ = = =

1

12 1

2

VT TV

γminus

= sdot


3

4



3





b


1

12 1

2

VT TV

γminus

= sdot

(1) 1p

5p

p

v

CC

γ =






2p


γ = = 1p



218

c1 1

111

1

1344 g

cmm


3p 3p

d

2 2

222

2

4368 g

cmm


1p

4p

3 3

333

3

08 g

cmm


1p

1 2 3 5792 gm m m m m= + + rArr = 1p

32

kg 29 m

mV

ρ = rArr ρ = 1p



a


micro ν 1p

3p1 2 3 4m m m m m= + + +

1 2 3 4ν = ν + ν + ν + ν1p


molmicro = 1p

b

4p 4p

c


5p


1 2 3 41 2 3 4v v v vv

C C C CC


ν

1p


1 1 11 1

2121 2

22

2

p V R TT TTV Tp R T


sdot = ν sdot sdot

1p

112 1

1 1 15

31 2

- 155 - 155

155 10 J

ced

ced

TQ Q R T

R T p VQ



= minus sdot

1p


219

d


1 1 1 1 123 2

2 1

2ln ln ln 22 2

V T V p VQ R T RV V


1p

3p5

23 035 10 JQ = sdot

( ) 131 1 2 1 1 121 105


1p

531 105 10 JQ = sdot


1p



220



1 1 2 2 3 31 2 3

1 1 2 2 3 31 1 2 2 3 3

1 2 3


S S S S


1

2

2775 n m222 n m


3

2

b

( ) ( )1

1 10 1

1 0 1

1 1

UI UR IR

R R


(1)

Analog ( )2

0 2

1

UIR

=sdot + α sdotθ

(2)


3



4


1 2 3

1 1 1 1 pR R R R




3

5

a2

U N SW tl

sdot sdot= sdot ∆


3



a

1K rarr

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

2p

4p1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

1p

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω 1p

b

2K rarr

2

2

2

154 A

EIR r

I

=sdot +

rArr =

1p

4p2

2

2

154 A

EIR r

I

=sdot +

rArr =1p



221

c2 2

3e eRR Rsdot

= rArr = Ω 3p 3p

d3 3 36 A2

3

EI IR r= rArr =

sdot+ 2p

4p




a


400 = 3

b b b b

bb b

b

P U I IUR RI

= sdot rArr

= rArr Ω2p

4p

bb c

EIR R r

=+ + 1p

485 1616 3cRrArr = Ω = Ω 1p


c

icircnchisK rarr b

t

PP



100 VR bU U= = 1 2R R R+ =


2 b RI I I= +

2 2b RU I R= sdot

2p

4p



η = = 1p

d


4p


1 1 1

e sR R R= +

prime

1p


r= rArr =

sdot


2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =



632 Wbb b

b

UP PR

= rArr =

2p



222

DOPTICAtilde


( )( )1


3

2 a 3

3

a1

2

50 cm1 80 cm

fC

ff=



1 2

1 2

1 22

1

2

16 cm

h hf f

h fhf

h

= rArr

sdot= rArr

=

3

4c

39 Vextrextr s s s

LL e U U Ue


5

a

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3



a 3p 3p

b




i i = rArr =

2p

c

2 5 2sin cos tg = 3 3 5

r r r= rArr = rArr

2p

4p1

1

1

tg i = = 50 cmtg = = 775 cmtg

tg

x hh i H H

r hxrH

rArr rArr=

2p


223

d

2p

4p

0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p



a

19min min

max

= 262 10 J = 163 eVR


λ λ 2p



1p

b


λ2p

4p max

min c

extrextr c

EL im

L E=




4p252 m = 66 10

2 sc

cm v EE v v

msdot sdot




224

TESTUL 5

A MECANICAtilde




a


1p



b






d





a


0


4p2 20

2G gFr v v df f


V = 9 ms 1p

b


2p


1 1 1 2( )m v m m vprime= + 1p

Rezultă 1 1

1 1

3 msm vvm v

prime = =+ 1p


225

c


4p

1 2( )0 m m g Fr Df

+minus = sdot 1p

1 2( )m m gFrf

+= 1p


d

21 2

21 1

( )cs

cl

E m m vxE m v



3p

Deci 103

x vprime= 1p



226





a

ν1 = 2Omicro

m 1p

3p1

2

116

νν

= 1p

2

2Hmicrom

ν = 1p

b


4pp 2

ν= ν1RT1

p 2ν

= ν2RT2

2p

Rezultă 21

12

16TT

νν

= = 1p

c


pV2prime= ν2RT1

1p

4p

1

22

1VV

=νν


1p


V2prime= 2

L x S +

1p


d


4p1 2

2m m m+ =


Obţinem 1 2

1 2

2micro micromicro =

micro + micro1p

Rezultă 1 2

1 2

2micro micromicro =




227


a


4p 4p

b


1p

4p


12

1

lnc T VQ RV



1 1 3

13 1

3

4

1 44

PV RTPV RT

T T TT

= ν= ν

= rArr =

1p


c


p

QQ

η = minus 1p


215η = 1p

d

minCarnot

max

1 TT


3pCarnot

3 754

η = = 1p



228





a


V

ERR r+

1p


2vR (2) 2 2 2 2

v v

v

R ERU IR r

= =+

2p


b


EIR r

=+ 1p


EIR r

=+

2p


c2 1

1 1VrR

EU U

=

minus

1p

4p

2 1

1 1AR E

I I

= minus

2p


d

RV rarr infin 1p

3p1V

VV

V

ER EU ErR rR

= = =+ + 1p



a


4p

1 1 3

2 2 4

S

S

R R RR R R

= += +

1p

1 3 2 4

1 2 1 2 3 4

1 1 1 ( )( )tot

tot S S


+ += + =gt =

+ + + 1p


1 2 3 4

( )( ) 4


+ += + rArr = Ω + + +

1p


3pW = 504 J 1p


229

c


4p1 2 3 4

1 2 3 4

EI R R R RrR R R R

=+ +

+ +2p


d


tot

RR r

η = =+

2p

4p


tot

RR r

η = =+ 2p



230

DOPTICAtilde




a


1 2

11 1( 1)

fn

R R

=

minus minus

2p

4p1

2 0RR

rarr infinlt

1p

R = 15 cm 1p

b

2


4p2

1 2

1 1 1( 1)l

fn

R R

=

minus minus

1p

|R1| = |R2| = 15 cmRezultă nl=137 1p

c

2 2

1 1

1y xy x

β = = = minus 1p

3p

2 1

1 1 1x x f

minus = 1p


d


21

1

xx

β = 2 1 1

1 1 1x x f

minus =

1p

4p


2 1 2

1 1 1x x fminus =

2

21

xx

β =


1 21 1 2

1 1

f xx f ff x

= minus minus+

1p


11 1

1 1

f fff x

f x


+

1p

2

1

ff

β = minus


1p



231



2kk Dx

lλ

=k = 5

2p3p



2lDi

l= 2p


c


1p

5p



2knx eD

lminus

=

xk = 04 m

1p


d



2 2k D k

l lλ λ

= 1p





232

TESTUL 6

A MECANICAtilde




a

2

2f

BG F cB cA



4 msBv = 1p

b

0cE∆ = 1p

3p20J

20JpE mgh

E

∆ = =

∆ =2p

c

2 0

2f

BF cB


4p2 2

2 2B Bv vgd d

gmicro = rArr =

micro1p

4md = 1p

d

( ) 0 0


4p


cos ( )sin


1p

40JL = 1p



233


a

1 1F t psdot ∆ = ∆

1p


1p

12mvF

t=

∆1p

1 2kNF = 1p

b


1p

4p

2

2 2

22

p p mv

F t pp mvFt t

∆ = =

sdot ∆ = ∆

∆= =

∆ ∆

2p

2 1 kNF = 1p

c

H F t= sdot ∆

H F t= sdot ∆1p

3p

1 1

2 2

2 N s1 N s

H F tH F t


2p

d

2p

4p

3 2 2 2N sp p mv∆ = = = sdot

2p



234





a 4p 4p

b

A BU U U∆ = ∆ = ∆ 1p

4p2 1 2 1 2 2 1 1


250 JU∆ = 1p

c


5p


650 JAQ = minus 1p


50 JBQ = minus 1p





a

12 34 0Q Q= =

123 3 2 3 2 3 2 2 2 2 3 2 3 2

5 5 5 5( ) ( ) ( ) ( ) ( )2 2 2 2V


2p

5p1 1 2 2 2 1

4 1 3 2 3 4

1 2

3 4


γ γ γ

γ γ γ



-11

23 4 1 4 1 15 5( ) ( )2 2primit


3p


235

b41 1 4


4p


c

1 cedat

primit

QQ

η = minus 1p

3p4 1 1

1-1

4 1 1

5 ( ) 121 15 ( )2

p p V

p p Vγminus

γ


εminus ε sdot2p

d

pL Q= ηsdot = 1p

3p-1 -1

4 1 1 4 1 1-1

1 5 5(1 ) ( ) ( 1) ( )2 2


ε2p



236





a

11

1 1 2 1 1 2 1 48 16 8 8 8 4 p

p

RR

= + = + = = = Ω 1p

5p

1 4 20 24sR = + = Ω 1p

22

1 1 1 3 1 69 18 18 6 p

p

RR

= + = = = Ω 1p

2 6 6 12sR = + = Ω 1p

1 1 1 3 1 824 12 24 8 AB

AB

RR

= + = = = Ω 1p

b

1 1 3

1 1 05 8 4 VABU I R IR

I R= += sdot =

2p

4p

1 11 1 2 2 2

2

4 1 025 A16 4

I RI R I R IR

= rArr = = = =

1 22 1 A4 1 20 4 20 24 VAB

I I IU

= + == + sdot = + =

2p

c

ABe

AB

UIR

= 2p3p

3AeI = 1p

d

ABAB e e

e

E UE U I R rI

minus= + rArr = 2p

3p

1r = Ω 1p



237


a

1 2

( )

( )


I

= + ++

= + +

3p

5p

2

6024 24

2 20 50 05A

II

I II

= +

minus + ==

2p

b( )e e e



43eR = Ω 1p

c


4p1

1

21

24 2A1236 3A12

b

b

PIUPIU

= = =

= = =2p

d

11

12 62

bURI

= = = Ω

22

12 43

bURI

= = = Ω2p

3p




238

DOPTICAtilde




a1 1 1 1

22 1 1 2 1 1 1 1


+= + hArr = hArr =

+ 2p3p


b



4p 2 1 2 12

2 1 2 2 2 1 2 1


+= + hArr = hArr =



c

2 2

1 2 1 1

sistemx xx x



d 5p 5p



239




b


2p3p


c



λ= 2p


relaţia 2

xD l∆ δ

= care devine ( 1) 45 ( 1) 45 ( 1)2 2 2 2



en

λ=

minus

2p


d


4p

( 1)nouke

nλ

=minus

1p



240

TESTUL 7

A MECANICAtilde


3



EE mgh m

gh= rArr = =


892 ms2

poco po c p p o

EmvE E E E E vm

+ = + rArr = rArr = =

3

4 c 3

5

d

0

0 sin 0 cos

f

x

f y f


+ + + =


cos 20 10 3 027

sin 10f

f

F mg FF NN F


sdot α

3



b


4p


1 1

2 1 2

l ml m m

∆=

∆ + 1p

1

2

25

ll

∆=

∆ 1p

c


4p2 2 1eF m g m g= + 3 3G m g= 1p


d


1m gm g kx k

x= rArr = 2p

3p

100 Nkm

= 1p



241


a


2p

4p


2 1 2 2 1 2

2

0 ( )80 N

N N G N m m gN


b





1p

5p


1 22 1 2

2 1 2 2 1 2


Oy N N G N m m g

minus minus minus =




1p


33 2 3

3 3 3 3

sin


Oy N m g N m g

α minus minus =




2p

Rezultă 3 2 1 1 2 1 2 1 2 3

3 2 1 1 2 1 22

1 2 3

(sin cos ) 2 ( ) ( )(sin cos ) 2 ( ) 342


m m m s


= =+ +

1p

c

1 1 1 1( ) 1626 NT m a g T= + micro = 1p

3p2 1 2 1 1 2 1 2

2

( ) 2 ( )4736



d

1

1 1

22 ( )

R TR m a g

== + micro 2p

3p3252 NR = 1p



242



4



υ ∆ = υ ∆ + 2 1


1 2 1 1 2 2 2 12 1 2 1

3( ) ( )2 2



2 2

p aVp aV

= =

1 11 2 2 1

2 2

p V p V p Vp V

rArr = rArr =

Obţinem 2 2 1 12 1 2 1

3( ) ( )2 2



2 2 2

p V RTp V RT

= υ= υ


3

5



2pV V p= rArr =

adiabat 11 1 2 2 2 2


2

2pp

γminus= 3



a


4p


1 1 1 1

2 22 2 2 2

1 21 2

( )( )


p V Vp V V RTRT

= ν rArr ν =

= ν rArr ν =

++ = ν rArr ν =

1p

Rezultă 51 2 1 11 2 1 1 2 2 2 2

2



b


2 1 1

p Vp V

ν=

ν 2p3p

1

2

5ν=

ν 1p

c


1 2

m m m= +

micro micro micro1p

4p1 2

1 21 2

1 2

2m m

m m m

=micro micro


2p

7gmolmicro = 1p

d


4p


1 2 1 2

1 2 1 2

1 21 2

1 2

2 1

1 2 1 2

( )22

V V V V V V

V V V VVV


C C C CC C




1p

1 22 1

1 2

134

7716

V VV

V

C CC R

mU C T KJ

micro + micro= =

micro + micro

∆ = ∆ =micro

1p



243


a


3p 3p

b



4p

1 1 1 33 1 2 1

3 3 1


= rArr = = rArr = =

1p



3 3 4 4p V p V= 4 1p p= 1 1 1 49 p V p VrArr =

4 19 45lV VrArr = =1p

c


p Vp V RT TR

= υ rArr =υ

1 12 2 2 1 1 2 2 1



1 13 3 3 1 1 3 3 1



4 3 19T T T= =

2p

5p

32vC R= rArr 1 1

1 3 3 1 1 13 8( ) 122v


υ2p

1 3 6000 J = 6 kJU rarr∆ = 1p

dmin

max

1cTT

η = minus

1

3

1cTT


88cη = 1p



244



3


rArr = = Ω


= rArr = Ω


η = = rArr η =+

3

4


182 31 1 1 1 11

2 3

k

ke

k

E E E Er Er r rE

r r r r

sum + += = =

sum + +

1 1 61 1 1 1 11

2 3

e

k

rr

r r r r

= = =sum + +

3

5

d Puterea maximă 2

max 4EP


2 22 2

2

2 6 04 ( )E E R R Rr r


+


3



a


1

1 1 1 22 3e

e

RRR R R

= + rArr = 1p

4p1 18V 2AU I= = 11 1 eU I R= sdot 1p

11 1

1

2 33 2R UU I R

I= sdot rArr = 1p

6R = Ω 1p

b


2

1 1 1 32e

e

RRR R R

= + rArr = = Ω 1p

5p

1 1 1 1

2 2 2 2



1 1 2 2U I r U I rrArr + = + 1p

1 2

2 1

1U UrI I

minus= = Ω

minus E = 10 V 2p

csc

EIr

= 2p3p

10AscI = 1p

d

2

max 4EP

r= 2p

3p

max 25 WPrArr = 1p



245


a

1R şi


lRS

= ρ

22

lRS

= ρ 1p

3p


1 2 2 1 1 2 1 1 2 2l lR R R R R R R l R l

S Sρ ρ


1p



1 2 1 2

1 1 1 ( )e

e

R R RRR R R R R R R

+ sdot= + rArr =

+ + + 3p4p

5eR = Ω 1p

c


3e e

nE EIR nr R r

= = =+ +

2p4p




e

EIr

= = 2p



246

DOPTICAtilde


2

a


2 2 2 2

2 2 2 1 2

1 1 1

62 4V

(155 )155055





ν minus

140 0 118 10 HzLL h


3

3

b



2

sintg tg1 sin

PB rr PB h r hh r

= rArr = =minus


2

sintg tg 3 07m1 sin

x rx PB hPB r


3

4


2r r

v v

D iil i

λ λ= rArr = =

λ


v

ii i fi i f fi


f = 30

3

5


4l

n= =

2

tg tg


rl r h lh

l lr h h ml l

= rArr =

= = =minus

3



a



1 2

1 20 cm1 1 1( 1)( )

Rfnn

R R


1p


2 1 1

1 1 1 ( 20)( 30) 12cm20 30

fxxx x f f x


+ minus minus 2p


1 1 1

( 12) 2 08 cm30

x y x yyx y x


minus


b


1 2

11 1( 1)( )

a

a

a

n Rf n n nn R R

minus= =

minusminus minus2p

3p

80 cmf = minus 1p


247

c


R ffnn

R R

minus= = = = minus


2p


1

2 2 1 1

1 1 1 ( 10)( 30) 15 cm10 10

f xxx x f f x


+ minus minus 1p


2 2

1 1

x yx y

β = = rezultă

2 12

1

( 15) 2 1cm30

x yyx

minus sdot= = =

minus 1p

d


a

a a

f fFF f f f f

sdot= + rArr =

+1p

4p

1 15 cm1 1 2( 1)( 1)( )( )

aa

a

Rfnn

R R


minus1p

( 20) 15 30 cm30 20


minus1p

12

2 1 1

1 1 1 ( 30)( 30) 15 cm30 30

F xxx x F F x


+ minus minus 1p




2Dil

λ= = 2p


i= rArr = = 1p

b


lλ

=


Dxl

λ= =

1p


λ= + sdot 1p


Dx x x xl


c


Dδ = minus =



∆ = minus minus

2p

5p


λ = minus minus



= minus rArr =minus

2p

0 33 45 mm2

Dx xl


dDin relaţia 1

2KK Dx K

lλ

= = rArr 22l x

Dsdot

λ = 2p3p

2 480nmλ = 1p



248

TESTUL 8

A MECANICAtilde




a

2

2cpEm

= 2p3p


b

pFt

∆=

∆ 1p

4pF ma= 1p

pam t∆

=∆

1p


c

amFFF fx

=++ 21 1p





d

2

2atd = 1p

4p2 1d d∆ = minus

21

1 2atd =

22

2 2atd = 1 1 st = 2 2st = 2p




249


a


4p1iE m gh= 2

1

2fm vE = 2p


b


2gth = 1p


2

2atd vt= + 1p


1p


c


1 1 112 2


1 1 1 2i fp p m v m m v= hArr = +

2p

4p( )1

1 2

2m g h dv

m mminus micro

=+

1p


v = 1p

d


1 2

2sistem

m m vE

+=

2

2sistemk lE ∆

=

1 2m ml vk+

∆ =

2p

3p


l∆ =

1p



250





a

pV RT= ν 1p

4p1

2

2

1

2

1

2

1

micromicro

νν

sdot==mm

pp

2p

rezultat final 47

2

1 =pp

1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0

2p V RT= ν

1p

4p0 finalp V RT= ν

1p


d


2p




b1 1L p V p V= ∆ sdot ∆ = 2p


c

min

max

1CTT

η = minus 1p




251

d

efectuat

primit

LQ

η = 1-2 2-3primitQ Q Q= + 2p

5p


p

V

CC

γ = p VC C R= + 1p

RC γ=

γ minus 1V

RC =γ minus

1p

12 1γ minus

η =γ +

1p

rezultat final 2 11

19η = = 1p



252





a

4 52

4 5s

R RR RR R

= ++ 2p

4p3

13 3

se

R RR RR R

= ++ 1p


b

1e

EIR

= 1


3 3

sp

R RRR R

=+

2p


c 1

EIR

=

2p3p


d3

1 3

EIR R

=+ 3p

4p




alR

Sρ

= 2p3p




c

becuri b

sursatilde

P UP U

η = =

3p4p


η = 1p

d

middotx

xRS

ρ= 1p

4p2 x

UIR

= 2p




253

DOPTICAtilde




b

2 1 2

1 1 1x x f

minus =

2 2

1 1

12

y xy x

β = = =

2p

4p012

21

=+ff 1p


c

( )1

1 11nf R

= minus 2p



d


4p2 1 1

1 1 1x x f


2 1 2



1 2

16 cm

x fx xx f

= = minus+

1p






b

2 1r rδ = minus 1p



enδ minus δ

=minus

1p


c 2Dil

λ= 2p

3p



254

d


δ = 1p

4p



δ = δ + 1p


= minus 1p

rezultat final( )12

ed nh

lminus

= 75 mh = micro 1p



255

TESTUL 9

A MECANICAtilde



5 20 205


= rArr = = ms 3

4

d [ ]

( )

21

2

0 0

m1s

2

sC

atx t x v t

C a

=

= + +

rarr

3

5

a ( )

1 1 2 2

1 2

1 2

1 2

1 2

1 2

N12m

s

s

s

s

F k F kF k

F Fk F Fk k

k kkk k

k

= ∆ = ∆

= ∆ + ∆

rArr = =∆ + ∆ +

rArr =+

=

3



a

1p

4p

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

= 1p

b2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+

2p4p

2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+ 2p


256

c


3p( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot =

1p

( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot = 1p

d( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

2p

4p( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

1p

( )1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

= 1p



a 05 JA

A

E mghE

==

2p3p

1p

b

2p

4p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=

1p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=1p

c


( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

4p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

= 1p

d


2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p

4p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

= 1p



257



b [ ] Jmol KSi

C =sdot

3

2 a 2100 kJ

Q m c tQ

= sdot sdot ∆=

3

3d 1 1 2 2

1 2


N NN Nmicro + micro

micro =+

micro = =

3

4

c 2

1

11 1

2

ln

ln

280 J

VL RTVpL p Vp

L

= ν

=

= minus

3



a0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p

0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

b

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

4p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot 1p

c

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

4p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot 1p

d

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

4p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= = 1p



258


a

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

3p

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

b

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

5p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

c( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

=

2p3p( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

= 1p

d

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p

4p

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p



259



d 2 WRtU

= [ ]SI1st = 3

2 b nEIne r

=+

3

3a

2

32 2 323 7221320

AB

ABAB

AB

RR RR R RR RR

R R RRR R

sdot sdot= + = +

+

sdot= =

+

3

4

c ( )0

00

02 1

1

10 grad

R R tR RR R A t

Rminus minus

= + α


rArr α =

3



a

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω

2p3p

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω 1p

b

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

2p3p

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

1p

c

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

5p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus 1p


260

d


( )( )

1 2 23

1 1 1 1 23 23

2 2 2 2 23 23

I I IE I R r I R

E I R r I R

+ =

= + + sdot

= + + sdot

1p

4p

cu 2 323

2 3

23

12

8 A3

R RRR R

I

= = Ω+

rArr =

1p

Din 2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr =

1p2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr = 1p



a


EIr R

=+

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus1p

4p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p

rArr r = 20 Ω 1p

b

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

3p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

= 1p

c

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

2p

4p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

1p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr = 1p

d

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

4p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

şi eR = r 2p



261

DOPTICAtilde



2

2c

mvE = 3



3

4

d 2 2 22 1

1 1 1

x y yx xx y y

β = = rarr =

Din grafic 1 2

2

1 2

1 2

40 cm 10 cm40 cm

20 cm

x yx

x xf fx x

rArr = =

rArr =

rArr = rArr =minus

3

5 b 9

63

550 10 1 275 10 m2 2 10

275igrave m

Di ie

i

minusminus

minus


sdot=

i = 275 μm 3



a


3pcorespunde cu 1 2

05


1p1 2

05


b

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

4p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr = 1p

c

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

4p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = = 1p

d

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

4p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p



262


a 4p 4p

b0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


2p3p0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


c

00

86

0 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot

2p

3p0

08

60 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot1p

d

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

2p

5p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot 1p



263

TESTUL 10

A MECANICAtilde




a



4p


1 12

2

m v m vm vvm

=

=

1p


b


112 2

v gtda

micro= =

2p3p


c


222

2

0375m s2vad

= = 1p

3p


aa gg



d


5p

21


22


1p

( )2 2 2

1 21 2 1 2 1 2( )





264


a






b


3p20

0(1 cos ) 2 (1 cos )2



c


m= = 1p

4p




m M= =

+


ecuaţia vitezei 2

2vd

g=

micro

1p


d



m M m M= minus =

+ + 2p

4p




265





a


pV RT

mpV RTm= ν


1p

4pm pV RT

microρ = =

1p


3128 kg mρ = 1p

b


1 1 11

p V RTmp V RTm


1p

3p1 1

1

p VmRT

micro=

1p


c


2 1 2 1

22

1

p V RTp VRT

= ν

ν =

1p



1




1



d


1p

5p

1 11

1 1 1 2 2

2 2 12

1



1p

1 2 1 1 1 2 2( )p V V RT p V p Vν+ = = + 1p

1 1 2 2

1 2final

p V p VpV V

+=

+ 1p




266


a


3p 3p

b



2 22 1 2 1 2 1 2 12 1

( )( ) ( )( ) ( )2 2 2


1p

3p

2 22 1

2LkV V

=minus

1p


c


21 1 1


R R= ν rArr = =

ν ν

1p

5p2

2 2 22 2 2 2

p V kVp V RT TR R

= ν rArr = =ν ν

1p

2 22 1 2 1( )kT T T V V


ν 1p

2 22 1

3 ( )2VU C T k V V∆ = ν ∆ = minus

1p


d


1 2

1 2

1 2 1 2 1 2

100 J300 J

400 J

LU

Q U L

minus

minus

minus minus minus

=∆ =

= ∆ + =

1p

4p


23 12 3 2 2

2 2

2 3 2 3

0

ln ln 245 J

245 J

UV VL RT kVV V

L Q

minus

minus

minus minus

∆ =

= ν = = minus

= = minus

1p


1p




267





3p 3p

b


1p

4p



4 4s

ee s

r rrr r

= rArr = =1p


015er = Ω 1p

c


1 23 ( )4

E i r I R RI i

= sdot + + = sdot


142A34

neIr R R

= =+ +

1p

4p




= 1p


d2R Sl =ρ

3p4p




268


a


1 2

1 2 1

( )

1

E E I r r RE EI A

r r R

+ = + ++

= =+ +

1p

3p


1

1 1

2 2

b

b

U IRU E IrU E Ir

== minus= minus

1p

rezultat final

1

2

8 V55V25V

b

b

UUU

===

1p

b


1 2

1 2

035Ass

E EIr r R

+= =

+ +

1p


1 2

266pR RR

R R= = Ω

+ respectiv 1 2

1 2

214 App

E EIr r R

+= =

+ +1p


1p


c



1

ext s

PfP

=

1p


d


2ext p pW R I t=

1p


1 2 1 2

057p pext

total p

R I RWW E E r r R

= = =+ + + 1p


total

WW

= 1p



269

DOPTICAtilde




a


2 2

1 1

12

y xy x

β = = = 1p

3p


1 1 1 125Cf x x

= = minus = δ 1p


b


1 1

10 cmf xxf x

= =+


4p




1 40 cmfC



2 1

f xxf x

=+



c


1

025xx

β = = respectiv 22

1

133

xx

β = = 1p




d


= = 1p


1 1 1F x x

= minus 1p


1

FxxF x

=+

1p




270


a


1 0108 mmxik

= =


Did

λ= se obţine 1

1i d x dD k D

λ = =

1p

3p

Din 22 2 2

D di id D

λ= rArr λ = 1p


b


3 1 13 3 Dx id

= = λ

1p


d= = λ 1p




c


1p


1 1 2 2 11


λ 1p


1p


d


1

nλ

λ = respectiv 2

2

nλ

λ =



2 12

k Dxn d

+ λ=

1p



n dλ

= = 1p






271

TESTUL 11

A MECANICAtilde


c 2 2 2 2

2 2 2 cp m v m v E

m msdot sdot

= = =sdot sdot


2 c 2


= + sdot + 2

55 2

g tx sdot=

24

4 2g tx sdot

= 5 4 45 mx x x= minus = 3


αmicro =

α3

4 b 1N100m

Fkl

= =∆

2N200m

k = 1

2

12

kk

= 3

5 b p F t∆ = ∆ F G= m100 kgs




a




2 2 1

1 2 1 2


= =+ +

1p

Rezultă 2

m375s

a = 1p

b




23 3F T T= sdot =

2p


c


4pRezultă


sin cosmM = =



d


V = a Δt2p

3p

Obţinem m5s

v = 1p



272


a


i fp p=

21 1 2 0

1 2

m v m vu

m msdot + sdot

=+

2p

3p

Rezultăm60s

u = 1p

b


4p




Obţinem Bp =m56kgs

1p

c



4pObtinem

21 2

1 2( ) ( )

2B

Bt


Rezultă 1700 JBt

E = 1p

d


2C f

Bc G F


4pDeci 2

2sin cos2C

t Bc t t




2p




273



b 12


2 b 201204 10A

A




4 b 1 2 1 2

1 2

m m m m+= +

micro micro micro 3

1 2


+micro micro

3


3 2 11mp R T m m mV


3



a


Np V R TN


11

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro

1p

4pSe obţine 1

11

AN P NnV R T

sdot= =

sdot2p

Rezultă 25

1 3

molec12 10m

n = sdot 1p

b


22

p VR T

sdotν =

sdot2p

3p


c


11

m R TpV

sdot= sdot

micro 1p

4p


11

pR T

sdotmicroρ =

sdot

22

2

pR T

sdotmicroρ =

sdot1p

Se obţine 1 1 2

2 2 1

p Tp T

ρ sdot=

ρ sdot1p

Rezultă 1

2

113

ρ=

ρ 1p


274

d


2 2 2 2

p V R Tp V R T


SF 1 1 1

2 2 2

p V R T

p V R T



1p



1 1 2 2

1 2

1 2

1 2

p V p VT Tp V V

T T

sdot sdot+

=+

1p




a


1

ln ln 2vL R T R Tv


34 3 4 3 3

1( )2


141 1 1

4

1ln ln2

vL R T R Tv


4p





3p

34 12465 JU∆ = minus 1p

c


4pObţinem

1

4 3 14





d


p

LQ

η = 1p


2p

Obţinem 505η = 1p



275



c 2

2 UW R I t tR


= 3

2 c 3


4 c eSC

e

EIr

= 65er = Ω 24 V

5eE = 4AeSC

e

EIr

= = 0 VU = 3

5b

2 2U UP RR P

= rArr =

2

10 mS UlP

sdot= =

ρ sdot3



a


e e

EIR r

rArr =+

1p


1 2

1 2

1 2

30 V1 1e

E Er rE

r r

+= =

+ 1

2err = = Ω 1p


1 2

4s se

s s

R RRR R

sdot= = Ω

+ 1p

Obţinem 6 AI = 1p

b



Obţinem 1

2

1II

rArr = 1p

c



10 A3

e

e

EIR r

prime = =+

2p






276


a




b




c

Randamentul va fi e

e

RR r

η =+

1p

4p




2p

Rezultă 886η = 1p

d


4p


12

21

8022 3

2

e

RR RR RR

sdot= + =

+

30 A43e

EIR r

= =+

2p




277

DOPTICAtilde




a


1 1 1 1 1 12x x f x x f


2p

4pRezultă 2

1 32

cf x

minus= =

minus1p

Obţinem 2

3 2015 10

c minus= = δsdot

1p

b

2 1

1 1 1x x f

minus = 1p

4pobţinem 2

1 21

1 22

x x xx



c


2 1

1 1 1x x f

minus = 1p



1

x fxx f

primeprime =

prime +1p

Rezultă 275 5 15 cm75 5


1p




D a ii Da

λ sdot= rArr =

λ2p

3p

Rezultă 3 3

9

2 10 1 10 5 m400 10

Dminus minus

minus

sdot sdot sdot= =

sdot1p

b


4pRezultă

axD

δ = 2p

Obţinem 3 3

72 10 16 10 8 10 m4


δ = = sdot 1p


278

c



4pObţinem

9

max 3

400 10 442 10

xminus

minus

sdot sdot= sdot

sdot1p


d


naλprime =

1p

4pDin Dia

λ= rezultă ii

n= 2p

Rezultă 3

3 31 10 3 10 075 10 m4 43

iminus




279

TESTUL 12

A MECANICAtilde




a


00

( )

05

kt M m m akta t

M m m

= + +

= =+ +

2p


2

2 2

m8 02s

Mg ma

a

micro =

= micro =

2p

b


0 1( )Mg ma

kt Mg m m amicro =

minus micro = +

2p

3p


m( 2)s

kt Mga tm m


+ 1p

c


4p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + =

2p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + = 1p

d


4p


( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus = 1p



a


30 5 10 J



Deci 0cE∆ = 1p


280

b




3p4p


c



4p


d


1p


2p




281



5

b

1 11

1

abs ced

abs

Q QQ

minusη = respectiv

2 22

2

abs ced

abs

Q QQ

minusη =


1 2

abs abs

L LQ Q

η = η = iar

2 1abs cedQ Q=


3



a


1p

3p0p VmRT

micro=

1p


b


4p


A

Np V RTN

= 1p


A

N RTpVN+ α

= 1p

Rezultă 0

1 14pp

= + α = 1p

c


kmolmicro = şi


kmolmicro =

1p

4p

012i ii

A i

N m iN

ν = = =micro

1p

1 2

1 2

m m m= +

micro micro micro

1p

0

1 2 1 2

1 2


A

A A

Nm N

m m N NN N

micromicro


micro micro

1p


282

d


4p


= ν ν = 1p


1 2Q Q Q= + 1 1 1 2 2V VQ C T C T= ν ∆ + ν ∆1p

( 5) 450 J2

pV TQT

α + ∆= sdot =

1p



a


2 1

p pT T T nTT T

= = = 1p

3p


3 2

V V T nkTT T

= = 1p


1 4

V V T kTT T

= =

1p

b


1p

4p


= minus minus


= + minus 1p



1p


283

c


Ccald

T T nkT nkT nk


Cη gt η adică 1 2 2 2 25 2 3

nk nk n knk nk n



adevărat

1p

4p



minusη = =



minusη = =

1p




minus minus


5n n DA q e d

nminus

lt gt minus




minus minus minus

1p



krarrinfinrarrinfin

η = lt

1p

d


cedQL

ϕ =

1p

4p


1 5 2 32 2 2 2

cedQ nk nL nk n k

minus minusϕ = = =

η minus minus +


absQL

ε =

1p


1 3 2 512 2 2 2

nk knk n k

+ minusε = minus =

η minus minus +



1p







284



2




ERIR r r x xr

=+ + +


ERIr R r

= =+

3

3 c 3

4




2 2

U UI R R R= =


R UUab I= sdot =


abR

U UIR R


2

4RUP RI

R= =

3

5


0 i D

i D

U U EI U U E

r R

le += minus minus +


1 5 V2

i

e ii

UU U U

le= +

gt


3



a


n

ki

k k k

I I

U E I rU IR

=

=

minus = minus =

sum


k k

E RI Ir r

= minus

2p

4p


1

11

nk

i kn

i k

ErI

Rr

=

=

=+

sum

sum și respectiv 1

1

11

nk

i kn

i k

ERrU IR

Rr

=

=

= =+

sum

sum2p

b


1 1

lim1 11

n nk k

i ik kgol n nR

i ik k

E ERr rU

Rr r

= =

rarrinfin

= =

= =+

sum sum

sum sum 1p

4prespectiv 1

1

lim 011

nk

i kgol nR

i k

ErI

Rr

=

rarrinfin

=

= =+

sum

sum 1p


0 0n n

ksc sc k sc

i ik

ER I I Ur= =



285

c


1

1

1

1

11

nk

i kn

i k echivalent

echivalentn

i k

Er

r EIR rR

r

=

=

=

= =++

sum

sum

sum

2p

4p

adică 1

1

1

nk

i kechivalent n

i k

ErE

r

=

=

=sum

sum 1p

respectiv

1

11echivalent n

i k

r

r=

=

sum

1p

d

echivalentE E= 1p

3pechivalent

rrn

= 1p

EI rRn

=+

1p



a


0 2

A B

A

B A

I I IE Ir RI

RI RI

= + = + = minus

1p

3pRezultă

3 22

3 23

3 2

B

A

EIr R

EIr R

EIr R

= + = +

= +

2p


286

b

( 2 )3 2CD B A

E R xU RI xIr R

minus= minus =


5p

Pentru max

min

03 2

3 2

2

CD

CD

CD

ERx Ur R

ERx R Ur R

RU o x

= = + = = minus +

= =

1p

Graficul

1p


CDdU Edx r R

α = = minus+

1p


tg CDU

R = minus = Ωα


2 mrE Uα

= minus + = 1p

c


2

1 2



1p

3p

Rezultă

2 2

(2 )E x RIRx R x

minus=

+ minus1p

Pentru

0 2A

2A

02

Ex IR

Ex R IR

Rx I


= = 2

max max 20 WP RI= =

1p



=


xEI cu xy nx rRn

= =+

1p

4p


= =+

1p



R= 1p




287

DOPTICAtilde




a


2 1

1 1n nx x R

minusminus = 1p

5p



1 1n nx x R

minusminus =

minus 1p

Se deduce

12

1( 1)Rxx

nR n x=

+ minus 1p


b


1 1

y nxy x

= 15p

4pLa a doua trecere

2 2 1 1

y xy x

= 15p

Dar 1 2y y= deci

2 03 cmy = minus 1p

c



1 1 2x x R

+ = Rezultă 12

1

2 214 cm2

xxx R

= =minus

1p


21

007 cmxyx

minus= = 1p


d


12 1

1

5 cm 7cm2Rxx R xx R


3p


2 x

Ryy yx R


1p




288


a



2 0 2 2 2 2( ) ( ) ( )

nn a l b n r e n e


minus 1p


b


D l∆

= 1p


Rezultă ( )2 2 2 20 2 2 1 1( ) ( ) ( ) ( )



c


4pRezultă

2Dinlλ

= 2p

d



0 1 1n n= = 1p



289

Cuprins





4

TESTUL 1

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 d 34 c 35 a 3



a



= 1p


b




deci 0fF F =- 2p3p

Rezultă F = 4 N 1p

d



2p

4pObţinem 1

2 fF F Fam m

= =- 1p






b






5

c



Obţinem 2Ahh = 1p


d





Rezultă 2 JGL = 1p



6


Subiectul I Punctaj

1 b 32 b 33 a 34 c 35 d 3




11

mν =

micro2p

3p


b


A

NN

ν = 1p


22

mν =

micro 1p


2

mN N= sdotmicro

1p


c


1 1

m R TpVsdot

= sdotmicro 1p

4pRezultă 5 2



2 2

m R TpVsdot

= sdotmicro

1p


d


4p

unde 1 2m m+ν =

micro 1p


1 2 1 2 2 1

1 2

( )m m m mm m m m



1p




7


a


1 0

311

1

300 K

831 831 10 Pa

= 3 dm

Tp p

R TVp

=


=

1p


32 1

32 1

2 600 K

831 10 Pa

=2 6 dm

T Tp pV V

= =

= = sdot

=

2p


313

33 2

600 K

4155 10 Pa2

=2 12 dm

T Tpp

V V

= =

= = sdot

=

2p

b


12 2 1 V 2 1 17( ) ( ) ( )2


4p


23 2 12




12 23 17 2ln 22


Q Q Q R T 1p


c


12 V 2 1 15( )2



d


4p


23 2 12







8


Subiectul I Punctaj

1 c 32 a 33 d 34 d 35 b 3




3p


b



S

EIR r

=+

2p

Rezultă 1 AI = 1p

c


torul R este 1

1

EIR r

prime =+

2p




1SC

EIr

= 3p4p




9


a


5p


1 1 1

PR R R= + 1p


S PS P

E ER RR r R r

sdot = sdot

+ + 1p



b


S

RR r

η =+

1p



P

RR r

η =+

1p


c


1 11

EP RR r

= sdot

+ 1p

3pDeci 1

11

( ) PE R rR

= + sdot 1p

Rezultă 6 VE = 1p


EIr

= 2p3p




10

DOPTICAtilde

Subiectul I Punctaj

1 b 32 a 33 d 34 a 35 c 3




f= 2p

3p


b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+1p


c


1

xx

β = 1p

5pRezultă 12

β = minus 1p




3p




11



1

cλ =

ν2p

3p


b




2 1

( )

S S


=minus

1p


c



Obţinem 10 1

Se Uh


0 2Se U

hsdot

ν = ν minus 1p


d



4pRezultă 22







12

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1a [ ] J Nm= =SIL 3

2 c 2

2 2 00 2 270 m

2vv v ad d

a= + rArr = = 3


2

10 6

6 ms

v v at v t

a

= + = +

rArr =

854 kgFma

= = 3

4 b 2

2

2

2

C

C

mvE p mv

pEm

= =

rArr =

2

2

2

2

C

C

mvE p mv

pEm

= =

rArr = 3

5 b 2

0max

0

20 m 10 m2

5 JC C p C

vh hg

E E E E

= = rArr =

= + rArr =

3



a


1 1

1

t f

n

G F ma

N G

minus =

= 2p

4p


21 25 msa = 1p

b


2 20 1 02 0v v a l v= + = 1p

3p1 12v a l= 1p

1 10 316 msv = = 1p

c


2 2

2

fF ma

G N

minus =

=

1p



22 01a

gmicro = minus =

1p


13

d


1t f

n

G F FN G

= +

=

2p

4p


5 NF = 1p



a

1 1f fL F d= minus

1p

4p1f

y

F N

N G F

= micro

= minus

1p


1p

1

75 JfL = minus 1p

bcosL Fd= α 2p

3p

255 JL = 1p

c

1C fE L L∆ = + 1p

4p

2

2CmvE =

1p

2 CEvm

=

1p

2 3 34 msv =

1p

d

2C fE L∆ =

2 2f fL F D= minus

1p

4p2f

F N

G N

= micro

=1p

C

f

EDF

=

1p

D = 6 m 1p



14


Subiectul I Punctaj

1b mpV RT=

micro

30 kPamp RT

V= =

micro

3



4c

3127 kgmpRT

microρ = = 3

5

b 1 1 2 23 32 2

U RT U RT= ν = ν

2 1 2 12 2U U T T= rArr =

11

2 200 J3UL R T RT= ν ∆ = ν = =

3



a


1 2


= = 1p

4p


1 2

1 2

V p pmR T T

micro∆ = minus

1p

1 2

1 2 2 1( )mRTTV

p T p T∆

=micro minus

33324 mV =

1p

b

11

1

p VRT

ν =

2p3p

1 026 kmolν = 1p

c

22 2

mp V RT=micro

2p

4p22

2

p VmRT

micro=

1p

2 512 kgm = 1p

d

11

1

pRT

microρ =

3p

4p

31 25 kgmρ = 1p



15


a


3p 3p

b


1 2

3V V V VT T

= rArr = 1p

4p12 1 2 1( )L p V V= minus 1p

121

2 1( )Lp

V V=

minus 1p

5 21 2 10 Nmp = sdot

1p

c


23 31cedQ Q Q= + 1p


1p

131 1 1 1

3

ln ln3VQ RT p VV

= ν = minus

1p


d


4pp VC C R= + 1p

1 1 15 5absQ RT p V= ν = 1p




16


Subiectul I Punctaj

1b 1 Wh = 3600 J 6720 10 JW = sdot

3

2 c 1 1 2

2

E I EIR r R R r

= =+ + +

1 1 2

2

E I EIR r R R r

= =+ + +

1 22

1

2 8R R r RR r+ +

rArr = rArr = Ω+

3

3 b 23ABRR = = Ω 3

4 c V1sc

EIr

= =Ω

3

5c

2

bec 90n

n

URP

= = Ω


3



a

1e pR R R= + 1p

3p2 3

2 3p

R RRR R

=+

1p

44eR = Ω 1p

b

1e

EIR r

=+

1p


88VU V= 1p

c

1 2 3I I I= + 1p

4p2 2 3 3I R I R= 1p

1 23

2 3

I RIR R

=+

1p

3 08I A= 1p


17

d


1 23

1 2e

R RR RR R

= ++

1p

4p125 A

e

EIR r

=+

1p

1 2

1 1 2 2

I I I I R I R

= +=

1p

12 083 A3II = =

1p



a


1 3 3E Ir U I R= + +

2p

4p

13

3

E Ir UIR

minus minus=

1p

3 15 AI = 1p

b


2 2 3 3I R I R= 1p

4p3 3

22

75I RRI

= = Ω

1p

22 2 2P R I= 1p

2 1875 WP = 1p

c

1e pR R R= + 1p

4p

11 30UR

I= = Ω

1p

2 3

2 3p

R RRR R

=+

1p

4875eR = Ω 1p

d

1 1

s

P UP E

η = =

2p3p

05 50η = = 1p



18

DOPTICAtilde

Subiectul I Punctaj

1 c 3

2 a 2

1

sin 3sin 60sin 2

i n r rr n

= rArr = rArr = 3

3 b 1 2 3 3 4C C C C C= + + rArr = δ 3


3

5b 2 2

1 1

2 1

3

3 150 cm

y xy x

x x

β = = = minus

rArr = minus =

2 1

1 1 1 375 cmff x x

= minus rArr =

3



a

2 21

1 1

3y xy x

β = = = minus

1p

4p2 13x x= minus 1p

11

43fx = minus

1p

1 40 cmx = minus 1p

b

32

1

3xx

β = =

1p

4p3 13x x= 1p

1 32

1 3

x xfx x

=minus

1p

2 60 cmf = 1p

c 3p 3p


19

d

1 2

1 1 1F f f

= + 1p

4p

20 cmF = 1p

14

1

FxxF x

=+

1p

43

1

1xx

β = = minus 1p





b

0L h= ν 1p

4p

150 055 10 Hzν = sdot

1p

151

1


λ 1p


c2

2

hcε =

λ 2p

3p


d

2 ex CL Eε = + 2p

4pCS

EUe

=

1p

079 VSU = 1p



20

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 a 34 a 35 d 3



a


0eF G+ =

1p


11

m glk

∆ = 1p


b

fF F= 1p

4pN G= 1p

fF N= micro


∆ =m glk 2 125 cm∆ =l 1p

c

e fF F ma+ =

1p

4peF F= 1p

2

2

minus micro=

F m gam 1p


d

22

2pk lE ∆

=

2p

4p2

Flk

∆ =

1p




21


a

F G ma+ =

1p



b

vat

∆=

∆1p

3p0 = +v v at 0 0v = 1p


c

total∆ =cE L 2p

4p2

2cmvE∆ = 1p


d


4p

2

1 2

= +mvE mgh

2

2 2PmvE =

1p

2( )2∆

=a th

1p


1p



22


Subiectul I Punctaj

1 a 32 d 33 b 34 a 35 c 3



a A

N mN

=micro

2p4p


b


4p0

0

mV

ρ =

mV

ρ = 1p

0

0middotmVV

V m=

∆ρ +1p



3p


d


4p0 p VRT

ν =


2p



a


2 2

1 1

=p Vp V

2p

3p

Rezultat final 2

1

2pp

= 1p


23

b



2 14T T= 1p


U RT∆ = ν rArr 12 10800U J∆ = 1p

c


( )=L A p V 2p

4p1 1 1

2 2p V RTL ν

= =


2p

d



2 14 =T T 3 12T T= 1p





24


Subiectul I Punctaj

1 d 32 b 33 a 34 c 35 c 3



aRS= R1+ R2 2p


b

UAB = R2 I 1p

4ps

EIR r

=+ 2p


c s A

EIR R r

=+ +

3p4p

I primecong 12 A 1p

d 1

EIR r

=+

3p4p




a


4pISC = 10 A 1p

ISC = scEIr

= 1p


b


4pEIR r

=+

1p


c

echivalent

echivalent

RR r

η =+

2p


91η 1p





25


1 a 32 b 33 b 34 d 35 a 3




b

2 1

1 1 1x x f

minus = 1p

4p1fC

= 1p


c

2 2

1 1

β = =x yx y 2p

3p


d

1 1 60 cm= + = minusx d x 1p

4p12

1

x fx x f

=+

1p





3p


bfW h= ν 2p

4p








26

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1c

5

8

km = 3 10 uas = 012

1ua min1 km = 15 10

cc

sdot rArrsdot

3


m

m

km = 200 m 48 h

= 15 s

dvt

d vt

= ∆ rArr =∆

3

3c

25

3

= 75 10 J2 = 208 kW h

11 J = 1 W s = 10 kW h3600

c c

c

m vE EE

minus


sdot sdot

3

4



= 75 Nm m

m v vF F

tsdot +

= rArr∆



5


0 0 02

0 0 2

0

4 1 3828 cm4

F l F lE lS l E S

d FS l l ld E

l l l


∆ = minus

3



a


4p2 = 2 m sxx

RR m a a am






27

d


5p


1p



1p



2p



a



= 11314 mdrArr

2p

5p2p

2 1 2

= 14 s

= 12 h 24 min 34 s

dv tt

t t t t

= rArr ∆∆

∆ = minus rArr1p

b

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr =

2p3p

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr = 1p

c


4p


2p

( )1 20

879 MJr

r

F

F

m g h h L

L


rArr = minus

1p

d

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

2p

3p

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

1p



28


Subiectul I Punctaj


3



3

b 1 1

2 2

1 1 11

22 2 2

5 cu 80 K 52 1673 3 cu 80 K2

V V

V V



ν

ν3

4

d 23 23 23

2323

40 J0

Q U LU

L= ∆ +

rArr ∆ = minus=


123112

1231 12 23 31

0110 J

UU

U U U U∆ =

rArr ∆ =∆ = ∆ + ∆ + ∆


3

5

c 22 72 10

VA

VA

Q C TQ NN NNC T

N


3



a

1 2 m m m= = 1p

3p1 2

2 1

125ν micro= =

ν micro2p

b1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


1p

4p1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


c

3

kg 157 m

mp V R TpR Tm

V


rArr ρ =

3p

4p

3

kg 157 m

mp V R TpR Tm

V


rArr ρ = 1p


29

d5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

4p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot 2p





c

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

4p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

2p

d

5 1 3

pp v

v v

p v

CC C R RC C

C C R


γ minus= +

1p

5p


( )( )

1 2 1 2

1 2

1 2

1 2

1

1v v v v v v


f f



2p

2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = =

1p2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = = 1p



30


Subiectul I Punctaj

1a 2

2

514 nC

U S qI U r tl t ql

S rq


=

3

2

a ( )( )

1 0 1 2 1

1 2 2 12 0 2

3 1

1 - 1

4 10


Kminus minus


α = sdot

10 0

1

9 1

RR Rt


3

3




3

4b

=12800 C

PIP tU q q

q UIt


3

5


= rArr = Ω


= rArr = =


3





1

9 E EI R r R


+ 2p

b



5p


1817 18

E EI R R rR R r I

R


rArr = Ω cong Ω1p




1p

2 05 mml lR S SS R



31

c


1p

4p1

1

105 A 11 s

EIR R r

I t


primeprime = rArr =

2p

1

1

105 A 11 s

EIR R r

I t



dI = 0 1p

3pU = E 1pU = 20 V 1p



a = 2400 mA h = 8640 C q sdot 2p


b1

06 AUI IR

= rArr = 3p 3p

c 14400 s = 4 hq qI t tt I


3p 3p

d


1p

5p


UI IR



2 22

6480 s = 18 hq qI t tt I


1p

2 22

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆

1p2 2

2

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆ 1p



32


1


3

2 c 3

3




3



D iil n n


3


0

2 4

s

s

h c h c e U

h c h c e U



3



a

22 1

12

1

2 4 cm

2 2 cm

xx x

xx

x


2p3p


b

2 1

1 1 1

1

25 4 cm

x x f

Cf

Cf

minus = =

rArr = δrArr =

2p4p


c

2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus

2p

4p2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus 1p



33

d


4p

2

1 2 1

1 1

2 1

22

8 cm 1875 1 1

xx x x

x x C

C Cx x


primeminus = +



3p




b

1

1

22

kDxD li

l


sdot= sdot

1p


2

R V Dx

lλ minus λ sdot

∆ =sdot

2p

076 mmx∆ = 1p

c2

2

2 2 192 mm

2

k x ixDi

l


= sdot 3p 3p

d


4p

1 21 22 2

D Dk kl l


sdot sdot1p


Soluţia 1

2

32

kk

= =

1p


0Cprime lt rArr


34

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj

1 a 32 b 3

3


0 sin cos 0sincos

Gt Ff mg mg

tg


micro = = αα

3



5



= + = + =

3



a


4p 4p

b


şi 2a


1p


2 2

T m g m aT m g m a

minus sdot = sdot



1 2



+ 1p

c T = m1(a + g) = 2222 N 3p 3p

d


1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p

4p

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p



35


a

2p

3p

cos5640 J

L F dL

= sdot α=

1p

b

NFfL Ff d

Ff= minus sdot

= micro

1p

4pN

FfL Ff dFf

= minus sdot

= micro 1p




4p4504 JfEc = 2p

d564 W

med

med

LPt

P

=∆

=

2p4p

564 W

med

med

LPt

P

=∆

= 2p



36


Subiectul I Punctaj

1 c 3

2 a 33 b 34 a 35 c 3



a3096 kgm

pRT

microρ =

ρ =

2p3p

3096 kgm

pRT

microρ =

ρ = 1p

b

a

a

N m N mNN

sdot= rArr =

micro micro1p

4p



a

a

N p VNp V RT NN RT

= rArr = 1p


RT= = sdot 1p

c

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus=

1p

4p

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus= 1p

1

2

11

pp f

=minus

1p

p2 = p1(1 ndash f ) = 96 ∙ 105 Pa 1p

d


4p


1p

21 O 2

1 2


ν + micro ν

ν + ν2p



37




3p 3p

b


2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= =

2p

4p2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= = 2p

c9kJ

p

p

LO

LO

η = rArr

= =η

2p

4p

9kJ

p

p

LO

LO

η = rArr

= =η

2p

d


4pT2 = T3 1p

ΔU = 0 1p



38


Subiectul I Punctaj

1 a 32 c 33 c 34 a 35 b 3



a


1 1 1

PR R R= + 1p



R = 11 Ω 1p

b




cE = I3r + UAB 2p

3pUAB = 22 V 1p

d


1

UIR

= 1p



2

UR


U1 = U2 = 96 V 1p




3pUb = 120 V 1p

b



Uf = 4 V 1p




W = 4464 kJ 1p



39


1 c 32 a 33 d 34 c 35 c 3





3p

Rezultă cvn

= v = 212 ∙108 ms 1p

b



r = 450 1p090 45rθ = minus = 1p

c




1p

4p1sin sin2

n l n π= 1p

1

2

sin nln

= 1p

l = 450 1p


3p 3p




1Cf

= 2p3p


c

2 1

1 1 1x x f

minus = 2p



1 1

x yx y

β = = 1p

y2= y1=10 cm 1p


2p3p




40

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj

1 c 32 a 33 c 34 c 35 b 3



a

2p

4p

2p

b





M m M m


= =+ +

1p

21msa = 1p


09 NT = 1p

d



17 NF = 1p



a


4pfc u FL L L= + 2p

2500JcL = 1p


41

b

u

c

LL

η = 2p

3p2000 08 802500

η = = = 1p

c

cossin sinfF f


1p

4p

u GL L mgh= = 1p

f

f

FF u

u

LL ctg L

L ctg


α1p

1 0254

micro = = 1p

d

2 60 40 s3

t∆ = sdot = 1p

4p

u cu c

L LP Pt t

= =∆ ∆

1p

2000 50 W40

2500 625 W40

u

c

P

P

= =

= =2p



42


Subiectul I Punctaj

1 d 32 d 33 b 34 c 35 c 3



a

1 1 1 1 1

2 2 2 2 2

( ) ( 2 )2 2 2

( - ) ( - 2 )2 2 - 2




+sdot


4p5

51

55

2

10 1 5 10 Pa18 9

10 1 5 10 Pa02

p

p

sdot= = sdot

sdot= = sdot

2p

b

F



133(3) NF = 1p

c

1 1 1

2 2 2

( 2 ) ( 2 )





2p

4p

51

52

510 Pa6625= 10 Pa

6

p

p

prime =

prime2p

d

1

2 p ll hpsdot

+ = 1p

4p1

1 12

p lhp

sdot= minus prime

1p

01m = 10 cmh = 1p




43



3p4p



c

212 1 1 1

1


= ν = = sdot =


1p

5p

5 51 11 1 2 2 2 1

2 1



3 523

5 2 10 3 10 (1 27) 1500 17 2550J2



1620 2550 1530 600 JQ = minus + = 1p

d

U Q L∆ = minus 1p

3pL Q= 1p

600JL = 1p



44


Subiectul I Punctaj

1 a 32 a 33 d 34 b 35 d 3



a

1 1 1 1 1AU I R I= rArr =

2 2 2 2 075AU I R I= rArr =2p

4p( )1 1 2 1 1 2

2 2 1 2

( )6V

( )E I R r R R I I

EE I R r I I


2p

b

1 11 1 2 2

2 2

( )( ) ( )

( )E I R r

I R r I R rE I R r

= +rArr + = + = +

2p

4p2 2 1 1 2 1

1 2 1 2

I R I R U UrI I I I

minus minus= =

minus minus 1p

2r = Ω 1p

c

66

15012 10 603 10fir

lRS


sdot2p

4p6 075A 075 60 45C8fir

EI q I tR r

= = = = sdot ∆ = sdot =+ 2p

d0

00 0

1(1 ) 1 1

fir

fir firfir



α2p

3p

40 Ct = 1p



a

ss

s

EIR r

=+ 1p

4p10sE E= 1p

10sr r= 1p

4 133 A3sI = = 1p


45

b

pp

p

EI

R r=

+ 1p

4p10prr = 1p

10

110p

ErE E

r

= = 1p

05ApI = 1p

c


4p2 pU I R= 1p

1 1 VU

1p

2 35VU = 1p

d1

s

RR r

η =+

2p

RR r

η =+

1p

3p

17 259227

η = 235 972236

η =

2p



46


1 d 32 c 33 c 34 c 35 a 3



a 4p 4p

b

sin sini n r= 1p

4psin sin n r i= 1p



c


4p

sin sin

3sin 12sin

23

i n r

irn

=

= = =1p

30r = deg 1p



d cosADAB

r= 2p

3p

115cmAB 1p



47


a


1

2

1

2

s

s

ch L eU

ch L eU

= +λ

= +λ

2p

5p


2

2 1

1 1 2

1 2


( )( )

s s

s s

hc e U U

e U Uh

c

minus = minusλ λ

minus λ λ=

λ minus λ

2p


b1

1s

cL h eU= minusλ 2p


c0 0

Lh Lh

ν = hArr ν = 2p3p

150 0924 10 Hzν = sdot 1p

d

2max

2smveU = 1p

4pmax2 seUv

m= 1p





48

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj


0 m 8 N4 m 0 N

x Fx F

= rArr == rArr =


(6 2) 2 8 J

2L + sdot

= =

3


22 2i f

kx mv kxE E vm

= rArr = rArr = = 3

3

a Viteza medie mdvt

= (1)





= şi 2 2dtv



1 22 2

mm m


+= rArr = rArr = =

++km h


3

4 a 35 d 3



a


1km20h

v =

2km40h

v =

1130min h2

t = =

2115min h4

t = = 2p4p


b

Viteza medie mdvt

= 2p


t = = 1p




d


21

2cmvE =

2

22

2cmvE =

Rezultă 2

1

222

1

c

c

E vE v

=2p

3p

2

1

4c

c

EE

= 1p



49


a




hl =α

iar 1

0NL =


2p

ECB = 32 J 1p

bLFf

= LFf1 + LFf2



c


1p

4p2




= = 1p

d



CDEE = 1 1

15





50


Subiectul I Punctaj

1 b 32 c 33 b 3





3

5 c 3



a



1 2

12p p p pT T

= rArr = 1p

2p = 5

2

N12 10m

sdot 1p


max1 max 1

p p p TTT T p

= rArr = 2p4p

max 12 600KT T= = 2p

c


1 00 1

91p Tp T

ρρ = = ρ 2p

3p

31 0819kgmρ = 1p

d


mp V RT=micro

şi 21 2



12 112





a


3p

2p


51

b


VVrArr = 1p


2 3

V VT T



2 12 2888KT T= = 1p


3

1 2p

vv

C RC RC


4p

1 3 1 1 1 1 13 3 3(2 ) 1800 J2 2 2


d

2

1

| |1 QQ

η = minus 1p

4p

1 2 3 3 2( )5

1 2p

Q Q Cp T TRC R

rarr= = υ minusγ

= =γ minus

Rezultă 1 1 1 15 252

Q RT p V= υ =

1p

12 1 2 3 1 1 3 1 1 1 1 1 1 1

2

3| | | | ln ( ) ln 2 (ln 2 15) 21932v



122η = 1p



52


Subiectul I Punctaj

1 c 32 c 33 d 34 d 3




a



1 21

PRI

= 2p

R1 = 5 Ω 1p

b



EIR r

= =+

1p



e

RR r

η =+

2p3p

80η = 1p

d


4pPuterea maximă

2

max 125 W4 EPr

= = 2p



a


5p


2

V

p

V

RRR RR

= = Ω+

32e pRR R= + = Ω 2p


EIR r

= =+

1p



3pP = 75 W 1p


3pPE = 100 W 1p


e

RR r

η =+ 2p

4pη = 75 2p



53



2max

max max2

2S

C S SmV eUE eU eU V

m= rArr = rArr = 3

2 b 33 a 34 c 3

5


1

2

R RR R

= minus= +

1

1 2

1 1 1 1 2( 1)( 1)( ) ( 1)( ) 5mnC n nR R R R R







3pDin 2

2 1 21

4 20cmy y y yy


c


4pCum 2

2 1 11

4 4 5x x x d xx


1

2 1

16cm54 64cm

dx

x x

= minus = minus

= minus =1p

d


2 1 1 2

1 1 1 ( 16) 64 128 cm16 64

x xfx x f x x


minus minus minus2p


1 1 2( 1)( 1) ( )

Rfnn

R R


2p

2( 1) 128 cmR n f= minus = 1p



54


a


3p 3p

b

302



sinsin irn

rArr =

3sin8

r = 1p

c


5p

tg hOI

α = rArr 1038mtg 3

h hOI = = =α 1p


= rArr2

sintg 0809m1 sin

rAB H r Hr

= sdot = =minus

2p


d


4p



53

i n r i n i

i i i n

i

= rArr = minus


rArr =

2p



55

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj

1 a 32 d 33 a 34 b 35 a 3



aG1 = m1g 1p


b




1 2

g m ma

m mminus

=+

1p


c


4p1 2

1 2

2m m gTm m

=+

2p


d

eF k l= sdot ∆ 2p

4p2Tlk

∆ = 1p




a





fFL = minus 1p

b

totalcE L∆ = 1p

4p

2

2cmvE∆ =




1p


56

c

util

consumat

LL



11 ctg

η =+ micro α

ctg xh

α = 1p


d

c totalE L∆ = 1p

4p

2

2cmvE∆ = minus

total fFL L= 1p

2

2mv mgd= micro

2

2vd

g=

micro 1p




57


Subiectul I Punctaj

1 d 32 d 33 d 34 a 35 c 3



a

1 1 1p V RT= ν 1p

3p

11

1

RTVp

ν=

1p


b


1 2

V VT T

= 2p

4p1 22

1

TVTV

=

1p


c


4p11

mV

ρ =

22

mV

ρ = 2p


d

( )12 1 2 1L p V R T T= ∆ = υ minus 1p


52VC R= 1p





58


a

11

NnV

= 1p

3p1 1 1p V RT= ν A

NN

υ = 1p

rezultat final 11

1

Ap NnRT


b

123 3 1U U U∆ = minus 1p

4p1 1VU C T= ν

3 3VU C T= ν 1p

( )123 3 1VU C T T∆ = ν minus 1p


c

43-4 3

3

ln VL RTV

= ν 1p


33-4 3

4

ln pL RTp

= ν 34 3 ln 2L RT= ν 1p


1p

d

cedat 4-1Q Q= 1p


1p

4p VC C R R= + = 1p




59


Subiectul I Punctaj

1 d 32 a 33 d 34 a 35 d 3



a

0

0p

R RRR R

=+

2p

4pe pR R R= + 1p


b


EIR r

=+

1p


( )0

A

E I r RI

Rminus +

= 1p


c

V eU R I= 1p

4pu r I= sdot 1p


d

A

EIr

= 1p

3p 0VU = 1p




a

21 1 1P R I= sdot


4p1

1

EIR r

=+

22

EIR r

=+

1p




60

b scEIr

=

2p3p


c

ext s

totala s

P RP R r

η = =+

2p

4p1 2sR R R= + 1p


η = = 1p

d rEP4

2

max =

2p




61


1 a 32 d 33 c 34 b 35 b 3



a1

1

1Cf

=

11

1fC

=

2p3p




c

112

111fxx

=minus

1p

4p2 2

1 1

y xy x

β = =

1p


d

1 2

1 1 1

sf f f= + 1p

4p2 1

1 1 1

sx x fminus = 1p







c0extL h= ν 2p

4prezultat final 14


dc sE e U= sdot 2p

4p




62

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj

1 a 0

0

FllE s

∆ =sdot

3

2 c 2

m4 ms 22 s s

va at

∆= rArr = =


3

3 b


= sdot =

3

4d

355 10 2750 N20

PP F v Fv

F

= sdot rArr =

sdot= =

3


cos ( sin )cos sin


α + micro α 3



a


3px24 = 3 ∙ (4 minus 2) 1p

x24 = 6 m 1p

b

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

=

2p

4p

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

= 2p

c

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

5p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1ptotal 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


d

total

total

9 m2254 s

m

m

xvA

v

=

= =

2p

3ptotal

total

9 m2254 s

m

m

xvA

v

=

= = 1p



63


a


3p2

0 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

=

1p20 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

= 1p

b


0

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus = 1p

4p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

c 056B

B

p

p

E mgh

E J

=

=

2p4p

056B

B

p

p

E mgh

E J

=

= 2p

d


4p024 J

f AB

f AB

pF B A

pF

L E E

L

= minus

= minus

2p

024 Jf AB

f AB

pF B A

pF

L E E

L

= minus

= minus 1p



64


Subiectul I Punctaj


2 d [ ]SI

J1mol K

C =sdot

3

3c 2

1

ln

1929 J

VL RTV

L

= ν

= minus

3


5 a A

NN

ν = 3



a

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

2p

4p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

1p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot 1p

b0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

c

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

2p

4p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

1p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr = 1p

d

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

4p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

= 1p



65


a

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

4p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

= 1p

b

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

2p

4p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

1p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr = 1p

c

L12341 = Aria12341 1p

4p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot 1p

d( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot

2p3p( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot 1p



66


Subiectul I Punctaj


2 a [ρ]SI = 1 Ω m 3

3 c q = I Δt 3

4 b 0 0 00

112 3 6eR R RR R= + + = 3

5 d 6 2003

UR RI

= rArr = = Ω 3



a2 3

12 3

11

e

e

R RR RR R

R

= ++

= Ω

2p3p

2 31

2 3

11

e

e

R RR RR R

R

= ++

= Ω 1p

b


4p


2

3

RR

1p

rArr 22

3

1 R

I RI =

+

1p

rArr 32

2 3

IRIR R

=+ rArr I = 04 A 1p

c

e

EIr R

= rArr+ 2p

4pe

Er RI

= minus 1p


d

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

2p

4p

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

1p2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =1p



67


a

Din legea Joule

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

3p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr = 1p

b

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1p

4p

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


c

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

2p

4p

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

1p2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω 1p

d

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

2p

4p

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr = 2p



68


1 c 1

SI

m ms s

cminus

= = ν sdot 3


3 c Negativă 3


3

5 a sin 2 141sin

in nr

= rArr = = 3



a 3p 3p

b

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

2p

4p

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p

c

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


2p

4p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

d

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

4p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1psistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr = 1p



69


a

00

814

0 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot

2p

3p0

08

140 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot1p

b 8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

2p

3p8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

1p

c

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

4p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ



d


34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

2p

5p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr = 1p



70

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj

1 a 32 a 33 b 34 c 35 d 3



a


2p4p


b


1 1 1 1

1

(2) ( ) ( )

G T m a G T m a


+ = rArr minus + =

+ + = + rArr + minus = +

1p

4p



1 2 1

( )2


minus += =

+ + +

1p


2 1 1 21 2

1 ( ) ( ) ( )2



+ +rArr 1 1

1

2 ( )2

m m m gTm m

+=

+ 1p

rezultat final 2

m02s

a = 25NT = 1p


1

4 ( )22

m m m gF Tm m

+= =

+ 2p

3p


d




2p

4p1

1

22

m mgNm m

=+

1p


1p


71



a




2p

4p2 2

90 m2 2

at F th gm

= = minus =

1p


1p

b


22

2

45 m2

2

FF g tv g t mm h hg

v gh


2p

4p



c



3pmax2v gh= 1p


d



4p2 maxL mgh= 1p





72


Subiectul I Punctaj

1 c 32 a 33 d 34 c 35 b 3



a


11

1

AA

A

pV RTNpV RTm NN

NN RTVpN


=

3p4p


b


11 A A

m N NmN N

microν = = rArr =

micro

1p2p


c


1A

NN

ν = 2 12 1

2 2A A

N NN N


3 3A A

N NN N

ν = = = ν 3p


1 2 3

2 36

amestecamestec

total


ν ν + ν + ν 1p


d



pν

= 1p


ν + ∆ + ∆= = 1p




73


a


3p 3p

b


4p


1

p Vp V RTRT

= ν rArr ν = 1p


2 11 2

3p p T TT T

= rArr = 1p


c


5p


2 22 2 3 3 3 2

3

32

p Vp V p V V Vp

= rArr = =

1p


2 3 2 1 12

3ln 3 ln 4800 J2



3

32

p TT Tp



1p


1p


1 2 2 3primit

L LQ Q Qminus minus

η = =+

1p3p




74


Subiectul I Punctaj

1 a 32 b 33 b 34 c 35 d 3




sc

E EI rr I

= rArr = 2p3p


b


4p


2 3 4

( )R R RRR R R

+=

+ + 1p



2 3 4

( )e

R R RR R RR R R

+= + +

+ +1p


c


EIR r

=+


eU IR= 1p

4p


1p

2 2 3 4 ( )U I R I R R= = +

1 2 5 1 5 2( ) U U U U I R R I R= + + = + + 1p

1 5

2

( ) eI R R RIR


d






75


a




2p

4p


W WIUt Ut

= = şi 22

23

W WIUt Ut

= =


2p

b


18URI

= Ω 22

9URI

= Ω 2p


1 2p

R RRR R

=+

1p



1 1 1WP R It

= = 2 21 2 2

WP R It

= = 2p4p



p

p

RR r

η =+ fie cu putil

consumat

R EPP EI

η = = 2p3p




76


1 d 32 c 33 a 34 d 35 a 3



a


1 1 1

0 4 4y x x x xy x x



2p4p


b



1 1 1 5Cf x x

= = minus = δ 1p

3p



c


= = 1p


2 1

1 1 1F x x

= minus 1p


1

FxxF x

=+

1p


d


FC

= = 1p


1

25 cm

F xXF x

= =+

1p





a



ν = = sdot 1p


1p


77

b



2extmvLε = + 1p


mε minus


c



LUe

ε minus= 1p


d



5p


1 4λ

λ = (1) 1p


exthcL h= ν =λ


hcL

λ = (2)


hcL

λ =

1p

1 1 11






78

TESTUL 11

A MECANICAtilde

Subiectul I Punctaj

1 b 2 2

2 22

m1kg kg m s2 sSI


2 b 3





m2 sin 10s



sdot sdot= = =

∆ ∆ 3



a


1p

4p



1 2 1 2



+ +

2p




c


1 2 0T Gminus + = 12 fG F= 2p


=micro

1p


d


1p

4p





79




Rezultă 42500 JAt

E = 1p

b


4p

2 2

2 2 f

B AG F


1p

Obţinem 2

2f

BF



c



sinhd =

α1p



1p


d




2p

4pObţinem 1 2

AtE

h sm g

=sdot sdot

1p




80


Subiectul I Punctaj

1d 1 1J kg K

SI

Q Jc cm T kg K


3


R Tp ρsdot sdot

=micro 3

kg8m

pR T

sdotmicroρ = =

sdot3



5 b 3



a


A

m NN

=micro

2p

4pRezultă 0A

mNmicro

= 1p



1p VT

Rsdot

=ν sdot

2p3p


c


4p


2

pR T

sdotmicroρ =

sdot

33

3

pR T

sdotmicroρ =

sdot 1p


3 1 1

2T TT T

ρ sdot= =

ρ1p

Rezultă 2

3

2ρ=

ρ 1p

d


Np V R TN

sdot = sdot sdot 1p

4p

Obţinem 1 12 4A

Np V R TN



1

24

Ap NnR T

sdot sdot=

sdot sdot1p

Rezultă 263

mol036 10m

n = sdot 1p



81


a


4pRezultă 3 11

22 R TV Vp



b


12 23tL L L= + 12 0L = 3

23 22

ln VL R TV






d







82


Subiectul I Punctaj


2 1J A sSI



sdot ∆= = = = Ω

∆

3

3 b 2 2

3 3eR R RR

Rsdot sdot sdot

= =sdot

3

2 3e

E EIR r R r

sdot= =

+ sdot + sdot3

4 d 3


WIR t

= =sdot ∆

50 mAI = 3



a





e e

EIR r

=+

= 5A 1p


b


2 32RI R I= sdot 1 2 3I I I= +2p

4p




1lR

Sρsdot

= 2

4dS πsdot

= 2

1 11 4


ρ sdotρ2p



4pJustificare

1

2 666Ate

EI I cresteR r

sdot= = rArr

+2p



83


a


4pRezultă 2bb

PRI

= 1p


b


1p

4pRezultă b

Rb

PU U sI

= minus 1p

RU R I= sdot

R

b

URI

= 1p


c




d

2R bP R I= sdot 2p

3p




84



= rArr = rArr =λ

3

2 d 3

3 c 22 1 1 1

1


2

i n in n n nr n r

= rArr = = sdot = 3

4 c 1 1 1 02 m = 20 cm5

C ff C

= rArr = = = 3

5 b 3




1 125 10

Cf mminus= =

sdot2p

3p

Rezultă 4C = δ 1p

b


2 1 2 1 1

1 1 1 1 1 1 fxxx x f x f x f x



4pobţinem 2

2 22

2fxx x ff xminus

= rArr =minus

1p



d


4pRezultă 1 2

1 1 100 10025 20

Cf f

= + = minus 2p




a


2p

4pRezultă 25 8

199

66 10 3 10 199 10450 10 4

minusminus

minus


sdot1p


1p


85


0

hcL h= ν =λ

2p3p


c

0eU L= ε minus 1p


eε minus

= 1p

Obţinem 19

19

(49 44) 1016 10

Uminus

minus

minus sdot=

sdot1P


d


2cmE eU ν

= =

1p


m mε minus

ν = = 2p




86

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj

1 b 32 c 33 b 34 a 35 a 3






12

m3 s

F mg maFa gm

minus micro =

= minus micro =

2p

4p1

12

m3 s

F mg maFa gm

minus micro =

= minus micro = 2p

c


2 2

2

cos 0

sin 0f

f

F FN F mgF N

α minus =

+ α minus == micro

2p4p


mgF micro= =

α + micro α 2p


4pRezultă 3 kgFm

g∆ = =

micro2p




2

2MvMgh = 2p

4p




2

0 cos1802c


4p


2p


2 2


2p

4p


2p



87


Subiectul I Punctaj

1 c 32 c 33 a 34 d 35 b 3




1p3p


micro= = 2p

b

Din pV RT= ν 2p

3p

rezultă 4molpVRT

ν = = 4 moli 1p

c


5p

1 2 2

1 1

2871 1 004(3) 4(3)300

U U TfU Tminus



2

1 2 2 2

11 1 1

1 1 1 4(3)

RTp p p TVk RTp p T

V

νminus


scade cu k 1p

d


4p1 0


micro= = =

micro2p



88


a


1 1 1

8atm 300K 1200 1200 273 927 degC2atm

p T pT T K tp T p

= = = = = minus = 1p

4p


Graficul

2p

b

1 1 1p V RT= ν 1p

3p1 1

1

p VRT

ν =

1P

v cong 012 moli 1p

c

323 2

2

ln VL RTV

= ν

1p



2 1


= = = = 1p

223 2 1

1

V ln 11088 JpL pp

= = 1P

d

1abs ced ced

abs abs abs

Q Q QLQ Q Q


4p2

12 23 2 1 21




025 25η cong = 1p



89


Subiectul I Punctaj

1 b 3

2


2 3E E EI I I

R r R r R r= = =

+ + +




1 23

1 2

3 3 2727A3 4

E I IIR r I I

= = =+ minus

3

3 d 3

4


RR R r

η =+ +



R rη =

+



3

5


2

8 8( ) 9 9 4m

RE EPR r r

= = sdot+


5 34

2

rr rR r

plusmn = =


3



a

12 3

12

1 AEI R Rr RR R

= =+ +

+1p

4p2 3

12

2 VBAR RU I

R R= =

+ 1p

22

2 A3

BAUIR

= = 1p

33

1 A3

BAUIR

= = 1p

b

22 2 2qI q I t

t= = sdot ∆

∆2p

4p2

2 A 30 s 20 C3

q = sdot = 2p

c



scEI

R r=

+ 1p


d

p

EIR r

=+ 1p

4p1 2 3

1 1 1 1 142pp

RR R R R

= + + rArr = Ω 1p




90


a


2LL

U UI RR R R=

+ + 2p


Randamentul este 2

2

1 05 50( ) I 2L

RIR R

η = = = =+

1p

b


2 290400J3U tW I Rt

R= = =



290400 003 kWh36 10

W = congsdot

1p


kWhC = sdot = = 1p

c


xxrS


lRS

= ρ


=


UI xRl

=+

1p

4p


2(1 )

UP xRl

=+

1p



max0 605Ux P WR

= = =


min 20163Ux l P W

R= = cong

1p


1p





91


1 c 32 d 33 c 34 d 35 c 3



a


2 1

n nx x

=



= minus

2p

3p


nminus

δ = = 1p

b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=



c


1 1

x yx y

β = = 1p

5p

Rezultă 22 1

1

xy yx

= 1p

Rezultă 2 11

fy yx f

=+

1p








lλ

= 2p3p



4kk Dx

l+ λ

= 2p4p



92

c



4pSe obţine

2Di

lλ

= 2p


d


i i∆ minus

ε = = 1p


ε = minus 1p

Rezultă DD

δε = minus 1p





94

TESTUL 1

A MECANICAtilde




aDeoarece N = Gn 1p


b


0

v vat t

minus=

minus 2p



c



d





a




f g= sdot =

sdot1p



max 2 2cm v pE

msdot

= = 2p3p


c




d


4obţinem rF

r

Ld

F= minus 1p




95





a



obţinem 0

0

p L SR Tsdot sdot

ν =sdot

1p


b


1 2

m m=

micro micro1p

4Dar 1 2 ν = ν + ν sau 1 2 1 2

1 2

m m m m+= +


Deci 1 2

2micro + micro

micro = 1p


c


0

m mm R TV

p

ρ = =sdot

sdotmicro

1p

3pDeci 0

0

pR T

sdot microρ =

sdot1p


d


4Deci 0max

0

p TTpsdot

= 1p




96


a


31

1 11

2 10 Pa

= 4 dm

9627 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

4


32

2 22

4 10 Pa

= 4 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p


33

3 33

4 10 Pa

= 8 dm

38508 K

p

Vp VT

R

= sdot

sdot= cong

ν sdot

1p


34

4 44

2 10 Pa

= 8 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

b


4

unde 34 V 4 3 4 35( ) ( )2


şi 41 P 1 4 V 1 4 1 47( ) ( ) ( ) ( )2



c




d


primit

LQ

η = 1p


Deci 12341

12341 cedat

LL Q

η =+ 1p




97





a


1 1 1

Pr r r= + 1p

3pObţinem 1 2

1 2P

r rrr r

sdot=

+1p


b


1 2P

E r E rEr r

sdot + sdot=

+1p


P

S P

EIR r

=+

1p


Rezultă 05 AI = 1p

c


2

P

P

EIR r

prime =+

3p4


d


4Deci 2 22

2

E I RIr

primeminus sdot= 1p




a






98

b


4

unde S

S

EIR r

=+

1p



RP




S

RR r

η =+

2p3p

Rezultă 50 η = 1p

d



P

EIR r

prime =+ 1p





99

DOPTICAtilde




a


1 1 1x x f

minus = 2p

4Obţinem 1 21

1 2

x xfx x

sdot=

minus1p



1

1Cf

= 2p3p


c


1 1 1 x x f

minus =prime 1p

4

cu 1 2

1 1 1 f f f

+ = 1p

obţinem 12

1

f fff f

sdot=

minus sau 2 2

22 2

x xfx x

primesdot=

primeminus 1p


d


1 1

y xy xprime prime

β = = 1p

4


1 1

y xy x

β = = 1p

Deci 2 2

2 2

y xy xprime prime

= 1p

Rezultă 2

2

12yyprime

= 1p



100


a


3Obţinem xiN

= 1p


b


Dil

λ sdot= 2p

4Obţinem

2l iD

sdotλ = 1p


c


2Dx

lλ sdot

=1p


42

Dxl

λ sdot= 1p



2Dx x x

lλ sdot

∆ = minus =1p


d


2Di

lλ sdot

= 1p

4unde 1

1 cn n

λλ = sdot =

ν1p

Deci 1 12

D inl i i

λ sdot= =

sdot1p




101

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj


2gt hh t

g∆

∆ = rArr = = 3p

2 b pF F t pt

∆= rArr ∆ = ∆

∆

3p

3


L = 22 J 3p



α 3p



b



2p

3p

2 2

1 1

G T m aT G m a

minus =minus = 1 2

g mam m

∆rArr =

+ 2p

c



+ = = 1p


2 1

08 kg01 kg

m mm m

+ =minus =

2p



2p

4p

1 2( ) 126 kg ms= + = sdotp m m v 2p



102




2p3p


b

GL mg h= ∆

1p


2

2atx = 1p


45 JGL = 1p

c



4p

21 2 2c f

h xE mg F= minus

1p

21 (sin 03)

2cmgxE = α minus

1p

1 12 JcE = 1p

d

22 2

1 12c

cc

p p EEm p E

= rArr =

1p

4p



2 24 JcE =

1p

2

1

2pp

=

1p



103



1 2

1 2

m m+micro =

ν + ν

1 1 2 21 2

1 2

A A

N Nm mN N

N N

micro micro= =

=

1 2

2micro + micro

micro = b

3p


1

1 1 ii i f f f i

f

VTV T V T TV

γminus

γminus γminus

= rArr =

1

1iV i

f

VL C TV



c

3p

41 1 2 2

1 32 2 5

1 1

3 3

p V p V

V VV V

γ γ

γ

=

= rArr = b

3p

52 2

1 1

22

1

1 1

065 260 K

Q TQ T

T TT

η = minus = minus

rArr = rArr =

b3p



a

1 2

1 1 2 2( )

m m m

m p V p VRT

= +micro

= +3p

4p


b


4pRezultă 1 1 2 2

1 2

p V p VpV V

+=

+2p

5 28 10 Nm3

p = sdot

1p

c

mpV RT=micro 2p

4p1 21 2pV pVm m

RT RTmicro micro

= = 1p



104

d

2 22

2

m RTp V

=micro 2p

3p5 2

2 241 10 Nmp = sdot 1p



a


3p 3p

b

3 31

1


11

186 10 NmmRTpV

= = sdotmicro 1p

4p

3 312 2 12 4 10 m

2V V minusρ

ρ = rArr = = sdot

1p

5 21 12

2

93 10 Nmp VpV

= = sdot 1p

2 1

212 1

1

ln 25915 J

T Tm VL RT

V

=

= =micro

1p

c

13 450 K

2TT = = 1p

5p

11

3 323 2

3

224 10 mTV VT

γminusminus

= = sdot

1p

5 233

3

083 10 NmmRTpV

= = sdotmicro 1p


4 44 40 Kp VT

mRmicro

= = 1p

d

min

max

1CTT

η = minus 1p

3p4

1

1CTT

η = minus 1p

95Cη = 1p



105


Subiectul I Punctaj


21

2

1 A

AA

A

EI Rr R

EIR R r

= rArr = Ω+

= =+ + a

3


c3

4 22

225 W4

ER r Ir

EPr

= rArr =

= = c

3

5( 1) 10 ka V a

V

UR R n n RU

= minus = rArr = Ω


3



a

e

UIR

= 3 4 15sR R R= + = Ω 1p

4p1

1

5sp

s

R RRR R

= = Ω+ 1p

2 10e pR R R= + = Ω 1p

1I A= 1p

be

EIR r

=+ 2p

4p

2eEr RI

= minus = Ω 1p

c

2 3

2 3

25pR RR

R R= = Ω

+ 1p

4p

1 10s pR R R= + = Ω 1p

4

4

5se

s

R RR R R

= = Ω+ 1p

12 A7e

EIR r

= =+

857 VeU IR = = 1p


106

d


1p

4p1 2

2 1 4

1 26 A7

s

I I I I R I R

I I

= +=

rArr = =

2 3 4

3 2 4 3

3 46 A

14

I I I I R I R

I I

= +=

rArr = =1p

1 4AI I I = + 1p

9 A7AI = 1p



a


4p1 3 1 1 2 3( )P P r R R R R= rArr = + + 1p

( )21 2 1 1 2 3( )R R R R R R+ = + + 1p


b

2

max 84EP r

r= = Ω 1p

3pPmax = 450 W 1p

1 2

1 2

05R RR R r

+η = =

+ + 1p

c

2 Ann

n

PIU

= = 1p

4p1( ) 6 AnE I R r U I= + + rArr = 1p


4 15n

R

URI

= = Ω 1p

d

P UI= 1p


432 WP = 1p



107

DOPTICAtilde

Subiectul I Punctaj

1

2 22 1

1 1

12

1

2

53

23 3

y x y yy x

fxx ff x

fy

β = = rArr = β

= =+


3

22 5π

∆ϕ = δ = πλ c 3

3 375 nmaan

λλ = = b 3

4 206 nmexex

hc hcL eUL eU

= + rArr λ = =λ +

a 3

5 00


c 3



a


4

2( 1)C nR


( 1)2

CRnrArr minus = 1p

15n = 1p

b

2 1 2 1

1 1 1 1 1 Cx x f x x


3p12

1

15 m1

xxCx

= =+


2

1

5xx

β = = minus 1p

c



4

12( 1)

l

nCn R

= minus 1p

1 12

lnn C R=

+ 1p

163ln = 1p


108

d


4p

13 1 1 3 1

1 1 1 1 1 Cx x f x x


13

1 1

025 m1

xxC x

= = minus+ 1p

3

1

083xx

β = = 1p



a




2kk ix x iminus

= rArr = 1p


b

2Dil

λ= 1p

3p2liD =λ

1p

142 mD = 1p

c


4p

2 ( 1)klx e n kD




Pentru 33

1kx i en

λ= rArr =

minus 1p

42 me = micro 1p

d

1 2r vk kδ = λ = λ 1p

4p2 12k k= 1p

1 21 2k k= = 1p



109

TESTUL 3

A MECANICAtilde




a

1

2

AI dtgIB d

α = = 1p

4p1 0 =d v t

2

2 2atd = 2p

Rezultat final 2 tg

=sdot α

vta

15s=t 1p

b

2 0v v at at= + = 1p

3p1

21 1 135 kJ2

= =cm vE 1p

2

22 2 90 kJ2

= =cm vE 1p

c1 1= ∆ = minus

fF c cL E E 2p3p


d







a

0T F G+ + = 1p

4pcossin

T mgT F

α =α =

1p


α=

αF mg

3 N 09 N2

= congF 2p

b cosmgT =

α2p

3p



110

c


3p=AE mgh

2

2BmvE = 1p



d


4

p mv= 1p

final2mv mv= 1p


= =vv 1p



111





a

pV RT= ν 1p

4pν =

microm m

Vρ =

1p

p RTmicro = ρ 1p


b

1 2 3 1 2 3( ) ( )p V V V RT+ + = ν + ν + ν 1p

4p

1 1 2 2 3 3

1 2 3

p V p V p VpV V V

+ +=

+ + 1p

173pp = 1p


c

1 1 1 ∆ = minusU U U 1 1 1 15 5 2 2

= ν =U RT p V 1 15 2

U RT= ν 1p

4p1 1 1 1

5 2

= ν =pV RT U pV

1 1 1 15 2

= ν =pV RT U pV 1p

1 1 1 1 15 10( )2 3



d

1 1 2 2 3 3

1 2 3


ν + ν + ν p

3p1 2 34 9


micro = 1p




112


a


4p

2 1 1 1 2 13 3= =p V p V p p 2 1 1 1 2 13 3= =p V p V p p 1p

1 1 1 1 12 2 4 4L p V p V RT= = = ν

14LRT

ν =

1p


b


4p

12 1 13 2 32


12 34LU∆ = 1p


c


3p23 1 15 6 152



= sdot =Q L 1p

d

P

LQ

η = 1p



92PQ L= 1p


η = = 1p



113








c


2 2

23

U rU r

= = 1p

5pK icircnchis 1 11

1 1

=+p

r RRr R

2 22

2 2

=+p

r RRr R

11 2p p

U R RI

= = 2p


d

1

1G

p

U IR

= 2p

4p1 11

1 1

12 k= = Ω+p

r RRr R 1p




a

2

b

UPR

= 2p3p


b


5p

2 A= =bPIU

1p



UnE I r

1p


c4 (4 )= + +b b aE I r R R 2p


d

2

max 4

=EP

r cacircnd R r= 2p

4p




114

DOPTICAtilde



b 2 1 1

1 1 1x x f

minus = 2p3p


c


4p2 1 2

1 1 1x x f

minus = 1p


d



1 12

1 1

x fx x f

=+ 1p





2Dil

λ= 2p

3p


b


4p2 2 4d i i i= + = 2p


c


4p

2 π∆∆ϕ =

λr 2

3sdot λ

∆ = =y lr

D

Rezultă 2 3π




II

= 1p


115

d


21

15 cm= =+

x fxx f




1

22 sdot=

minusl xl

x

2p

4p


λ= 1p




116

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1

b8

3

m 3 10 s

1 al 631 10 ua

1 an

c

dc dt

t


∆ ∆ =

3

2 d 3

3

a

m m

vF m a m

t∆

= sdot = sdot∆

m5s

v v∆ = =

= 25 NF

3

4

c

fR N F= +




θ = = micro

3

5


2

22 2

2 42

m v F dv v

m v F d


3



a

1

1

0const2m

d va vt

+= rArr = =

∆ 2p

4p11

1

2 100 m sdv vt


∆ 2p

b

22

2

50 sdv tt

= rArr ∆ =∆ 2p

4p

1 2 110 st t t t∆ = ∆ + ∆ rArr ∆ = 2p


d11 2 727 m sm m

d dv vt

minus+= rArr = sdot

∆ 4p 4p



117



b


0

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

2p

4p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

1p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p 1p

c


1 1 231 msin

hd d= rArr =α

1p


( )

2

2 2

2

2 2

0 sin - cos 2

237 m2 sin cos

ABAB R

m vEc L m g d m g d

vd dg



2p

1 2 868 mD L d d D= + + rArr = 1p

d


4p


2 2 1p

2 = 09 m 1p

pp

Ek xE xk

x


1p2

= 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr =

1p



118



11

1 1 2

2 2 12

2

1 8

A

A

m NN N

m N NN


micro

3

2

d

0 V

Q U LL U C T

Q= ∆ +



= 12465 JL

3

3

c

12 12

12 12

1213 13 13

13

12 13

31 2 32

5 61 3 80 1202 5

V

V

RC C RQ R T

Q C TQQ C T R T KQ

T T


3


3

5


2

1

1C

p

LQ

TT

η =

η = minus


3



119



3p 3p

b51

1 1 1 1 15 10 PaR Tp V R T p pV


c


1p

5p

0const 0 0 1i f

QV L U U U pQ U L

= = rArr = rArr ∆ = rArr == ∆ +

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p 1p

d ( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

2p

4p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

1p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p 1p



a

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

1p

4p


1

2 3 3 23

53 51 1 1 1

1

2 = 8 = 15

5 3

p

v

p V V pVp V V pVC

C

γγ

= rArr = rArr =

γ = =

1p

13 1 3

1

15 15 374 lR TV V Vp



120

b

2p

4p

( )12 2 1

12

37395 J

VQ C T TQ


=2p

c

1231 12 23 31 1pL L L L= + +

12 0 L =1p

4p

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

( )

( )

23 23 3 2

23 1 1 13 315 22 4

vL U C T T

L R T T R T



1p

23 186975 J 1pL =


31 12465 J 1pL = minus

1231 62325 JL =1p

d


2 3 2 3

2 3 2 3

A A A

B B B

Q Q QQ Q Q

= += +

32 2

2

lnAVQ R TV

= ν sdot sdot sdot


32 2 2

2 3

ln ln BB

V VQ R T R TV V





( )2 3 23

ln 0BVR T R T TV


( )2 3 2 2 3 23

1

2 3 2 3

4ln ln 3

42 ln 15 2 0 3

B

B A

VR T R T T R T T TV

R T

Q Q



rArr gt

3p 3p



121


Subiectul I Punctaj

1

b

1

2

21

2

156

lRl LLR

l lR lR L

= ρ sdotsdot

= ρ sdotsdot

= =

3

2

a U R I= sdot

2

lRmS R

m m d Sd ll S d S


sdot sdot

2 mU Id S

= ρ sdot sdotsdot

= 3 VU

3

3


2 WRI t

=sdot ∆

1 R = Ω

3

4



02 AI =

3

5

b2

max 4

p

p

EP

r=

sdot

1 2

1 2

6 12 5p pr rr r

r rsdot

= rArr = Ω = Ω+ 1 2

1 2

26 Vpp

p

E E E Er r r

= + rArr =

max 14083 WP =

3



a

1

3 1 1

075 A

EI THORN IR R r

= =+ + 1p

3p1

1

1

45 C 1p

R

R

qI p

tq

=∆

=

1p1

1

1

45 C 1p

R

R

qI p

tq

=∆

= 1p

b

2

3 2 2

06 A

EI IR R r


4p2 2 1

84 1U E I r pU V p

prime= minus sdot=

1p2 2 1

84 1U E I r pU V p



122

c


( ) ( )( )

1 1 2 2

1 1 2 2

90 19p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1p

4p1 2

1 1 2 2

18 V 19

pp

p

E E E Er R r R r


3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= =

1p3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= = 1p

d



( )3

3

2 1

1 1 1 1

1

2

115 106

R

R

I I I

E I R r I

I AI A

+ =

= sdot + +

==

2p

4p

1 1 2 2 1 1883 1

AB

AB

U I R I R pU V p


1p1 1 2 2 1

1883 1AB

AB

U I R I R pU V p




a


1 2 1 2

085 A

E EI IR R r r

minus= rArr =

+ + +3p 3p

b


3 2 2

2 A 1p

EI IR R r


1p




123

c

[ ] 015 mint isin ( ) 1 1 1

685 V

AB

AB

U E I R rU

= minus sdot +

=1p

6p


1p


( ) ( )( )

1 1 2 2

1 1 2 2

079

p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1 2

1 1 2 2

7 V

pp

p

E E E Er R r R r

= + rArr =+ +

3

25 A 1p

p

p

EI I


+

1p


2p

d 3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


=

2p3p3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


= 1p



124

DOPTICAtilde

Subiectul I Punctaj

1



02

2

sinsin n irnsdot

=

2sin 04r =


3

2


1 0

1 1ln gf n

= minus sdot

2

1 1l

a

n gf n

= minus sdot


( )2

1

14a l

l a

n nff n n

sdot minus= =

minus

2 1 4f f= sdot

3

3 c 3

4

d


1

22

2

2

2

extr

extr

m vL

m vL

sdotε = +

sdotε = +


3

5

c

2

11 2

2k

Dx kl

λ sdot= sdot sdot

sdot 2

2 2k

Dx kl

λ sdot= sdot

sdot

22 1 k k

d x x= minus

( )1 2 2 - 02k Dd

lsdot


3




b


4p



0 01 1

1

sin 2 cmsin

r rr d dd r

= rArr = rArr =

1p



125

c

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot

3p4p

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot 1p

d


4p





2n = = 1p




1

13 1 3 3

2Dx ilλ


sdot 2p

3px3λ1

= 36 mm 1p

b

x4λ2

= 36 mm 1p

4p

x4λ2

=i2 1p


12

2

3 4

450 nm 1p

sdot λλ =

λ =1p

c

3

1

3 1

34

13 3

4 3

4 2

3 360 nm2

DxlDxl

x x

λ

λ

λ λ


sdot =

3p4p


d1 1

14 4 4 6 mm

08 2Dx x

lλ λ


4p 4p



126

TESTUL 5

A MECANICAtilde




a


0N Gn Gt Ff+ + + =

1p


2p

1p

b




1p




c


N T T= + 1p

4p( ) 306BT m a g N= + = 1p

( )( )2 22 1 cos 90N T= + minus α 1p

5911N N= 1p

d


3p( )2

2B B


072m sBv 1p



127


a




b




2B



c


2

cc

mvEc p mv= = 2p


2 2C B

Ffmv mv Lminus = 1p

Rezultă

6 m sCv = 12 m sCp = 36 JCEc = 1p

d


max

2 2Cmv kx

=


4p

max cmx vk

= 1p




128






HemNamicro

= 2p3p


b



c


1p

4p


1p

Obţinem

1 1 1

2 2 2 1 2

1 2 1 2


ν= ν =gt ν = ν =

ν + ν = ν + ν


d





a


1 2

p pT T

= 1p


3 4

p pT T

= 1p


T2 = T4 = 1 3 360KTT = 1p

b

LTOT = L12 + L23 + L34 + L41 1p



129

c

TOT

p

LQ

η = 1p



d

min

max

1 TT

η = minus 1p

3p1

3

1 TT

η = minus 1p

30η = 1p



130





a


1 2 3

1 2 3

1 2 3

36V1 1 1

E E Er r rEp

r r r

+ += =

+ +

1p

3p

1 2 3

1 1 1 1 1pp

rr r r r

= + + =gt = Ω 1p


p

EI

R r= =

+A 1p

b


1 2

1 1 1

2 2 2


= +


2p

5p


11 2 1 2

E r E E RI


=sdot + sdot + sdot

2 1 2 12

1 2 1 2


+ minus=

+ + 2p


c

Deoarece

ABU I R= sdot

1 2I I I= + 1p

4p1 2

1 2

1 2 3

1 1 1AB

E Er rU

r r r

+=

+ +2p

UAB = 40 V 1p

d

Icircn relaţia

1 2

1 2

1 2

1 1 1AB

E Er rU

r r R

+=

+ +1p

3p


vRrarr 1p




131


a


4p



sdot=

sdot


U tminus sdot

=sdot

2p

5AQIU t

= =sdot

1p

b


4pRezultă 2pQRI t

= 1p

Rp = 22 Ω 1p


3pE = 120 V 1p

d


x p

x p

R Rr

R R=

+

2p

4p

px

p

R rR

R r=

minus 1p



132

DOPTICAtilde




a



2 smv e U= sdot 1p

4p11

22

S ex

S ex

ch eU L

ch eU L

= + λ = + λ

2p


b


11

S exch eU L= +λ

2p

4p

11



c


4p0exLh

υ = 1p


d2

2

cW Nh Nh= υ =λ 2p

3p




a1 2

11 11e

m

fnn R R

=

minus minus

2p3p


b


2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2p


minus x1 = 80 cm 1p


133

c


1 22 1

1 1 1d x xx x f

= minus + minus =

1 22 1

1 1 1d x xx x f



11

1y xy x

β = = = minusminus

2p


d


1 2

2

1

2 1

1 1 1

2

1 1 1

F f fxx

x x F

= +

minusβ = =

minus

minus =

2p4p



134

TESTUL 6

A MECANICAtilde




atg 1

3tg ϕ = micro = 2p

3p

30ϕ = 1p

b

t fG Fa

mminus

= 1p

4psin cosmg mga

mα minus micro α

= 1p


2310 =577ms3

a = 1p

c



1p

4pt fG F F= + 1p


310 3 10 100 N3

F = sdot = 1p

d



1p

4p

t fG F F+ = 1p

sintG mg= α 1p

3 1 1 100 3 100 3 200 N2 23




135


a

sin 5

1000m54 =15ms3600s

hpl

v

= = α =

= sdot1p


cosUcirc α


rArr sintg cos


α1p



4 352 sin 2 10 15 15 10 W=15kW100


b


4psincos


α1p

7500W=75kWP 1p


2p3p



3p25kJpE E∆ = ∆ = 1p



136





a

1

1 1 2 1 21

22 1 2 2 1

2

2 540 2 54 4 083 450 3 45 5

pVV T V TRT

pV T V V TRT


ν2p

3p

1

2

4 085

ν= =

ν 1p

b

1

1 1

22 2

p VVRT

p V VRT

ν= =

ν 2p

3p

1

2

08VV

= 1p

c

1 1 11 2 1 1 2 2

2 2 2

( ) ( )pV RT

p V V R T TpV RT

= ν rArr + = ν + ν= ν

1p

5p

1 1

1 2 1 22 2

( ) ( )

p V RTp V V RT

p V RT

= ν rArr + = ν + ν= ν

1p

1 1 2 2 2 2

1 2 2

5 08 450 540 ( ) 3 18

p T Tp T T


ν + ν sdot ν 2p


d

1 36lV =

12

2

2 3 36 3 18 54l3 2

V VV


1p

4p 1 2 1 2

1

1 22

9

08 08

V V V V lV V VV

+ = + =

= rArr =2p




137


a 4p 4p

b

212 1

1

41 1 4 1 4

ln 0

5( - ) ( ) 02V

VQ RTV

RQ C T T T T

= ν gt


4p

12 41 229356JprimitQ Q Q= + = 2p

c

23 4 1 4 1

134 4

2

5( ) ( ) 02

ln 0

VRQ C T T T T

VQ RTV


= ν = lt2p

4p


d1 cedat

primit

QQ

η = minus 2p3p

3043η = 1p



138





a

2AP R I= sdot 2p

3p25 WP = 1p

b

0

0 0

0

A

A

A

I I IxI R I Rl

l xE I R I Rl

= + = minus

= +

3p5p

05 mx = 2p

c

ee

E EI RR I

= rArr = 1p

4p0

00

05A

A

A

I I IR lI Ix R

= +sdot

= =sdot

2p

151 151eI A R= = = Ω 1p


3p2250Wh 225 kWhW = = 1p



139


a


3p200 4A50

totaltotal

PP U I IU


b



5p



1p

21

1 1 1 2

22

2 2 2 2

40 102 4

2

60 152 4

2

I PP R RI

I PP R RI

= rArr = = = Ω

= rArr = = = Ω

2p

10VABU = 1p



d

1

1 01 1 3

0

191 1800 C

5 10

RR Rt tR minus


α sdot

2p

4p

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot= t2 =

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot

2p



140

DOPTICAtilde




a

1 2


n ff R R


2p

4p

1 2


n ff n R R


2p

b

2 1

1 1 1

aerx x f= + 1p

4p2

24 cm 48cm5


2

1

xx

β = 1p

04β = 1p

c

2 1

1 1 1

apax x f= + 1p

4p

2

96 cm 192cm5


21

xx

β = 1p

04β = 1p

d2p

3p




141


a

412 10 m 12mm2Dil


3p


b

λapăapatilde nλ

λ = 1p

3piapă

12 3 09mm4apatilde

iin

= = sdot = 1p


c



2p




2p

10nouk = 1p

d






142

TESTUL 7

A MECANICAtilde


3


md dvt t t

= =+

Cum 112 6

d dtv v

= = şi 222 2

d dtv v

= =

6 154

6 2

md vv vd d

v v

rArr = = =+

a

3

4


∆= = minus

∆






Icircnlocuind 35

rArr micro =c

3

5


11 2

2mvmgh v gh= rArr =




∆ b

3



a


3p 3p

b


4p



Randamentul este 2

u

c

L mghL Fl

η = = 1p


l= =

η1p


143

c


0fG T N F+ + + =


2p


lα =

Rezultă 2fhF F mgl

= minus

2p

600 NfF = 1p

dcos

cosf

f f

FF N F mg


α 2p3p

065rArr micro = 1p



a


cE L∆ = 1p

4p



1p

Ff = μN = μmg

Rezultă 2 2

200 2 21 ms 458ms


2p

b



1p


1

2 2mv mv mgR= + 1p


1 11 ms 331msv = = 1p

c



max max2 2mv vmgh h

g= rArr =

2p3p

max 105msh = 1p

d


max 081 m2vh fh fg

= = = 1p



222 2


2p

2 204 v m s= 1p



144




1 2 2 1 1 21 2 2 3 2 3 2

2 3

( )( ) 150 J ( ) 200J 0752

p p V V LL L p V VL

rarrrarr rarr

rarr


c

3

5


max

1CTT

η = minus


1 max 1

1L T LQ T Q

η = rArr minus =


3 150 K8

TT = =d

3



a


1 1

0 1

V VT T

= şi

2 2

0 2

V VT T

= 1p

4p

dar 1 2V V V+ = şi 2 1 12


23VV = iar

1 2 2VV V= = 1p


32TT = şi 0

234TT = 1p

Rezultă 1 4095KT = şi 2 20475KT = 1p

b


4p

1 1 ( )

2 2l lV S V x S= = +


2 2 ( )2 2l lV S V x S= = minus 1p

Obţinem 50 00 1 1 2

500 2 2 0 2

2 N( ) 066 102 2 2 3 m

N( ) 2 2 102 2 2 m





1p

51 2

52 2

N066 10m

N2 10m

p

p

= sdot

= sdot1p


536 NF = 1p

d


mp V RT=micro

şi 2 2 2 20 2 2

1 1 1

m V m Tp V RTV m T

= rArr =micro 2p

5p

Dar 0 01 2

3 32 4T TT T= = şi 1 2

23 3V VV V= =

1p

2 2

1 1

2m Vm V

rArr = 1p

2

1

2mm

rArr = 1p



145


a


1p

3p


2p

b


1 2 1 2


= rArr = rArr = 1p


2 3

p pT T

=


31 13 3

p p ppT T

rArr = rArr =

2p

5 23 10 Nmp = 1p

c


4p1 1

32 32



1 2 2700JU rarr∆ = 1p

d

1

LQ

η =



1p



44 1 4 1 3

p p p p TTT T T T

= hArr = rArr =


2p


1 1

4 2 02218 9

p Vp V

η = = = 22rArr η = 1p



146



41 2sR R= 1 1 1 2

2 3pp

RRR R R

= + rArr = 22 53 3sR RR R= + =

1 1 23AB

RR R

= +


b

3



a

1 2R R R R= + = 1p

4p

22 1

1 1 1 3 22 3

RRR R R R

= + = rArr = 1p

3 22 53 3R RR R R R= + = + = 1p

3

1 1 1 3 1 85 5

5 6258

AB

AB

R R R R R RRR

= + = + =

= = Ω1p

b


UU IR IR

= rArr = = 2p


( ) 264ABAB

EI E I R r VR r

= rArr = + =+

2p

c




AB

EIR r

= =+

1p





d

2

max 4EP

r= 2p

3p

max 8712WP = 1p



147


a


2 4 1

3 4 5

00


minus minus =

2p


1 3 1 1 3 1 2 2

2 3 2 4 4 2 2

2 3 2 5 5



3p



4p19MNU V= minus 1p

c

24 4 4P I R= 2p

3p4 8P W= 1p

d2

1 1 1W I R t= 2p3p

1 10800 J 108 kJW = = 1p



148

DOPTICAtilde


680nma a aa

nnλ

λ = rArr λ = λ =

d 3

2 c 3

3



2MO PO AOMOAO PO

= rArr =

2ON PO O BONO B PO

= rArr =


=09 m2 2


b

3

4


k Dxl

λ=


2

1 1 mm222 2 mm2

Dk xl

Dk xl

λ= rArr = =

λ= rArr = =

c3

5


2 1 1 2

1 1 1 x xfx x f x x

minus = rArr =minus


1

x x xx

β = rArr = β 1 1

11 1

x xf βrArr = =

minus β minusβ


20 10 cm1 1

f minusrArr = =

minus minusb

3



a



1 2

1 1 1 1( 1)( ) n nC n RR R R C


R = 10 cm 1p

b

Din 1 1 20C f cmf C

= rArr = = cm 1p


2 1 1

1 1 1 fxxx x f f x

minus = rArr =+

2p


x minus= =

minus2p


149

c1 2d x x= minus +

2p3p


d


1 1

y xy x

β = = 1p

3p22 1

1

xy yx

= 1p

2 2 cmy = minus




a


1 11 1

2 12 2

( )(1 )

c ch L eU h L eU

c ch L eU h L eU f


2p

5p

11

1

2

08

ch LeU

c eUh L

minusλ

=minus

λ

1 2

1 2

0802


λ λ2p



LL hh


1410 HzL = 1p


1

ch hε = ν =λ

2p3p

191 011 10 J 006eVminusε = sdot = 1p

d


11

22

2

2

2

e

e

m v eU

m v eU

=

=

2p

4p1 1 1

2 2 108v U Uv U U

= = 1p

1

2

54

vv

rArr = 1p



150

TESTUL 8

A MECANICAtilde




a


4p2

1 2atx = 2 0x v t= 2p


= = 1p

b0v v at= + 2p

3p16msv = 1p

c


3prezultat final

-=fracircnare

fracircnare

va


d

1 2= +d d d 2p

5p1 0=d v t 1 1 2= = = 64d x x m 64 m 2p


2 2 fracircnare

vdaminus

= 2 = 32d m




aE mgh= 2p


b

c totalE L∆ = 1p

4p

2

2cmvE∆ = = +

ftotal G FL L L 1p



1p



151

c

c totalE L∆ =

1p

4p 0cE∆ = GL mgh=

f


rezultat final = -f

totalFL mgh = -4500

f

totalFL J 1p

d

= +f f f


4p2f


F oprire

F FL d

+= minus 1p







152





a

pV RT= ν 1p

4p1 1 1 2

2 2 2 1

p mp m

ν micro= = sdot

ν micro2p

rezultat final 1

2

74

pp

= 1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0


4pVf = 2 V 1p


d




4p 4p



a1 1

1 1 11

p Vp V RTRT

= ν rArr ν = 2p3p


b


4p

1 2

1 2 12 1

2 1

34

V

p pV ctT T U C T

T T


=2p




153

c

min

max

1 TT

η = minus 1p

4pmin 1=T T 1p

max 2 1= = 4T T T 1p


η = = 1p

d

P

LQ

η = 1p

4p


1


1-2 2-3 3-1L L L L= + + 1-2 0L = 32-3 2

1

ln VL RTV

= ν 3-1 1 3( )L R T T= ν minus 1p


minusη =

+ 25η 1p



154





a


( )1 1 1 3E I R R= + de unde 1 3AI =2p


b



2422

131 )()( 1p


c


2p4p


d


2p3p




a



bP1 + P2 = Rp I

2 2p3p


c

echivalent

echivalent

RR r

η =+ 2p



d


4p


EIR r

=+

1p




155

DOPTICAtilde




b

2 1

1 1 1x x f

minus = 1p

4p2

1

xx

β = 1p


c

21 CCCsistem += 1p


2 1

1 1sistemC

x xminus = 1p


d

1 2= -x d x 2p

4p

1 22

1 2

x fxx f

=+

1p




a


3p11 10

yi = 1p


b


2i 1p

4p1

12 22id i= + rArr d = 3i1

2p



156

c

22 29

2iy i= + 2

22 072 mm19yi = = 1p


1 2Dil

λ= 2

2 2Dil

λ=

1 22

1

ii

λλ = 1p


d


iin

= 2p


1 1

1

-025 i ii

= 1p


n = 1p



157

TESTUL 9

A MECANICAtilde


5

21 1 2 2

12 1 2

2

2 2m 1156s

m v m v

mv v vm

=

= =3



a 3p 3p

b

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

4p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

c2 2

08 NT m g m aT

= minus=

2p3p2 2

08 NT m g m aT

= minus= 1p


158

d

1 0 1 1 0

1 1 0

1 1 0

( ) 0

( )g 0 ( )


+ + + + + =

minus + == +

1p

5p


1p

2 2

2

2

0 0

m G TOy G Tm g T

+ =minus =

=

1p

2 1 0

2 10

( )m g m m gm mm


=micro

1p

m0 = 03 kg 1p



aEA = m1gh 2p

3pEA = 1 J 1p

b

ndashf AB

f AB

B A F

B A F

E E L

E E L

=

= +2p

4p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p

c


4p

( )( )

1 1 1 2 1

1 1 1 2

1 1

1 2

m v m m v

m v m m vm vv

m m

= +

= +

=+

1p

unde 1m25s

v = 1p

01 255 m08503 s

v sdot= = 1p

d


( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

2p

4p( ) 22

1 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

1p

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

= 1p



159



VU VC T∆ = ∆ b 3


3pRT

microρ =

[ ] 3

kgmSi

ρ = c

3

41 1

1 1

33 22

L p VLU p VU

=

∆ = rArr =∆

a

3

5 36 gA A

m N Nm mN N


a 3



a15 mol

mν =

microν =

2p3p

15 mol

mν =

microν = 1p

b

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

2p

4p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

1p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ = 1p

c

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

3p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot 1p

d

2 2

3 3

mp V RT

m mp V RT

=micro

minus ∆=

micro

1p

5p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot 1p



160


a

L1231=Aria1231 1p

3p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot

1p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot 1p

b

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

5p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot05p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot 05p

c

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p

3p

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot 1p

d

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= =

1p

4p

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= = 1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η =

1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η = 1p



161




d 3

2 2W RI t= ∆ b 3

3 [ ] 2 1SI

V1 11 N m A sA


4

( )

( )

Aria trapez2

200 300 2500 500 mC

2

b B hQ

Q Q

+ sdot= =

+ sdot= = rArr = a

3

5

( )0

2

0

0

1

288

2110 C

R R t

UR RP

R Rt tR

= + α

= rArr = Ω

minus= rArr = deg

α


3



a


3p

1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω

1p1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω 1p

b

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

4p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω 1p

c

Potrivit legii Ohm

22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

2p

4p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

1p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

= 1p


162

d

1p


( ) ( )2 4

2 1 2 4 3 4

2 4

I I II R R I R R

I I

= +

+ = +

rArr =

1p( ) ( )

2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr =

1p

( ) ( )2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr = 1p



a

1 2

1 2

1 2

1 1 1

07 A

E Er rI

RR r r

I

+=

+ +

=

3p

4p

1

2

2E EE E

==

1

2

2r rr r

==

1p

bU = R I 3p

4pU = 7 V 1p

cP = RI2 3p

4pP = 49 W 1p

d88

e

RR r

η =+

η =

2p3p

88e

RR r

η =+

η = 1p



163

DOPTICAtilde

Subiectul I Punctaj

12

2 smv eU= rArr

2

2SI

2 ms

S

e

eUm

=

d 3

2 2

1

sinsin

i nr n

=

a 3


4 0 00

275 nmextext

hc hcLL


b 3




a

1

1 01 m10

fC

f

=

= =

2p

3p

1

1 01 m10

fC

f

=

= = 1p

b

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

4p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr = 1p

c

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

4p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


d

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

2p

4p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

1p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus1p



164


a

Din grafic15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot

1p2p15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot 1p

b

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

5p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot1p

c


4p

rArr

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot1p

d

( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p

4p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot= 1p



165

TESTUL 10

A MECANICAtilde




a



2

7 msdvt



1

2 msvv a t at

= rArr = =

1p

4p


2 11 1 1 1

1

2 9m2vv a d da

= rArr = =


1 3 3 21 3 3

1 3 31 3 3

22

0

final

final

final

v v a dv a d

v v a tv a t

v

= + = minus= + rArr

= minus= 21

33

21

33

15ms

12m2

vat

vda


1p



b


4 1 3 41 3 4

4 1 4 113 3 3

3


t


3p4p


c



1p


1p


1p



166

d


2 20G N N G+ = rArr = 1p


2p


1p



a


2

2 20

45 m2

2 2

gth

v v g h v g h

∆ = =

= + ∆ rArr = ∆

2p

4p2

2cmvE mg h= = ∆

1p


b


( )c

p


∆ ∆= =

minus ∆ minus ∆

2p

3p

rezultat final 016c

p

EE

= 1p

c


2

22

c

c

L E Fd

mvE

v gh

= ∆ =

∆ =

=

2p

4p

2

2 2mv mghF

d= = 1p


d


2

( 1)

2

f i

i f i

f


mvE fmgh


2p

4p

( 1)( 1) ( 1)



1p




167





a


mpV RTm= ν


micro=

2p3p



A

N N pVpV RT NN RT

= rArr = 2p3p


c


1

p VmRT

micro= respectiv 2

22

p VmRT

micro= 2p


2 12 1

V p pm m mR T T


1p


d


2p



2 3 2






a


primit primit

QLQ Q

η = = minus 2p



1p


b


1 1

1 T TT T

∆η = minus = 1

TT ∆=

η1p


η1p


2 300 KT = p


168


2 22

m m RTp V RT Vp

= rArr =micro micro

2p3p


d


4p


2 3 2

p p pT TT T p

= rArr = 1p


2

1V Vm pQ C T

p

= minus micro

1p




169





a


= = Ω 1p


232 3

R RRR R

=+ 1p


2 3

R RR R R RR R

= + = ++

1p


b




s

EIR r

= =+

1p



c


2 2 3 3

I I IR I R I

= +=

2p3p


1p

d


1 3R R R= +1p


1 2

093As

e

EIR R r

= =+ +

1p


3 3U R I= 1p




170


a


1 1 1 11

33 3 3 3

3

125 A

05 A

PP U I IU

PP U I IU

= rArr = =

= rArr = =

2p


1p

Din legea lui Ohm R

R

URI

= 1p


b



Iminus

= + = + rArr = 1p




d




consumat

095PP

η = = 1p




171

DOPTICAtilde




a


1 20 cmfC

= =


1 1

30 cmf xxf x

= =+




1 25 cmfC



2 1

f xxf x

=+


1p


2p

b



= 1p


c


1

1346 cmFxxF x

= =+

1p


1

y xy x

β = = 1p


1

xy yx

= 1p


d


1

4xx

β = = minus 1p


2 1

1 1 1F x x

= minus 1p





172


a


1ext s

hc L eU= +λ

(1) 1p

4prespectiv 2

2ext s

hc L eU= +λ

(2) 1p


1 21 2 1 2

1 1 s ss s

e U Uhc e U U h

cminus λ λ


1p


b




se obţine 10

1

sc eUh

ν = minusλ

1p


c



4p

Respectiv 22


Se obţine 1 1

2 2

s

s

v Uv U

= 1p

rezultat final 1

2

073vv

= 1p

d


11

ext shc L eU= +λ

(1) respectiv 33

ext shc L eU= +λ

(2)


1p

3p

3 1

1 1s

hcUe

∆ = minus

λ λ 1p




173

TESTUL 11

A MECANICAtilde


c 2

2 2mkg sN s s=

kg m kg m


= adimensional 3

2 c 3

3


5h = 0cE =2

0

2m v m g hsdot

= sdot sdot 0m2 10s

v g h= sdot sdot =

2

2 05 kgc

o

Emvsdot

= =

3

4 c 3

5 d ( ) 60 kW


= = = 3



a


3p 3p



= =+ +

3p

4p

Rezultă 2

m75s

a = 1p

c


sdot=

sdot ∆


4dS πsdot

=

3p4p


d



4pRezultă 1 2





174


a


ti tfE E= 2

0

2 cf pfm v E Esdot

= +

20


= =

2p

3p


b


4pRezultă 1 1



c




h h= sdot =

1p


2 20


= + sdot sdot


2p

Rezultă m1038 kgs

p = sdot 1p

d



Rezultă 0GL = 1p



175



d 2 3

2 3


mol k

minussdot


sdot

3

2 c p vRc c= +micro

p vRc cminus =micro

3


2 8VV = 12

1

8TT

γminus= 53

p

v

cc

γ = =

2 232 3 3

2 11

8 (2 ) 4 4T T TT

= = = rArr = sdot 3

4 b 3





a



3 311

1

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro 1p

3pSe obţine 3 31

11

025 10 mm R TVp


sdotmicro1p


b


NN

ν = 1p


mν =

micro1p


mNmicro

= 1p


c


Np V R TN

sdot = sdot sdot 1p


V= 1p

Se obţine 1

1

Ap NnR T

sdot=

sdot1p



176

d


1 2

( )m R T m m R TV V



1 1 2

1 2

( )m R T m m R TV V



2p

4punde 1

11

m R TpV

sdot sdot=


1 22

2

( )m m R TpV

minus sdot sdot=


1p


1 2

( )m R T m m R TV V






a



5p



4 1

34 3 3 33 1

4ln ln 2 2 ln 2 4 ln 22


= = = =


2p


b






c


max

1 TT

η = minus 1p

4p1 1

min 1p VT TvR

= =

3 3max 3

p VT TvR

= =

1p

Obţinem 1 1

3 3

1 12 2

p V pVp V p V


Rezultă 75η = 1p

d


3p 3p



177



c SI

V RA


SI

WR RI t


3

2 d 11

lRS

ρsdot= 2

2

lRSρsdot

= 1

2

14

RR

= 3

3 b 34 d 3




a


1p

4p





c


3

13 AABUIR

= = 2 24 AI = 2p




1 2 1 2 1 2( )I R R r r E Esdot + + + = + 3p4p




a


4extPP =

22

16ER I

rsdot =

sdot1p


2

P

EIR r

= +

1p


Rezultă 12 2784 016R = Ω Ω 1p


178

b



r= =

sdot


Rezultă 15 kJW =

c



Rezultă 32 WP = 1p

d


1 1 1 11

8E r IE R I r I RI


1

1

RR r

η =+

2p3p

Rezultă 80η = 1p



179

DOPTICAtilde


a 22


2 c 3


3

i rir

= rArr = = 3

4 c 15

2 1214

1 2

1

5 10 252 10

c

cλ ν sdotν

= = = =λ ν sdot

ν

3

5 c 9

43

600 10 2 6 10 06 mm2 2 10Dil

minusminus

minus


sdot3



a


1 1 1x x f

minus = 1p

4Rezultă 1

21

20 ( 30)20 30

fxxf x

sdot minus= =

+ minus2p


b

Distanţa focală 2

1 1 2

1 1 11nf n R R

= minus minus

unde 1 1n =

( )1 1 11nf R R

= minus minus minus

2p

4p

obţinem ( )2 11 1

2n Rn

f R fminus

= rArr minus = 1p

Rezultă 20 1 152 20

n = + =sdot

1p

c


3p 3p

d


1 2 1

2( 1)

nCf n R

C nR

= = minus

= minus

2p


1 15 1151 112

C nnCn

minus minus= =

minus minus1p

Rezultă 2CC

= 1p



180



8

1 91

3 10600 10

cminus

sdotν = =

λ sdot2p

3p


b


16 152 9

2

3 10 1 10 055 10 Hz54 10 18

cminus


λ sdot 1p



2 1

( )( )

h e UhU

e


∆ = unde 152

2


λ1p


c


cL h h= ν =λ

1p

4pObţinem 17


Dar 17

190 19

0044 101 16 1016 10

eV J Lminus

minusminus

sdot= sdot rArr =

sdot1p


d


11 0 ch h Eν = ν +1p


1p





181

TESTUL 12

A MECANICAtilde




a


1 1

1 1

0

f

F T m aN m gF N


2p


2 2

2 2

0f

f

T F m aN m gF N

minus =

minus == micro

2p


1 2

( ) m125s

F g m mam m


+1p


c


1 1 2

m2s


m1s


4p






2 21 cos180


5p

Rezultă 2 2

1 0482

v vgdminus

micro = = 2p


3pp = 8000 Ns 1p

c


21

112 2

MvW =

2p3p

1 8000JW = 1p

d


4 2 2Mv kx

= 2p

4p

Rezultă 212

kN3204 mMvk

x= = 2p



182





a


1 2


1p

3p1 1 2 21 2p V p Vm m

RT RTmicro micro

= = 1p

1 1 1

2 2 2

35m pm p

micro= =

micro1p

b


1 2


1p

4p

1 2

21 2

2 2

( )

2

m m RTm RTp V

V p

+micro micro

= =micro

1p


12

2 2 1 12 1 21 2

2 1 2 2 2

2 2

1 1( )1 1 1( ) (p p )

2 22

mmp pmp m RT p

p



micro

1p


p = + =


1p

c


1 2

1 2


+micro= =

+micro micro

2p


12

2

12

2 1 2

( 1)

1 1( )

mmm

mmm

+micro =

sdot +micro micro

1p


1 2

kg12kmol

p pp p


+1p

d

0 2 rmp V RT=micro

1p

4p02r

p VmRTmicro

= 1p

0 0

1 2 1 1 2 2 1 2

2 2 2 0(6)3

rm p pm m p p p p

micro= = = =

micro + micro +2p



183


a

1 (273 27)K 300KT = + =

1p

3p


2 1

2 2 600Kp p T TT T

= = = 1p


3 2

V VT T


3 1

p pV V

=


1p

b



1p

4p


13 3 3 3 1 1 3 1 1 3 11 1( ) ( )2 2





sdot1p

c

123 12 23L L L= +



123 1 1 13 3 300JL RT p V= ν = = 1p

d

1 ced

abs

QQ

η = minus 1p

4p31 3 1( )( )



1 1

1 1 1 1

2 (4 ) 11 773 5 13(2 ) (4 2 )2 2

R T TR RT T T T



1p



184



2

b


=+

+


2

2

( )( )

E Rx R xRx rR rx

+=

+ +


=

Rezultă RrxR r

=minus

3

3

c


=





3

4

c


2EI

R r=

+


22

EI rR=

+


2

22 2122 7

rRI EI R r E

+= sdot =

+

3

5


nal 1

2nn n

E EIR r R r

= =+ +

Rezultă 174

E E=3





1

171AEIr R

= cong+

2p

b




22 ( ) 22ABx R xR x x

R Rminus

= = minus

2p

4p

22 ( ) 22ABx R xR x x

R Rminus

= = minus


2p


185

c


max 2ABRR =

2p

4pRezultă min

1 2

EI Rr R=

+ + 1p


dsc

EIr

= 2p3p

12AscI = 1p



a


R r=

+1p

4p


22( )

REP RIR r

= =+

1p


2 2

2 4

( ) 2 ( )0 0( ) ( )



1p


4mEP

r= De aici

2

14 m

ErP

= = Ω 1p

b


2p3p


c


02

0( ) 4R E Ef

R r r=

+ 2p


2

02 4(1 )

4 m

f fERP f



2p


186

d

1

2

22

2

22

RR r

RrR

η =+

η =+

2p

4p1

2

1223

η =

η =1p

2

1

43

η=

η 1p



187

DOPTICAtilde




a



1p

5p

2 1 2 1

1 1

n n n nx x R

minusminus = 1p

3 2 3 2

2 2

n n n nx x a R

minusminus =

minus 1p


2 1 1 2

n n n n n nx x R R


b


1 2 1

y n xy n x

= 1p


1 3 1

y n xy n x

= 1p



3 1 1

n x yn x y

β = = 1p

c


2 1 3 2

1 2

imnx f n n n n

R R


+

15p

3p


2 1 3 2

1 2

obnx f n n n n

R R

minusrarr infin =

minus minus+ 15p

d


3pRezultă

22( 1)Rf

n=

minus 2 2

1 1

y xy x

β = = 2 1

1 1 1x x f

minus = 2p



188


a




b

Putem scrie 2

kx rD l










2Dx a l b a l bl


c


4pRezultă

2lDi

lλ

= 1p


Rezultă 2icircDi

lλ

= 1p

d


D D fi il f l f

λ λ∆ = minus =

minus minus2p

4p


2 2D g Di gi

l l+ λ λ

∆ = minus = 2p




190

TESTUL 1

A MECANICAtilde




a


4punde 1 2 1 2( ) sin ( )t

HG m m g m m gL



= + sdot sdot minus

1p


b


5p

unde 2 2

1 2 1 2( ) cos ( )nL HG m m g m m g

Lminus




1 2( )fF L

m m g L H

sdotmicro =

+ sdot sdot minus1p


c


4p




Lminus





d


3pDeci 2 2


L+ micro sdot minus

= + = sdot sdot 1p

Rezultă 28 NF = 1p



191


a


4p

Unde 2 2

0 2 2

M MM cM pM



Obţinem 2

max 2Mvhg

= 1p


b



4 52

QQ cQ pQ cQ cQ cQ

m vE E E E E E

sdot= + = + = =

1p

4pDar M QE E= 1p


5Q Mv v= 1p


c



Deci 2

2M

rm vF m g

dsdot

= sdot + 1p


d


2p3p




192





a


3pObţinem 1 11

1

p VR T

sdotν =

sdot1p


b


4pObţinem 2 2 2

22 A

p V NR T N

sdotν = =

sdot1p

Dar 22

A

NN

ν = 1p


c




1 2


ν + ν1p


d


4p



Rezultă 2 2

1 1

05VV

prime ν= =

prime ν1p



193


a



Dar 12341 0U∆ = 1p


b






c




d


4p






194





a


PR R R R R= + + + 1p

3p



b



P S

EIR r

=+

1p


Rezultă 02 AI = 1p

c




Rezultă 4 Vu = 1p

d


S

EIr

= 2p

4pObţinem SC

EIr

= 1p




195


a


EIR r

=+

1p

4pUnde 12 34

12 34BD

R RRR R

sdot=

+ 1p

Iar 12 1 2R R R= + 34 3 4R R R= + 1p

Rezultă 1 AI = 1p

b


5p









EIr

= 2p3p




196

DOPTICAtilde




a



= 1p


b


1 1 1x x f

minus = 1p

3pObţinem 1

21

x fxx f

sdot=

+1p


c


1 1 1x x f

minus =prime prime

1p


1

12

xxprime


1p


β1p


d


1 1 1x x f


1p


1

xxprimeprime


1p

Obţinem 1

fx f


1p




a


Dil

λ sdot= 1p

3pObţinem

2l iD

sdotλ = 1p



197

b


1p




∆α = = 1p


c


D iil

λ sdot= = 1p

4pDeci 1 2

DD = 1p


Rezultă 1 md = 1p

d


Dil

λ sdot= 1p

4pUnde 2

1 cn n

λλ = sdot =

ν1p

Deci 12 2 2

D iil n n

λ sdot= =

sdot1p




198

TESTUL 2

A MECANICAtilde



∆ SIpF Ft 3

2 c 15 mgiT G F= + = 3

3 a 2

0max2 2 10 m 0

2vd h r

g= = = ∆ = 3

4

b 2 2

5 45 4

4 5

452 2

4 s 5 s

at atx x x a

t t


= =22 msarArr =

2 210 9

10 9

9 10

19 m2 2

9 s 10 s

at atx x x x

t t


= =

3

5


3



a


1 1

2

0

0f

f

F T FT F

minus minus =

minus =

2p

4p

1 1 1 2( )F m m g= micro + 1p

2 2 kgm = 1p

b


1p


2 025micro = 1p


199

c


3 1

2

0

0e f

e f

F F FF F

minus minus =

minus =

2p

4p

3 2 1 2( )F m m g= micro + 1p

3 15 NF = 1p

d


4 1 1

2 2

e f

e f

F F F m aF F m a

minus minus =

minus =

1p

4p24 2 1 2

1 2


minus micro += =

+ 1p

eF k l= ∆ 1p

2 2( ) 1333 Nmm a gkl

+ micro= =

∆1p



a

00 50 mspv

m= =

1p

3p0u

vtg

= 1p

5 sut = 1p

b


4p0

32 2pA pA

cA c

E EE E= rArr = 1p

20

3vh

g= 1p

833 mh = 1p

c

0 3c cAE E= 1p

4p0

3vvrArr = 1p

29 msv = 1p


d

20

max 125 m2vh

g= = 1p


0 104 mc

f

Ehmg F

= =+ 1p

max 12hh

= 1p



200



1

2


= ν ∆ = ν ∆

∆rArr = γ

∆

3

2c

11

1 1 ii i f f f i

f

TTV T V V VT


= rArr =

44

if f i

VV = rArr ρ = ρ

3

3


2 2

V pV p

=


2 1 2

V p TV p T

=


2 2

3p Tp T

= =

3

4c

11

1 21 1 2 2

2 1

03V pp V p VV p

γγ γ γ

= rArr = =

2 1

1 2

N n VnV n V

= rArr =

3

5b 2 2

1 1

22 1

1

1 1

12 kJ

Q TQ T

TQ QT

η = minus = minus

rArr = =

3



a

1 2

1 2

m m+micro =

ν + ν 1p

4p

1 21 2

1 2

pV pVm mRT RT

micro micro= = 1p

1 2 2 1

1 2

T TT T


+ 1p

165 gmolmicro = 1p

b

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν2p

4p1 2

2 1

83 5

TTTT T

=+ 1p

2844 KT = 1p

c

pRT

microρ = 2p

3p30699 kgmρ = 1p


201

d

1 21 2

22 fpp pT T T

ν + ν = ν rArr + =

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν

2p

4p1 2

2 1

116 5

TTTT T

=+ 1p

5 22 1

2 1

11 (2 ) 153 10 Nm2(6 5 )f

p T TpT T

+= = sdot

+ 1p



a

2 3T T= 1p

4p

11 1 1

1 1 2 2 2 12

VTV T V T TV

γminus


1p

1 11 300 Kp VT

R=

ν 1p

2 14 1200 KT T= = 1p

b

323 2

2

ln VL RTV

= ν 1p


2 3

V pV p

= 1p

22 1

2

32RTp pV

ν= = 1p

23 2441 JL = 1p

c

31 31 31Q L U= + ∆ 1p

4p

1 3 1 331

( )( )2


83

V V= 1p

31 1 3( )VU C T T∆ = ν minus 1p

31 1314 JQ = minus 1p

d

min

max

1CTT

η = minus 1p

3p1

2

1CTT

η = minus 1p

75Cη = 1p



202





4 a 2

1 2max

1 2

( ) 4 W4( )

+= =

+E EP

r r 3

5 c 00

0 0

90 mA(1 ) (1 )

U U II IR R t t

= = = =+ α + α

3



a


e e

EIR r

=+

1p

4p1 3 1 3e eE E E r r r= + = + 1p

2 31

2 3e

R RR RR R

= ++

1p

1 AI = 1p

b

1 1 1U I R= 1p

3p1

1

e

e

EIR r

=+ 1p

1 154 VU = 1p

c

2V pU I R=

1p

4p

2 1

e

p e

EIR R r

=+ + 1p

2 3

1 1 1 1

p VR R R R= + + 1p

11 VVU = 1p

d


e e

E IR r

=+

1p

4p

1 23 3 3

1 2


r r= + = + = +

+ 1p

1 2

1 2

1 2

8 V1 1 3p

E Er rE

r r

minus= =

+1p

061 AI = 1p



203


a

22 2

2

PRI

=

2p3p

2 10R = Ω 1p

b


5p

1 21 3

1 2e

R RR RR R

= ++

pentru K deschis 1p

1 22

1 2e

R RRR R

=+


1 2er r r= + 1p

2 45r Ω 1p

c

22 int 2P r I= 1p

4p1 2I I I= + 1p

1 1 2 2I R I R= 1p

22 6 AI I= = 2 int 162 WP = 1p

d2

2

e

e e

RR r

η =+ 2p

3p

037η = = 37 1p



204

DOPTICAtilde


b 2

1

ff

β = 3

2 a 2 2 22

1 1 1

1 45 cma

x n x xx n x n

= rArr = rArr = 3

3 c 15 mmaa

iin

= = 3

4


c

1 2

1 1 11

l l

l

f n nnn R R

= =

minus minus

lfrArr rarr infin 3

5a 1 1 2 2

192 61 10 J

ex c ex c

ex

L E L E

L minus

ε = + = +

rArr = sdot

3



a2p

3p


b

2p

4p


2 25 cmf = 1p

c

1 12

2 1 1 1 1



4p1 2 3 3 35 cmd x x x= + rArr =

1p

4 3 2

1 1 1x x f

minus = 1p

4 875 cmx = 1p


205

d

3

1

yy

β = 1p

4p1 2β = β β 1p

21

1

15xx

β = = minus 42

3

25xx

β = = minus

1p

3 1 5625 cmy y= β = 1p



a

W N= ε 1p

4p

hcε =

λ1p

W Pt= 1p


= = sdot 1p

b

ex cL Eε = + 1p

4p

c sE eU= 1p

ex shcL eU= minusλ

1p


c

1s

N eIt

=

2p

3p18


= = sdot 1p

d1p

4p




206

TESTUL 3

A MECANICAtilde




a

2 0yG F N+ + =

1p



b

1 2 0x x fF F F+ + =


4p


1 2

2

cossinm

F Fmg F

+ θmicro =

minus θ 1p


c

1 2x fF F F ma+ + =

1p



F F F mgam

1p

2412 msa 1p

d



vv = 1p

2 cos2matP F= θ 1p

1854 W mP 1p



207


a

2

2PkxE = 2p

3p


b


4p2

2

= =i pkxE E sistem

1 2= + =f c c cE E E E

2p


c

2 2 21 1 2 2


1p


1p

21

1 1 2

( )

=+

kmv xm m m

12

2 1 2( )kmv x

m m m=

+1p


d

fc FE L∆ = 1p

4p

21 1


21

2vd

g=

micro1p




208





a1 0 2

lp p g= + ρ 2p3p

51 15 atm 15∙10 Pa p = 1p

b

1 1 2 2p V p V= 1p

4p

1 0 2

= + ρlp p g 1

2=


2 00( ) 0


1p

10 cm x 1p

c

1 1p V RT= ν 1p

4p1p ShRT

ν = 1p


d

2iU RT= ν 1p


2

5 078 2

= νNU RT2

5 02 2


aer 138 J U 1p



b

1 1 2 2 amestec

1 2

V VV

C CC ν + ν=

ν + ν 2p

4p2

=ViC R 1

3 2

=VC R 252VC R= 1p



209

c

min

max

1 ndashCTT

η = 1p



d

cedat

primit

1 ndashQQ

η =

1p

5p



3ndash1 1 13


= ν = ν 1p

p VC C R= + 1p


η = =V

p

C RC




210





a2

12

Ve A

V

R RR R RR R

= + ++ 2p

3p


b


2 2 3

1 2 3

( )A

V


= + + + = = +

111 16

=I A

25 8

=I A

31

16I A=

2p

4p


069 AAI 6875 V VU 1p

c 1

AA

EI R r R

=+ +

0VU = 2p4p


d

er R = 2p

4p1e AR R R= + 1p



a


=E

2p

4p2

max 4EP

r= 1p


b

2middotP R I= 2p

4pEIR r

=+

1p


c

ext

totala

p

p

RPP R r

η = =+

2p

4p2pRR = 1p



211

d

lRSρ

= 1p

3p2

4dS π

= 1p




212

DOPTICAtilde




a2 1 2

1 1 1 minus =x x f

2 22

1 1

y xy x

β = =

2p



b2 1 1

1 1 1ndash x x f

=

2 2

1 1

y x y x

β = = 2p



c

2 1 1

1 1 1 minus =x x f 2 120 cm=x 1p


2 1 2

1 1 1ndash x x f

=

1p


d 1 2

1 1Cf f

= + 2p3p




a

11

2λ

=Dil

2

2Dil

λ=

2p

4p1

1

cλ = rArr

ν1 2 1

12id

cλ ν

=

1p


b

1 1 2 2k i k i= rArr 1 1 2 2k kλ = λ 2p


1p



213



d


δ =l yD

2p


k λδ = = 1p



ν= = 1p



214

TESTUL 4

A MECANICAtilde


1

1

01 kgpm mv

= rArr =

m m

vF m a m

t∆

= sdot = sdot∆

m 2 2 2 s


2 NF =

3

2

a 0N F G+ + =


cosm gN sdot

=α

3




12 2 1 v v v= minus 12km 216 h


5





a




3p 3p


215

c


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α



20

1 1 307 m 2 sin cos

vd dg


2p

5p



u

d vv tt

+= = rArr ∆ =

∆


1p


( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot



c

d vv tt

+= = rArr ∆ =

∆


1p


1p

d

Notăm cu R



1p

4p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




2p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




1p




b


( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot

3p4p

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot1p


216

c




( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

2p

4p


2p

d


11 1 1 1

1

= 14 m s


minus



1p

4p

( )1 1 1 2

11 1 1 1

1

= 14 m s


minus



( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr =

1p( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr = 1p



217



2 2CO C Og = + 44


5

29 10 Pa = 29 atm

m R Tp V R T pV

p


micro sdot

rArr = sdot

3

2

a

12

1213

13

12 13

cu 22

5 cu 082

V

p pp

RQ C T C C R

Q CQ C T C RQ C

T T T


∆ = ∆ = ∆

3

3

d-1 -1


1 1 1

2 2 2 1 2

2

V S hV S h S h


2 75

p

v

C iC i

+γ = = =

1

12 1

2

VT TV

γminus

= sdot


3

4



3





b


1

12 1

2

VT TV

γminus

= sdot

(1) 1p

5p

p

v

CC

γ =






2p


γ = = 1p



218

c1 1

111

1

1344 g

cmm


3p 3p

d

2 2

222

2

4368 g

cmm


1p

4p

3 3

333

3

08 g

cmm


1p

1 2 3 5792 gm m m m m= + + rArr = 1p

32

kg 29 m

mV

ρ = rArr ρ = 1p



a


micro ν 1p

3p1 2 3 4m m m m m= + + +

1 2 3 4ν = ν + ν + ν + ν1p


molmicro = 1p

b

4p 4p

c


5p


1 2 3 41 2 3 4v v v vv

C C C CC


ν

1p


1 1 11 1

2121 2

22

2

p V R TT TTV Tp R T


sdot = ν sdot sdot

1p

112 1

1 1 15

31 2

- 155 - 155

155 10 J

ced

ced

TQ Q R T

R T p VQ



= minus sdot

1p


219

d


1 1 1 1 123 2

2 1

2ln ln ln 22 2

V T V p VQ R T RV V


1p

3p5

23 035 10 JQ = sdot

( ) 131 1 2 1 1 121 105


1p

531 105 10 JQ = sdot


1p



220



1 1 2 2 3 31 2 3

1 1 2 2 3 31 1 2 2 3 3

1 2 3


S S S S


1

2

2775 n m222 n m


3

2

b

( ) ( )1

1 10 1

1 0 1

1 1

UI UR IR

R R


(1)

Analog ( )2

0 2

1

UIR

=sdot + α sdotθ

(2)


3



4


1 2 3

1 1 1 1 pR R R R




3

5

a2

U N SW tl

sdot sdot= sdot ∆


3



a

1K rarr

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

2p

4p1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

1p

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω 1p

b

2K rarr

2

2

2

154 A

EIR r

I

=sdot +

rArr =

1p

4p2

2

2

154 A

EIR r

I

=sdot +

rArr =1p



221

c2 2

3e eRR Rsdot

= rArr = Ω 3p 3p

d3 3 36 A2

3

EI IR r= rArr =

sdot+ 2p

4p




a


400 = 3

b b b b

bb b

b

P U I IUR RI

= sdot rArr

= rArr Ω2p

4p

bb c

EIR R r

=+ + 1p

485 1616 3cRrArr = Ω = Ω 1p


c

icircnchisK rarr b

t

PP



100 VR bU U= = 1 2R R R+ =


2 b RI I I= +

2 2b RU I R= sdot

2p

4p



η = = 1p

d


4p


1 1 1

e sR R R= +

prime

1p


r= rArr =

sdot


2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =



632 Wbb b

b

UP PR

= rArr =

2p



222

DOPTICAtilde


( )( )1


3

2 a 3

3

a1

2

50 cm1 80 cm

fC

ff=



1 2

1 2

1 22

1

2

16 cm

h hf f

h fhf

h

= rArr

sdot= rArr

=

3

4c

39 Vextrextr s s s

LL e U U Ue


5

a

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3



a 3p 3p

b




i i = rArr =

2p

c

2 5 2sin cos tg = 3 3 5

r r r= rArr = rArr

2p

4p1

1

1

tg i = = 50 cmtg = = 775 cmtg

tg

x hh i H H

r hxrH

rArr rArr=

2p


223

d

2p

4p

0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p



a

19min min

max

= 262 10 J = 163 eVR


λ λ 2p



1p

b


λ2p

4p max

min c

extrextr c

EL im

L E=




4p252 m = 66 10

2 sc

cm v EE v v

msdot sdot




224

TESTUL 5

A MECANICAtilde




a


1p



b






d





a


0


4p2 20

2G gFr v v df f


V = 9 ms 1p

b


2p


1 1 1 2( )m v m m vprime= + 1p

Rezultă 1 1

1 1

3 msm vvm v

prime = =+ 1p


225

c


4p

1 2( )0 m m g Fr Df

+minus = sdot 1p

1 2( )m m gFrf

+= 1p


d

21 2

21 1

( )cs

cl

E m m vxE m v



3p

Deci 103

x vprime= 1p



226





a

ν1 = 2Omicro

m 1p

3p1

2

116

νν

= 1p

2

2Hmicrom

ν = 1p

b


4pp 2

ν= ν1RT1

p 2ν

= ν2RT2

2p

Rezultă 21

12

16TT

νν

= = 1p

c


pV2prime= ν2RT1

1p

4p

1

22

1VV

=νν


1p


V2prime= 2

L x S +

1p


d


4p1 2

2m m m+ =


Obţinem 1 2

1 2

2micro micromicro =

micro + micro1p

Rezultă 1 2

1 2

2micro micromicro =




227


a


4p 4p

b


1p

4p


12

1

lnc T VQ RV



1 1 3

13 1

3

4

1 44

PV RTPV RT

T T TT

= ν= ν

= rArr =

1p


c


p

QQ

η = minus 1p


215η = 1p

d

minCarnot

max

1 TT


3pCarnot

3 754

η = = 1p



228





a


V

ERR r+

1p


2vR (2) 2 2 2 2

v v

v

R ERU IR r

= =+

2p


b


EIR r

=+ 1p


EIR r

=+

2p


c2 1

1 1VrR

EU U

=

minus

1p

4p

2 1

1 1AR E

I I

= minus

2p


d

RV rarr infin 1p

3p1V

VV

V

ER EU ErR rR

= = =+ + 1p



a


4p

1 1 3

2 2 4

S

S

R R RR R R

= += +

1p

1 3 2 4

1 2 1 2 3 4

1 1 1 ( )( )tot

tot S S


+ += + =gt =

+ + + 1p


1 2 3 4

( )( ) 4


+ += + rArr = Ω + + +

1p


3pW = 504 J 1p


229

c


4p1 2 3 4

1 2 3 4

EI R R R RrR R R R

=+ +

+ +2p


d


tot

RR r

η = =+

2p

4p


tot

RR r

η = =+ 2p



230

DOPTICAtilde




a


1 2

11 1( 1)

fn

R R

=

minus minus

2p

4p1

2 0RR

rarr infinlt

1p

R = 15 cm 1p

b

2


4p2

1 2

1 1 1( 1)l

fn

R R

=

minus minus

1p

|R1| = |R2| = 15 cmRezultă nl=137 1p

c

2 2

1 1

1y xy x

β = = = minus 1p

3p

2 1

1 1 1x x f

minus = 1p


d


21

1

xx

β = 2 1 1

1 1 1x x f

minus =

1p

4p


2 1 2

1 1 1x x fminus =

2

21

xx

β =


1 21 1 2

1 1

f xx f ff x

= minus minus+

1p


11 1

1 1

f fff x

f x


+

1p

2

1

ff

β = minus


1p



231



2kk Dx

lλ

=k = 5

2p3p



2lDi

l= 2p


c


1p

5p



2knx eD

lminus

=

xk = 04 m

1p


d



2 2k D k

l lλ λ

= 1p





232

TESTUL 6

A MECANICAtilde




a

2

2f

BG F cB cA



4 msBv = 1p

b

0cE∆ = 1p

3p20J

20JpE mgh

E

∆ = =

∆ =2p

c

2 0

2f

BF cB


4p2 2

2 2B Bv vgd d

gmicro = rArr =

micro1p

4md = 1p

d

( ) 0 0


4p


cos ( )sin


1p

40JL = 1p



233


a

1 1F t psdot ∆ = ∆

1p


1p

12mvF

t=

∆1p

1 2kNF = 1p

b


1p

4p

2

2 2

22

p p mv

F t pp mvFt t

∆ = =

sdot ∆ = ∆

∆= =

∆ ∆

2p

2 1 kNF = 1p

c

H F t= sdot ∆

H F t= sdot ∆1p

3p

1 1

2 2

2 N s1 N s

H F tH F t


2p

d

2p

4p

3 2 2 2N sp p mv∆ = = = sdot

2p



234





a 4p 4p

b

A BU U U∆ = ∆ = ∆ 1p

4p2 1 2 1 2 2 1 1


250 JU∆ = 1p

c


5p


650 JAQ = minus 1p


50 JBQ = minus 1p





a

12 34 0Q Q= =

123 3 2 3 2 3 2 2 2 2 3 2 3 2

5 5 5 5( ) ( ) ( ) ( ) ( )2 2 2 2V


2p

5p1 1 2 2 2 1

4 1 3 2 3 4

1 2

3 4


γ γ γ

γ γ γ



-11

23 4 1 4 1 15 5( ) ( )2 2primit


3p


235

b41 1 4


4p


c

1 cedat

primit

QQ

η = minus 1p

3p4 1 1

1-1

4 1 1

5 ( ) 121 15 ( )2

p p V

p p Vγminus

γ


εminus ε sdot2p

d

pL Q= ηsdot = 1p

3p-1 -1

4 1 1 4 1 1-1

1 5 5(1 ) ( ) ( 1) ( )2 2


ε2p



236





a

11

1 1 2 1 1 2 1 48 16 8 8 8 4 p

p

RR

= + = + = = = Ω 1p

5p

1 4 20 24sR = + = Ω 1p

22

1 1 1 3 1 69 18 18 6 p

p

RR

= + = = = Ω 1p

2 6 6 12sR = + = Ω 1p

1 1 1 3 1 824 12 24 8 AB

AB

RR

= + = = = Ω 1p

b

1 1 3

1 1 05 8 4 VABU I R IR

I R= += sdot =

2p

4p

1 11 1 2 2 2

2

4 1 025 A16 4

I RI R I R IR

= rArr = = = =

1 22 1 A4 1 20 4 20 24 VAB

I I IU

= + == + sdot = + =

2p

c

ABe

AB

UIR

= 2p3p

3AeI = 1p

d

ABAB e e

e

E UE U I R rI

minus= + rArr = 2p

3p

1r = Ω 1p



237


a

1 2

( )

( )


I

= + ++

= + +

3p

5p

2

6024 24

2 20 50 05A

II

I II

= +

minus + ==

2p

b( )e e e



43eR = Ω 1p

c


4p1

1

21

24 2A1236 3A12

b

b

PIUPIU

= = =

= = =2p

d

11

12 62

bURI

= = = Ω

22

12 43

bURI

= = = Ω2p

3p




238

DOPTICAtilde




a1 1 1 1

22 1 1 2 1 1 1 1


+= + hArr = hArr =

+ 2p3p


b



4p 2 1 2 12

2 1 2 2 2 1 2 1


+= + hArr = hArr =



c

2 2

1 2 1 1

sistemx xx x



d 5p 5p



239




b


2p3p


c



λ= 2p


relaţia 2

xD l∆ δ

= care devine ( 1) 45 ( 1) 45 ( 1)2 2 2 2



en

λ=

minus

2p


d


4p

( 1)nouke

nλ

=minus

1p



240

TESTUL 7

A MECANICAtilde


3



EE mgh m

gh= rArr = =


892 ms2

poco po c p p o

EmvE E E E E vm

+ = + rArr = rArr = =

3

4 c 3

5

d

0

0 sin 0 cos

f

x

f y f


+ + + =


cos 20 10 3 027

sin 10f

f

F mg FF NN F


sdot α

3



b


4p


1 1

2 1 2

l ml m m

∆=

∆ + 1p

1

2

25

ll

∆=

∆ 1p

c


4p2 2 1eF m g m g= + 3 3G m g= 1p


d


1m gm g kx k

x= rArr = 2p

3p

100 Nkm

= 1p



241


a


2p

4p


2 1 2 2 1 2

2

0 ( )80 N

N N G N m m gN


b





1p

5p


1 22 1 2

2 1 2 2 1 2


Oy N N G N m m g

minus minus minus =




1p


33 2 3

3 3 3 3

sin


Oy N m g N m g

α minus minus =




2p

Rezultă 3 2 1 1 2 1 2 1 2 3

3 2 1 1 2 1 22

1 2 3

(sin cos ) 2 ( ) ( )(sin cos ) 2 ( ) 342


m m m s


= =+ +

1p

c

1 1 1 1( ) 1626 NT m a g T= + micro = 1p

3p2 1 2 1 1 2 1 2

2

( ) 2 ( )4736



d

1

1 1

22 ( )

R TR m a g

== + micro 2p

3p3252 NR = 1p



242



4



υ ∆ = υ ∆ + 2 1


1 2 1 1 2 2 2 12 1 2 1

3( ) ( )2 2



2 2

p aVp aV

= =

1 11 2 2 1

2 2

p V p V p Vp V

rArr = rArr =

Obţinem 2 2 1 12 1 2 1

3( ) ( )2 2



2 2 2

p V RTp V RT

= υ= υ


3

5



2pV V p= rArr =

adiabat 11 1 2 2 2 2


2

2pp

γminus= 3



a


4p


1 1 1 1

2 22 2 2 2

1 21 2

( )( )


p V Vp V V RTRT

= ν rArr ν =

= ν rArr ν =

++ = ν rArr ν =

1p

Rezultă 51 2 1 11 2 1 1 2 2 2 2

2



b


2 1 1

p Vp V

ν=

ν 2p3p

1

2

5ν=

ν 1p

c


1 2

m m m= +

micro micro micro1p

4p1 2

1 21 2

1 2

2m m

m m m

=micro micro


2p

7gmolmicro = 1p

d


4p


1 2 1 2

1 2 1 2

1 21 2

1 2

2 1

1 2 1 2

( )22

V V V V V V

V V V VVV


C C C CC C




1p

1 22 1

1 2

134

7716

V VV

V

C CC R

mU C T KJ

micro + micro= =

micro + micro

∆ = ∆ =micro

1p



243


a


3p 3p

b



4p

1 1 1 33 1 2 1

3 3 1


= rArr = = rArr = =

1p



3 3 4 4p V p V= 4 1p p= 1 1 1 49 p V p VrArr =

4 19 45lV VrArr = =1p

c


p Vp V RT TR

= υ rArr =υ

1 12 2 2 1 1 2 2 1



1 13 3 3 1 1 3 3 1



4 3 19T T T= =

2p

5p

32vC R= rArr 1 1

1 3 3 1 1 13 8( ) 122v


υ2p

1 3 6000 J = 6 kJU rarr∆ = 1p

dmin

max

1cTT

η = minus

1

3

1cTT


88cη = 1p



244



3


rArr = = Ω


= rArr = Ω


η = = rArr η =+

3

4


182 31 1 1 1 11

2 3

k

ke

k

E E E Er Er r rE

r r r r

sum + += = =

sum + +

1 1 61 1 1 1 11

2 3

e

k

rr

r r r r

= = =sum + +

3

5

d Puterea maximă 2

max 4EP


2 22 2

2

2 6 04 ( )E E R R Rr r


+


3



a


1

1 1 1 22 3e

e

RRR R R

= + rArr = 1p

4p1 18V 2AU I= = 11 1 eU I R= sdot 1p

11 1

1

2 33 2R UU I R

I= sdot rArr = 1p

6R = Ω 1p

b


2

1 1 1 32e

e

RRR R R

= + rArr = = Ω 1p

5p

1 1 1 1

2 2 2 2



1 1 2 2U I r U I rrArr + = + 1p

1 2

2 1

1U UrI I

minus= = Ω

minus E = 10 V 2p

csc

EIr

= 2p3p

10AscI = 1p

d

2

max 4EP

r= 2p

3p

max 25 WPrArr = 1p



245


a

1R şi


lRS

= ρ

22

lRS

= ρ 1p

3p


1 2 2 1 1 2 1 1 2 2l lR R R R R R R l R l

S Sρ ρ


1p



1 2 1 2

1 1 1 ( )e

e

R R RRR R R R R R R

+ sdot= + rArr =

+ + + 3p4p

5eR = Ω 1p

c


3e e

nE EIR nr R r

= = =+ +

2p4p




e

EIr

= = 2p



246

DOPTICAtilde


2

a


2 2 2 2

2 2 2 1 2

1 1 1

62 4V

(155 )155055





ν minus

140 0 118 10 HzLL h


3

3

b



2

sintg tg1 sin

PB rr PB h r hh r

= rArr = =minus


2

sintg tg 3 07m1 sin

x rx PB hPB r


3

4


2r r

v v

D iil i

λ λ= rArr = =

λ


v

ii i fi i f fi


f = 30

3

5


4l

n= =

2

tg tg


rl r h lh

l lr h h ml l

= rArr =

= = =minus

3



a



1 2

1 20 cm1 1 1( 1)( )

Rfnn

R R


1p


2 1 1

1 1 1 ( 20)( 30) 12cm20 30

fxxx x f f x


+ minus minus 2p


1 1 1

( 12) 2 08 cm30

x y x yyx y x


minus


b


1 2

11 1( 1)( )

a

a

a

n Rf n n nn R R

minus= =

minusminus minus2p

3p

80 cmf = minus 1p


247

c


R ffnn

R R

minus= = = = minus


2p


1

2 2 1 1

1 1 1 ( 10)( 30) 15 cm10 10

f xxx x f f x


+ minus minus 1p


2 2

1 1

x yx y

β = = rezultă

2 12

1

( 15) 2 1cm30

x yyx

minus sdot= = =

minus 1p

d


a

a a

f fFF f f f f

sdot= + rArr =

+1p

4p

1 15 cm1 1 2( 1)( 1)( )( )

aa

a

Rfnn

R R


minus1p

( 20) 15 30 cm30 20


minus1p

12

2 1 1

1 1 1 ( 30)( 30) 15 cm30 30

F xxx x F F x


+ minus minus 1p




2Dil

λ= = 2p


i= rArr = = 1p

b


lλ

=


Dxl

λ= =

1p


λ= + sdot 1p


Dx x x xl


c


Dδ = minus =



∆ = minus minus

2p

5p


λ = minus minus



= minus rArr =minus

2p

0 33 45 mm2

Dx xl


dDin relaţia 1

2KK Dx K

lλ

= = rArr 22l x

Dsdot

λ = 2p3p

2 480nmλ = 1p



248

TESTUL 8

A MECANICAtilde




a

2

2cpEm

= 2p3p


b

pFt

∆=

∆ 1p

4pF ma= 1p

pam t∆

=∆

1p


c

amFFF fx

=++ 21 1p





d

2

2atd = 1p

4p2 1d d∆ = minus

21

1 2atd =

22

2 2atd = 1 1 st = 2 2st = 2p




249


a


4p1iE m gh= 2

1

2fm vE = 2p


b


2gth = 1p


2

2atd vt= + 1p


1p


c


1 1 112 2


1 1 1 2i fp p m v m m v= hArr = +

2p

4p( )1

1 2

2m g h dv

m mminus micro

=+

1p


v = 1p

d


1 2

2sistem

m m vE

+=

2

2sistemk lE ∆

=

1 2m ml vk+

∆ =

2p

3p


l∆ =

1p



250





a

pV RT= ν 1p

4p1

2

2

1

2

1

2

1

micromicro

νν

sdot==mm

pp

2p

rezultat final 47

2

1 =pp

1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0

2p V RT= ν

1p

4p0 finalp V RT= ν

1p


d


2p




b1 1L p V p V= ∆ sdot ∆ = 2p


c

min

max

1CTT

η = minus 1p




251

d

efectuat

primit

LQ

η = 1-2 2-3primitQ Q Q= + 2p

5p


p

V

CC

γ = p VC C R= + 1p

RC γ=

γ minus 1V

RC =γ minus

1p

12 1γ minus

η =γ +

1p

rezultat final 2 11

19η = = 1p



252





a

4 52

4 5s

R RR RR R

= ++ 2p

4p3

13 3

se

R RR RR R

= ++ 1p


b

1e

EIR

= 1


3 3

sp

R RRR R

=+

2p


c 1

EIR

=

2p3p


d3

1 3

EIR R

=+ 3p

4p




alR

Sρ

= 2p3p




c

becuri b

sursatilde

P UP U

η = =

3p4p


η = 1p

d

middotx

xRS

ρ= 1p

4p2 x

UIR

= 2p




253

DOPTICAtilde




b

2 1 2

1 1 1x x f

minus =

2 2

1 1

12

y xy x

β = = =

2p

4p012

21

=+ff 1p


c

( )1

1 11nf R

= minus 2p



d


4p2 1 1

1 1 1x x f


2 1 2



1 2

16 cm

x fx xx f

= = minus+

1p






b

2 1r rδ = minus 1p



enδ minus δ

=minus

1p


c 2Dil

λ= 2p

3p



254

d


δ = 1p

4p



δ = δ + 1p


= minus 1p

rezultat final( )12

ed nh

lminus

= 75 mh = micro 1p



255

TESTUL 9

A MECANICAtilde



5 20 205


= rArr = = ms 3

4

d [ ]

( )

21

2

0 0

m1s

2

sC

atx t x v t

C a

=

= + +

rarr

3

5

a ( )

1 1 2 2

1 2

1 2

1 2

1 2

1 2

N12m

s

s

s

s

F k F kF k

F Fk F Fk k

k kkk k

k

= ∆ = ∆

= ∆ + ∆

rArr = =∆ + ∆ +

rArr =+

=

3



a

1p

4p

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

= 1p

b2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+

2p4p

2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+ 2p


256

c


3p( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot =

1p

( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot = 1p

d( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

2p

4p( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

1p

( )1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

= 1p



a 05 JA

A

E mghE

==

2p3p

1p

b

2p

4p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=

1p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=1p

c


( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

4p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

= 1p

d


2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p

4p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

= 1p



257



b [ ] Jmol KSi

C =sdot

3

2 a 2100 kJ

Q m c tQ

= sdot sdot ∆=

3

3d 1 1 2 2

1 2


N NN Nmicro + micro

micro =+

micro = =

3

4

c 2

1

11 1

2

ln

ln

280 J

VL RTVpL p Vp

L

= ν

=

= minus

3



a0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p

0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

b

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

4p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot 1p

c

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

4p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot 1p

d

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

4p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= = 1p



258


a

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

3p

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

b

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

5p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

c( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

=

2p3p( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

= 1p

d

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p

4p

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p



259



d 2 WRtU

= [ ]SI1st = 3

2 b nEIne r

=+

3

3a

2

32 2 323 7221320

AB

ABAB

AB

RR RR R RR RR

R R RRR R

sdot sdot= + = +

+

sdot= =

+

3

4

c ( )0

00

02 1

1

10 grad

R R tR RR R A t

Rminus minus

= + α


rArr α =

3



a

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω

2p3p

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω 1p

b

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

2p3p

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

1p

c

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

5p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus 1p


260

d


( )( )

1 2 23

1 1 1 1 23 23

2 2 2 2 23 23

I I IE I R r I R

E I R r I R

+ =

= + + sdot

= + + sdot

1p

4p

cu 2 323

2 3

23

12

8 A3

R RRR R

I

= = Ω+

rArr =

1p

Din 2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr =

1p2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr = 1p



a


EIr R

=+

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus1p

4p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p

rArr r = 20 Ω 1p

b

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

3p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

= 1p

c

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

2p

4p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

1p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr = 1p

d

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

4p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

şi eR = r 2p



261

DOPTICAtilde



2

2c

mvE = 3



3

4

d 2 2 22 1

1 1 1

x y yx xx y y

β = = rarr =

Din grafic 1 2

2

1 2

1 2

40 cm 10 cm40 cm

20 cm

x yx

x xf fx x

rArr = =

rArr =

rArr = rArr =minus

3

5 b 9

63

550 10 1 275 10 m2 2 10

275igrave m

Di ie

i

minusminus

minus


sdot=

i = 275 μm 3



a


3pcorespunde cu 1 2

05


1p1 2

05


b

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

4p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr = 1p

c

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

4p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = = 1p

d

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

4p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p



262


a 4p 4p

b0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


2p3p0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


c

00

86

0 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot

2p

3p0

08

60 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot1p

d

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

2p

5p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot 1p



263

TESTUL 10

A MECANICAtilde




a



4p


1 12

2

m v m vm vvm

=

=

1p


b


112 2

v gtda

micro= =

2p3p


c


222

2

0375m s2vad

= = 1p

3p


aa gg



d


5p

21


22


1p

( )2 2 2

1 21 2 1 2 1 2( )





264


a






b


3p20

0(1 cos ) 2 (1 cos )2



c


m= = 1p

4p




m M= =

+


ecuaţia vitezei 2

2vd

g=

micro

1p


d



m M m M= minus =

+ + 2p

4p




265





a


pV RT

mpV RTm= ν


1p

4pm pV RT

microρ = =

1p


3128 kg mρ = 1p

b


1 1 11

p V RTmp V RTm


1p

3p1 1

1

p VmRT

micro=

1p


c


2 1 2 1

22

1

p V RTp VRT

= ν

ν =

1p



1




1



d


1p

5p

1 11

1 1 1 2 2

2 2 12

1



1p

1 2 1 1 1 2 2( )p V V RT p V p Vν+ = = + 1p

1 1 2 2

1 2final

p V p VpV V

+=

+ 1p




266


a


3p 3p

b



2 22 1 2 1 2 1 2 12 1

( )( ) ( )( ) ( )2 2 2


1p

3p

2 22 1

2LkV V

=minus

1p


c


21 1 1


R R= ν rArr = =

ν ν

1p

5p2

2 2 22 2 2 2

p V kVp V RT TR R

= ν rArr = =ν ν

1p

2 22 1 2 1( )kT T T V V


ν 1p

2 22 1

3 ( )2VU C T k V V∆ = ν ∆ = minus

1p


d


1 2

1 2

1 2 1 2 1 2

100 J300 J

400 J

LU

Q U L

minus

minus

minus minus minus

=∆ =

= ∆ + =

1p

4p


23 12 3 2 2

2 2

2 3 2 3

0

ln ln 245 J

245 J

UV VL RT kVV V

L Q

minus

minus

minus minus

∆ =

= ν = = minus

= = minus

1p


1p




267





3p 3p

b


1p

4p



4 4s

ee s

r rrr r

= rArr = =1p


015er = Ω 1p

c


1 23 ( )4

E i r I R RI i

= sdot + + = sdot


142A34

neIr R R

= =+ +

1p

4p




= 1p


d2R Sl =ρ

3p4p




268


a


1 2

1 2 1

( )

1

E E I r r RE EI A

r r R

+ = + ++

= =+ +

1p

3p


1

1 1

2 2

b

b

U IRU E IrU E Ir

== minus= minus

1p

rezultat final

1

2

8 V55V25V

b

b

UUU

===

1p

b


1 2

1 2

035Ass

E EIr r R

+= =

+ +

1p


1 2

266pR RR

R R= = Ω

+ respectiv 1 2

1 2

214 App

E EIr r R

+= =

+ +1p


1p


c



1

ext s

PfP

=

1p


d


2ext p pW R I t=

1p


1 2 1 2

057p pext

total p

R I RWW E E r r R

= = =+ + + 1p


total

WW

= 1p



269

DOPTICAtilde




a


2 2

1 1

12

y xy x

β = = = 1p

3p


1 1 1 125Cf x x

= = minus = δ 1p


b


1 1

10 cmf xxf x

= =+


4p




1 40 cmfC



2 1

f xxf x

=+



c


1

025xx

β = = respectiv 22

1

133

xx

β = = 1p




d


= = 1p


1 1 1F x x

= minus 1p


1

FxxF x

=+

1p




270


a


1 0108 mmxik

= =


Did

λ= se obţine 1

1i d x dD k D

λ = =

1p

3p

Din 22 2 2

D di id D

λ= rArr λ = 1p


b


3 1 13 3 Dx id

= = λ

1p


d= = λ 1p




c


1p


1 1 2 2 11


λ 1p


1p


d


1

nλ

λ = respectiv 2

2

nλ

λ =



2 12

k Dxn d

+ λ=

1p



n dλ

= = 1p






271

TESTUL 11

A MECANICAtilde


c 2 2 2 2

2 2 2 cp m v m v E

m msdot sdot

= = =sdot sdot


2 c 2


= + sdot + 2

55 2

g tx sdot=

24

4 2g tx sdot

= 5 4 45 mx x x= minus = 3


αmicro =

α3

4 b 1N100m

Fkl

= =∆

2N200m

k = 1

2

12

kk

= 3

5 b p F t∆ = ∆ F G= m100 kgs




a




2 2 1

1 2 1 2


= =+ +

1p

Rezultă 2

m375s

a = 1p

b




23 3F T T= sdot =

2p


c


4pRezultă


sin cosmM = =



d


V = a Δt2p

3p

Obţinem m5s

v = 1p



272


a


i fp p=

21 1 2 0

1 2

m v m vu

m msdot + sdot

=+

2p

3p

Rezultăm60s

u = 1p

b


4p




Obţinem Bp =m56kgs

1p

c



4pObtinem

21 2

1 2( ) ( )

2B

Bt


Rezultă 1700 JBt

E = 1p

d


2C f

Bc G F


4pDeci 2

2sin cos2C

t Bc t t




2p




273



b 12


2 b 201204 10A

A




4 b 1 2 1 2

1 2

m m m m+= +

micro micro micro 3

1 2


+micro micro

3


3 2 11mp R T m m mV


3



a


Np V R TN


11

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro

1p

4pSe obţine 1

11

AN P NnV R T

sdot= =

sdot2p

Rezultă 25

1 3

molec12 10m

n = sdot 1p

b


22

p VR T

sdotν =

sdot2p

3p


c


11

m R TpV

sdot= sdot

micro 1p

4p


11

pR T

sdotmicroρ =

sdot

22

2

pR T

sdotmicroρ =

sdot1p

Se obţine 1 1 2

2 2 1

p Tp T

ρ sdot=

ρ sdot1p

Rezultă 1

2

113

ρ=

ρ 1p


274

d


2 2 2 2

p V R Tp V R T


SF 1 1 1

2 2 2

p V R T

p V R T



1p



1 1 2 2

1 2

1 2

1 2

p V p VT Tp V V

T T

sdot sdot+

=+

1p




a


1

ln ln 2vL R T R Tv


34 3 4 3 3

1( )2


141 1 1

4

1ln ln2

vL R T R Tv


4p





3p

34 12465 JU∆ = minus 1p

c


4pObţinem

1

4 3 14





d


p

LQ

η = 1p


2p

Obţinem 505η = 1p



275



c 2

2 UW R I t tR


= 3

2 c 3


4 c eSC

e

EIr

= 65er = Ω 24 V

5eE = 4AeSC

e

EIr

= = 0 VU = 3

5b

2 2U UP RR P

= rArr =

2

10 mS UlP

sdot= =

ρ sdot3



a


e e

EIR r

rArr =+

1p


1 2

1 2

1 2

30 V1 1e

E Er rE

r r

+= =

+ 1

2err = = Ω 1p


1 2

4s se

s s

R RRR R

sdot= = Ω

+ 1p

Obţinem 6 AI = 1p

b



Obţinem 1

2

1II

rArr = 1p

c



10 A3

e

e

EIR r

prime = =+

2p






276


a




b




c

Randamentul va fi e

e

RR r

η =+

1p

4p




2p

Rezultă 886η = 1p

d


4p


12

21

8022 3

2

e

RR RR RR

sdot= + =

+

30 A43e

EIR r

= =+

2p




277

DOPTICAtilde




a


1 1 1 1 1 12x x f x x f


2p

4pRezultă 2

1 32

cf x

minus= =

minus1p

Obţinem 2

3 2015 10

c minus= = δsdot

1p

b

2 1

1 1 1x x f

minus = 1p

4pobţinem 2

1 21

1 22

x x xx



c


2 1

1 1 1x x f

minus = 1p



1

x fxx f

primeprime =

prime +1p

Rezultă 275 5 15 cm75 5


1p




D a ii Da

λ sdot= rArr =

λ2p

3p

Rezultă 3 3

9

2 10 1 10 5 m400 10

Dminus minus

minus

sdot sdot sdot= =

sdot1p

b


4pRezultă

axD

δ = 2p

Obţinem 3 3

72 10 16 10 8 10 m4


δ = = sdot 1p


278

c



4pObţinem

9

max 3

400 10 442 10

xminus

minus

sdot sdot= sdot

sdot1p


d


naλprime =

1p

4pDin Dia

λ= rezultă ii

n= 2p

Rezultă 3

3 31 10 3 10 075 10 m4 43

iminus




279

TESTUL 12

A MECANICAtilde




a


00

( )

05

kt M m m akta t

M m m

= + +

= =+ +

2p


2

2 2

m8 02s

Mg ma

a

micro =

= micro =

2p

b


0 1( )Mg ma

kt Mg m m amicro =

minus micro = +

2p

3p


m( 2)s

kt Mga tm m


+ 1p

c


4p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + =

2p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + = 1p

d


4p


( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus = 1p



a


30 5 10 J



Deci 0cE∆ = 1p


280

b




3p4p


c



4p


d


1p


2p




281



5

b

1 11

1

abs ced

abs

Q QQ

minusη = respectiv

2 22

2

abs ced

abs

Q QQ

minusη =


1 2

abs abs

L LQ Q

η = η = iar

2 1abs cedQ Q=


3



a


1p

3p0p VmRT

micro=

1p


b


4p


A

Np V RTN

= 1p


A

N RTpVN+ α

= 1p

Rezultă 0

1 14pp

= + α = 1p

c


kmolmicro = şi


kmolmicro =

1p

4p

012i ii

A i

N m iN

ν = = =micro

1p

1 2

1 2

m m m= +

micro micro micro

1p

0

1 2 1 2

1 2


A

A A

Nm N

m m N NN N

micromicro


micro micro

1p


282

d


4p


= ν ν = 1p


1 2Q Q Q= + 1 1 1 2 2V VQ C T C T= ν ∆ + ν ∆1p

( 5) 450 J2

pV TQT

α + ∆= sdot =

1p



a


2 1

p pT T T nTT T

= = = 1p

3p


3 2

V V T nkTT T

= = 1p


1 4

V V T kTT T

= =

1p

b


1p

4p


= minus minus


= + minus 1p



1p


283

c


Ccald

T T nkT nkT nk


Cη gt η adică 1 2 2 2 25 2 3

nk nk n knk nk n



adevărat

1p

4p



minusη = =



minusη = =

1p




minus minus


5n n DA q e d

nminus

lt gt minus




minus minus minus

1p



krarrinfinrarrinfin

η = lt

1p

d


cedQL

ϕ =

1p

4p


1 5 2 32 2 2 2

cedQ nk nL nk n k

minus minusϕ = = =

η minus minus +


absQL

ε =

1p


1 3 2 512 2 2 2

nk knk n k

+ minusε = minus =

η minus minus +



1p







284



2




ERIR r r x xr

=+ + +


ERIr R r

= =+

3

3 c 3

4




2 2

U UI R R R= =


R UUab I= sdot =


abR

U UIR R


2

4RUP RI

R= =

3

5


0 i D

i D

U U EI U U E

r R

le += minus minus +


1 5 V2

i

e ii

UU U U

le= +

gt


3



a


n

ki

k k k

I I

U E I rU IR

=

=

minus = minus =

sum


k k

E RI Ir r

= minus

2p

4p


1

11

nk

i kn

i k

ErI

Rr

=

=

=+

sum

sum și respectiv 1

1

11

nk

i kn

i k

ERrU IR

Rr

=

=

= =+

sum

sum2p

b


1 1

lim1 11

n nk k

i ik kgol n nR

i ik k

E ERr rU

Rr r

= =

rarrinfin

= =

= =+

sum sum

sum sum 1p

4prespectiv 1

1

lim 011

nk

i kgol nR

i k

ErI

Rr

=

rarrinfin

=

= =+

sum

sum 1p


0 0n n

ksc sc k sc

i ik

ER I I Ur= =



285

c


1

1

1

1

11

nk

i kn

i k echivalent

echivalentn

i k

Er

r EIR rR

r

=

=

=

= =++

sum

sum

sum

2p

4p

adică 1

1

1

nk

i kechivalent n

i k

ErE

r

=

=

=sum

sum 1p

respectiv

1

11echivalent n

i k

r

r=

=

sum

1p

d

echivalentE E= 1p

3pechivalent

rrn

= 1p

EI rRn

=+

1p



a


0 2

A B

A

B A

I I IE Ir RI

RI RI

= + = + = minus

1p

3pRezultă

3 22

3 23

3 2

B

A

EIr R

EIr R

EIr R

= + = +

= +

2p


286

b

( 2 )3 2CD B A

E R xU RI xIr R

minus= minus =


5p

Pentru max

min

03 2

3 2

2

CD

CD

CD

ERx Ur R

ERx R Ur R

RU o x

= = + = = minus +

= =

1p

Graficul

1p


CDdU Edx r R

α = = minus+

1p


tg CDU

R = minus = Ωα


2 mrE Uα

= minus + = 1p

c


2

1 2



1p

3p

Rezultă

2 2

(2 )E x RIRx R x

minus=

+ minus1p

Pentru

0 2A

2A

02

Ex IR

Ex R IR

Rx I


= = 2

max max 20 WP RI= =

1p



=


xEI cu xy nx rRn

= =+

1p

4p


= =+

1p



R= 1p




287

DOPTICAtilde




a


2 1

1 1n nx x R

minusminus = 1p

5p



1 1n nx x R

minusminus =

minus 1p

Se deduce

12

1( 1)Rxx

nR n x=

+ minus 1p


b


1 1

y nxy x

= 15p

4pLa a doua trecere

2 2 1 1

y xy x

= 15p

Dar 1 2y y= deci

2 03 cmy = minus 1p

c



1 1 2x x R

+ = Rezultă 12

1

2 214 cm2

xxx R

= =minus

1p


21

007 cmxyx

minus= = 1p


d


12 1

1

5 cm 7cm2Rxx R xx R


3p


2 x

Ryy yx R


1p




288


a



2 0 2 2 2 2( ) ( ) ( )

nn a l b n r e n e


minus 1p


b


D l∆

= 1p


Rezultă ( )2 2 2 20 2 2 1 1( ) ( ) ( ) ( )



c


4pRezultă

2Dinlλ

= 2p

d



0 1 1n n= = 1p



289

Cuprins





5

c



Obţinem 2Ahh = 1p


d





Rezultă 2 JGL = 1p



6


Subiectul I Punctaj

1 b 32 b 33 a 34 c 35 d 3




11

mν =

micro2p

3p


b


A

NN

ν = 1p


22

mν =

micro 1p


2

mN N= sdotmicro

1p


c


1 1

m R TpVsdot

= sdotmicro 1p

4pRezultă 5 2



2 2

m R TpVsdot

= sdotmicro

1p


d


4p

unde 1 2m m+ν =

micro 1p


1 2 1 2 2 1

1 2

( )m m m mm m m m



1p




7


a


1 0

311

1

300 K

831 831 10 Pa

= 3 dm

Tp p

R TVp

=


=

1p


32 1

32 1

2 600 K

831 10 Pa

=2 6 dm

T Tp pV V

= =

= = sdot

=

2p


313

33 2

600 K

4155 10 Pa2

=2 12 dm

T Tpp

V V

= =

= = sdot

=

2p

b


12 2 1 V 2 1 17( ) ( ) ( )2


4p


23 2 12




12 23 17 2ln 22


Q Q Q R T 1p


c


12 V 2 1 15( )2



d


4p


23 2 12







8


Subiectul I Punctaj

1 c 32 a 33 d 34 d 35 b 3




3p


b



S

EIR r

=+

2p

Rezultă 1 AI = 1p

c


torul R este 1

1

EIR r

prime =+

2p




1SC

EIr

= 3p4p




9


a


5p


1 1 1

PR R R= + 1p


S PS P

E ER RR r R r

sdot = sdot

+ + 1p



b


S

RR r

η =+

1p



P

RR r

η =+

1p


c


1 11

EP RR r

= sdot

+ 1p

3pDeci 1

11

( ) PE R rR

= + sdot 1p

Rezultă 6 VE = 1p


EIr

= 2p3p




10

DOPTICAtilde

Subiectul I Punctaj

1 b 32 a 33 d 34 a 35 c 3




f= 2p

3p


b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=

+1p


c


1

xx

β = 1p

5pRezultă 12

β = minus 1p




3p




11



1

cλ =

ν2p

3p


b




2 1

( )

S S


=minus

1p


c



Obţinem 10 1

Se Uh


0 2Se U

hsdot

ν = ν minus 1p


d



4pRezultă 22







12

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj

1a [ ] J Nm= =SIL 3

2 c 2

2 2 00 2 270 m

2vv v ad d

a= + rArr = = 3


2

10 6

6 ms

v v at v t

a

= + = +

rArr =

854 kgFma

= = 3

4 b 2

2

2

2

C

C

mvE p mv

pEm

= =

rArr =

2

2

2

2

C

C

mvE p mv

pEm

= =

rArr = 3

5 b 2

0max

0

20 m 10 m2

5 JC C p C

vh hg

E E E E

= = rArr =

= + rArr =

3



a


1 1

1

t f

n

G F ma

N G

minus =

= 2p

4p


21 25 msa = 1p

b


2 20 1 02 0v v a l v= + = 1p

3p1 12v a l= 1p

1 10 316 msv = = 1p

c


2 2

2

fF ma

G N

minus =

=

1p



22 01a

gmicro = minus =

1p


13

d


1t f

n

G F FN G

= +

=

2p

4p


5 NF = 1p



a

1 1f fL F d= minus

1p

4p1f

y

F N

N G F

= micro

= minus

1p


1p

1

75 JfL = minus 1p

bcosL Fd= α 2p

3p

255 JL = 1p

c

1C fE L L∆ = + 1p

4p

2

2CmvE =

1p

2 CEvm

=

1p

2 3 34 msv =

1p

d

2C fE L∆ =

2 2f fL F D= minus

1p

4p2f

F N

G N

= micro

=1p

C

f

EDF

=

1p

D = 6 m 1p



14


Subiectul I Punctaj

1b mpV RT=

micro

30 kPamp RT

V= =

micro

3



4c

3127 kgmpRT

microρ = = 3

5

b 1 1 2 23 32 2

U RT U RT= ν = ν

2 1 2 12 2U U T T= rArr =

11

2 200 J3UL R T RT= ν ∆ = ν = =

3



a


1 2


= = 1p

4p


1 2

1 2

V p pmR T T

micro∆ = minus

1p

1 2

1 2 2 1( )mRTTV

p T p T∆

=micro minus

33324 mV =

1p

b

11

1

p VRT

ν =

2p3p

1 026 kmolν = 1p

c

22 2

mp V RT=micro

2p

4p22

2

p VmRT

micro=

1p

2 512 kgm = 1p

d

11

1

pRT

microρ =

3p

4p

31 25 kgmρ = 1p



15


a


3p 3p

b


1 2

3V V V VT T

= rArr = 1p

4p12 1 2 1( )L p V V= minus 1p

121

2 1( )Lp

V V=

minus 1p

5 21 2 10 Nmp = sdot

1p

c


23 31cedQ Q Q= + 1p


1p

131 1 1 1

3

ln ln3VQ RT p VV

= ν = minus

1p


d


4pp VC C R= + 1p

1 1 15 5absQ RT p V= ν = 1p




16


Subiectul I Punctaj

1b 1 Wh = 3600 J 6720 10 JW = sdot

3

2 c 1 1 2

2

E I EIR r R R r

= =+ + +

1 1 2

2

E I EIR r R R r

= =+ + +

1 22

1

2 8R R r RR r+ +

rArr = rArr = Ω+

3

3 b 23ABRR = = Ω 3

4 c V1sc

EIr

= =Ω

3

5c

2

bec 90n

n

URP

= = Ω


3



a

1e pR R R= + 1p

3p2 3

2 3p

R RRR R

=+

1p

44eR = Ω 1p

b

1e

EIR r

=+

1p


88VU V= 1p

c

1 2 3I I I= + 1p

4p2 2 3 3I R I R= 1p

1 23

2 3

I RIR R

=+

1p

3 08I A= 1p


17

d


1 23

1 2e

R RR RR R

= ++

1p

4p125 A

e

EIR r

=+

1p

1 2

1 1 2 2

I I I I R I R

= +=

1p

12 083 A3II = =

1p



a


1 3 3E Ir U I R= + +

2p

4p

13

3

E Ir UIR

minus minus=

1p

3 15 AI = 1p

b


2 2 3 3I R I R= 1p

4p3 3

22

75I RRI

= = Ω

1p

22 2 2P R I= 1p

2 1875 WP = 1p

c

1e pR R R= + 1p

4p

11 30UR

I= = Ω

1p

2 3

2 3p

R RRR R

=+

1p

4875eR = Ω 1p

d

1 1

s

P UP E

η = =

2p3p

05 50η = = 1p



18

DOPTICAtilde

Subiectul I Punctaj

1 c 3

2 a 2

1

sin 3sin 60sin 2

i n r rr n

= rArr = rArr = 3

3 b 1 2 3 3 4C C C C C= + + rArr = δ 3


3

5b 2 2

1 1

2 1

3

3 150 cm

y xy x

x x

β = = = minus

rArr = minus =

2 1

1 1 1 375 cmff x x

= minus rArr =

3



a

2 21

1 1

3y xy x

β = = = minus

1p

4p2 13x x= minus 1p

11

43fx = minus

1p

1 40 cmx = minus 1p

b

32

1

3xx

β = =

1p

4p3 13x x= 1p

1 32

1 3

x xfx x

=minus

1p

2 60 cmf = 1p

c 3p 3p


19

d

1 2

1 1 1F f f

= + 1p

4p

20 cmF = 1p

14

1

FxxF x

=+

1p

43

1

1xx

β = = minus 1p





b

0L h= ν 1p

4p

150 055 10 Hzν = sdot

1p

151

1


λ 1p


c2

2

hcε =

λ 2p

3p


d

2 ex CL Eε = + 2p

4pCS

EUe

=

1p

079 VSU = 1p



20

TESTUL 3

A MECANICAtilde

Subiectul I Punctaj

1 d 32 b 33 a 34 a 35 d 3



a


0eF G+ =

1p


11

m glk

∆ = 1p


b

fF F= 1p

4pN G= 1p

fF N= micro


∆ =m glk 2 125 cm∆ =l 1p

c

e fF F ma+ =

1p

4peF F= 1p

2

2

minus micro=

F m gam 1p


d

22

2pk lE ∆

=

2p

4p2

Flk

∆ =

1p




21


a

F G ma+ =

1p



b

vat

∆=

∆1p

3p0 = +v v at 0 0v = 1p


c

total∆ =cE L 2p

4p2

2cmvE∆ = 1p


d


4p

2

1 2

= +mvE mgh

2

2 2PmvE =

1p

2( )2∆

=a th

1p


1p



22


Subiectul I Punctaj

1 a 32 d 33 b 34 a 35 c 3



a A

N mN

=micro

2p4p


b


4p0

0

mV

ρ =

mV

ρ = 1p

0

0middotmVV

V m=

∆ρ +1p



3p


d


4p0 p VRT

ν =


2p



a


2 2

1 1

=p Vp V

2p

3p

Rezultat final 2

1

2pp

= 1p


23

b



2 14T T= 1p


U RT∆ = ν rArr 12 10800U J∆ = 1p

c


( )=L A p V 2p

4p1 1 1

2 2p V RTL ν

= =


2p

d



2 14 =T T 3 12T T= 1p





24


Subiectul I Punctaj

1 d 32 b 33 a 34 c 35 c 3



aRS= R1+ R2 2p


b

UAB = R2 I 1p

4ps

EIR r

=+ 2p


c s A

EIR R r

=+ +

3p4p

I primecong 12 A 1p

d 1

EIR r

=+

3p4p




a


4pISC = 10 A 1p

ISC = scEIr

= 1p


b


4pEIR r

=+

1p


c

echivalent

echivalent

RR r

η =+

2p


91η 1p





25


1 a 32 b 33 b 34 d 35 a 3




b

2 1

1 1 1x x f

minus = 1p

4p1fC

= 1p


c

2 2

1 1

β = =x yx y 2p

3p


d

1 1 60 cm= + = minusx d x 1p

4p12

1

x fx x f

=+

1p





3p


bfW h= ν 2p

4p








26

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1c

5

8

km = 3 10 uas = 012

1ua min1 km = 15 10

cc

sdot rArrsdot

3


m

m

km = 200 m 48 h

= 15 s

dvt

d vt

= ∆ rArr =∆

3

3c

25

3

= 75 10 J2 = 208 kW h

11 J = 1 W s = 10 kW h3600

c c

c

m vE EE

minus


sdot sdot

3

4



= 75 Nm m

m v vF F

tsdot +

= rArr∆



5


0 0 02

0 0 2

0

4 1 3828 cm4

F l F lE lS l E S

d FS l l ld E

l l l


∆ = minus

3



a


4p2 = 2 m sxx

RR m a a am






27

d


5p


1p



1p



2p



a



= 11314 mdrArr

2p

5p2p

2 1 2

= 14 s

= 12 h 24 min 34 s

dv tt

t t t t

= rArr ∆∆

∆ = minus rArr1p

b

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr =

2p3p

2

296 MJ

c

c

m vE

E

sdot= rArr

rArr = 1p

c


4p


2p

( )1 20

879 MJr

r

F

F

m g h h L

L


rArr = minus

1p

d

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

2p

3p

77644 N r rF

r

L F d

F

= minus sdot rArr

rArr =

1p



28


Subiectul I Punctaj


3



3

b 1 1

2 2

1 1 11

22 2 2

5 cu 80 K 52 1673 3 cu 80 K2

V V

V V



ν

ν3

4

d 23 23 23

2323

40 J0

Q U LU

L= ∆ +

rArr ∆ = minus=


123112

1231 12 23 31

0110 J

UU

U U U U∆ =

rArr ∆ =∆ = ∆ + ∆ + ∆


3

5

c 22 72 10

VA

VA

Q C TQ NN NNC T

N


3



a

1 2 m m m= = 1p

3p1 2

2 1

125ν micro= =

ν micro2p

b1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


1p

4p1 2

1 2

2 g 356

molm m m

ν = ν + νsdot


c

3

kg 157 m

mp V R TpR Tm

V


rArr ρ =

3p

4p

3

kg 157 m

mp V R TpR Tm

V


rArr ρ = 1p


29

d5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

4p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot

1p

5

5 10 Pa

mp V R T

m R TpV

p


sdot sdot= rArr

micro sdot

rArr = sdot 2p





c

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

4p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

1p

16

p

v

p

v

Q C TU C T

C QC U

= ν sdot sdot ∆


γ = = rArr γ =∆

2p

d

5 1 3

pp v

v v

p v

CC C R RC C

C C R


γ minus= +

1p

5p


( )( )

1 2 1 2

1 2

1 2

1 2

1

1v v v v v v


f f



2p

2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = =

1p2

1 2

5 083 83 6

v v

v v

C Cf

C C

f

minus= rArr

minus

rArr = = = 1p



30


Subiectul I Punctaj

1a 2

2

514 nC

U S qI U r tl t ql

S rq


=

3

2

a ( )( )

1 0 1 2 1

1 2 2 12 0 2

3 1

1 - 1

4 10


Kminus minus


α = sdot

10 0

1

9 1

RR Rt


3

3




3

4b

=12800 C

PIP tU q q

q UIt


3

5


= rArr = Ω


= rArr = =


3





1

9 E EI R r R


+ 2p

b



5p


1817 18

E EI R R rR R r I

R


rArr = Ω cong Ω1p




1p

2 05 mml lR S SS R



31

c


1p

4p1

1

105 A 11 s

EIR R r

I t


primeprime = rArr =

2p

1

1

105 A 11 s

EIR R r

I t



dI = 0 1p

3pU = E 1pU = 20 V 1p



a = 2400 mA h = 8640 C q sdot 2p


b1

06 AUI IR

= rArr = 3p 3p

c 14400 s = 4 hq qI t tt I


3p 3p

d


1p

5p


UI IR



2 22

6480 s = 18 hq qI t tt I


1p

2 22

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆

1p2 2

2

222 22

2

04 A

162 10

UI IRN e I tI N N

t e

= rArr =


∆ 1p



32


1


3

2 c 3

3




3



D iil n n


3


0

2 4

s

s

h c h c e U

h c h c e U



3



a

22 1

12

1

2 4 cm

2 2 cm

xx x

xx

x


2p3p


b

2 1

1 1 1

1

25 4 cm

x x f

Cf

Cf

minus = =

rArr = δrArr =

2p4p


c

2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus

2p

4p2 1 1

2 1

1

1 12 2

8 cm

Cx x x f

Cx xx


prime = minus 1p



33

d


4p

2

1 2 1

1 1

2 1

22

8 cm 1875 1 1

xx x x

x x C

C Cx x


primeminus = +



3p




b

1

1

22

kDxD li

l


sdot= sdot

1p


2

R V Dx

lλ minus λ sdot

∆ =sdot

2p

076 mmx∆ = 1p

c2

2

2 2 192 mm

2

k x ixDi

l


= sdot 3p 3p

d


4p

1 21 22 2

D Dk kl l


sdot sdot1p


Soluţia 1

2

32

kk

= =

1p


0Cprime lt rArr


34

TESTUL 5

A MECANICAtilde

Subiectul I Punctaj

1 a 32 b 3

3


0 sin cos 0sincos

Gt Ff mg mg

tg


micro = = αα

3



5



= + = + =

3



a


4p 4p

b


şi 2a


1p


2 2

T m g m aT m g m a

minus sdot = sdot



1 2



+ 1p

c T = m1(a + g) = 2222 N 3p 3p

d


1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p

4p

1 2

1 2

42 444 N

F T Tm m gF T

m m

= +

= =+

2p



35


a

2p

3p

cos5640 J

L F dL

= sdot α=

1p

b

NFfL Ff d

Ff= minus sdot

= micro

1p

4pN

FfL Ff dFf

= minus sdot

= micro 1p




4p4504 JfEc = 2p

d564 W

med

med

LPt

P

=∆

=

2p4p

564 W

med

med

LPt

P

=∆

= 2p



36


Subiectul I Punctaj

1 c 3

2 a 33 b 34 a 35 c 3



a3096 kgm

pRT

microρ =

ρ =

2p3p

3096 kgm

pRT

microρ =

ρ = 1p

b

a

a

N m N mNN

sdot= rArr =

micro micro1p

4p



a

a

N p VNp V RT NN RT

= rArr = 1p


RT= = sdot 1p

c

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus=

1p

4p

1

2(1 )

a

a

Np V RTNN fp V RT

N

=

minus= 1p

1

2

11

pp f

=minus

1p

p2 = p1(1 ndash f ) = 96 ∙ 105 Pa 1p

d


4p


1p

21 O 2

1 2


ν + micro ν

ν + ν2p



37




3p 3p

b


2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= =

2p

4p2

1 2

1 2

2

1 1

53

p pT T

pTT p

= rArr

= = 2p

c9kJ

p

p

LO

LO

η = rArr

= =η

2p

4p

9kJ

p

p

LO

LO

η = rArr

= =η

2p

d


4pT2 = T3 1p

ΔU = 0 1p



38


Subiectul I Punctaj

1 a 32 c 33 c 34 a 35 b 3



a


1 1 1

PR R R= + 1p



R = 11 Ω 1p

b




cE = I3r + UAB 2p

3pUAB = 22 V 1p

d


1

UIR

= 1p



2

UR


U1 = U2 = 96 V 1p




3pUb = 120 V 1p

b



Uf = 4 V 1p




W = 4464 kJ 1p



39


1 c 32 a 33 d 34 c 35 c 3





3p

Rezultă cvn

= v = 212 ∙108 ms 1p

b



r = 450 1p090 45rθ = minus = 1p

c




1p

4p1sin sin2

n l n π= 1p

1

2

sin nln

= 1p

l = 450 1p


3p 3p




1Cf

= 2p3p


c

2 1

1 1 1x x f

minus = 2p



1 1

x yx y

β = = 1p

y2= y1=10 cm 1p


2p3p




40

TESTUL 6

A MECANICAtilde

Subiectul I Punctaj

1 c 32 a 33 c 34 c 35 b 3



a

2p

4p

2p

b





M m M m


= =+ +

1p

21msa = 1p


09 NT = 1p

d



17 NF = 1p



a


4pfc u FL L L= + 2p

2500JcL = 1p


41

b

u

c

LL

η = 2p

3p2000 08 802500

η = = = 1p

c

cossin sinfF f


1p

4p

u GL L mgh= = 1p

f

f

FF u

u

LL ctg L

L ctg


α1p

1 0254

micro = = 1p

d

2 60 40 s3

t∆ = sdot = 1p

4p

u cu c

L LP Pt t

= =∆ ∆

1p

2000 50 W40

2500 625 W40

u

c

P

P

= =

= =2p



42


Subiectul I Punctaj

1 d 32 d 33 b 34 c 35 c 3



a

1 1 1 1 1

2 2 2 2 2

( ) ( 2 )2 2 2

( - ) ( - 2 )2 2 - 2




+sdot


4p5

51

55

2

10 1 5 10 Pa18 9

10 1 5 10 Pa02

p

p

sdot= = sdot

sdot= = sdot

2p

b

F



133(3) NF = 1p

c

1 1 1

2 2 2

( 2 ) ( 2 )





2p

4p

51

52

510 Pa6625= 10 Pa

6

p

p

prime =

prime2p

d

1

2 p ll hpsdot

+ = 1p

4p1

1 12

p lhp

sdot= minus prime

1p

01m = 10 cmh = 1p




43



3p4p



c

212 1 1 1

1


= ν = = sdot =


1p

5p

5 51 11 1 2 2 2 1

2 1



3 523

5 2 10 3 10 (1 27) 1500 17 2550J2



1620 2550 1530 600 JQ = minus + = 1p

d

U Q L∆ = minus 1p

3pL Q= 1p

600JL = 1p



44


Subiectul I Punctaj

1 a 32 a 33 d 34 b 35 d 3



a

1 1 1 1 1AU I R I= rArr =

2 2 2 2 075AU I R I= rArr =2p

4p( )1 1 2 1 1 2

2 2 1 2

( )6V

( )E I R r R R I I

EE I R r I I


2p

b

1 11 1 2 2

2 2

( )( ) ( )

( )E I R r

I R r I R rE I R r

= +rArr + = + = +

2p

4p2 2 1 1 2 1

1 2 1 2

I R I R U UrI I I I

minus minus= =

minus minus 1p

2r = Ω 1p

c

66

15012 10 603 10fir

lRS


sdot2p

4p6 075A 075 60 45C8fir

EI q I tR r

= = = = sdot ∆ = sdot =+ 2p

d0

00 0

1(1 ) 1 1

fir

fir firfir



α2p

3p

40 Ct = 1p



a

ss

s

EIR r

=+ 1p

4p10sE E= 1p

10sr r= 1p

4 133 A3sI = = 1p


45

b

pp

p

EI

R r=

+ 1p

4p10prr = 1p

10

110p

ErE E

r

= = 1p

05ApI = 1p

c


4p2 pU I R= 1p

1 1 VU

1p

2 35VU = 1p

d1

s

RR r

η =+

2p

RR r

η =+

1p

3p

17 259227

η = 235 972236

η =

2p



46


1 d 32 c 33 c 34 c 35 a 3



a 4p 4p

b

sin sini n r= 1p

4psin sin n r i= 1p



c


4p

sin sin

3sin 12sin

23

i n r

irn

=

= = =1p

30r = deg 1p



d cosADAB

r= 2p

3p

115cmAB 1p



47


a


1

2

1

2

s

s

ch L eU

ch L eU

= +λ

= +λ

2p

5p


2

2 1

1 1 2

1 2


( )( )

s s

s s

hc e U U

e U Uh

c

minus = minusλ λ

minus λ λ=

λ minus λ

2p


b1

1s

cL h eU= minusλ 2p


c0 0

Lh Lh

ν = hArr ν = 2p3p

150 0924 10 Hzν = sdot 1p

d

2max

2smveU = 1p

4pmax2 seUv

m= 1p





48

TESTUL 7

A MECANICAtilde

Subiectul I Punctaj


0 m 8 N4 m 0 N

x Fx F

= rArr == rArr =


(6 2) 2 8 J

2L + sdot

= =

3


22 2i f

kx mv kxE E vm

= rArr = rArr = = 3

3

a Viteza medie mdvt

= (1)





= şi 2 2dtv



1 22 2

mm m


+= rArr = rArr = =

++km h


3

4 a 35 d 3



a


1km20h

v =

2km40h

v =

1130min h2

t = =

2115min h4

t = = 2p4p


b

Viteza medie mdvt

= 2p


t = = 1p




d


21

2cmvE =

2

22

2cmvE =

Rezultă 2

1

222

1

c

c

E vE v

=2p

3p

2

1

4c

c

EE

= 1p



49


a




hl =α

iar 1

0NL =


2p

ECB = 32 J 1p

bLFf

= LFf1 + LFf2



c


1p

4p2




= = 1p

d



CDEE = 1 1

15





50


Subiectul I Punctaj

1 b 32 c 33 b 3





3

5 c 3



a



1 2

12p p p pT T

= rArr = 1p

2p = 5

2

N12 10m

sdot 1p


max1 max 1

p p p TTT T p

= rArr = 2p4p

max 12 600KT T= = 2p

c


1 00 1

91p Tp T

ρρ = = ρ 2p

3p

31 0819kgmρ = 1p

d


mp V RT=micro

şi 21 2



12 112





a


3p

2p


51

b


VVrArr = 1p


2 3

V VT T



2 12 2888KT T= = 1p


3

1 2p

vv

C RC RC


4p

1 3 1 1 1 1 13 3 3(2 ) 1800 J2 2 2


d

2

1

| |1 QQ

η = minus 1p

4p

1 2 3 3 2( )5

1 2p

Q Q Cp T TRC R

rarr= = υ minusγ

= =γ minus

Rezultă 1 1 1 15 252

Q RT p V= υ =

1p

12 1 2 3 1 1 3 1 1 1 1 1 1 1

2

3| | | | ln ( ) ln 2 (ln 2 15) 21932v



122η = 1p



52


Subiectul I Punctaj

1 c 32 c 33 d 34 d 3




a



1 21

PRI

= 2p

R1 = 5 Ω 1p

b



EIR r

= =+

1p



e

RR r

η =+

2p3p

80η = 1p

d


4pPuterea maximă

2

max 125 W4 EPr

= = 2p



a


5p


2

V

p

V

RRR RR

= = Ω+

32e pRR R= + = Ω 2p


EIR r

= =+

1p



3pP = 75 W 1p


3pPE = 100 W 1p


e

RR r

η =+ 2p

4pη = 75 2p



53



2max

max max2

2S

C S SmV eUE eU eU V

m= rArr = rArr = 3

2 b 33 a 34 c 3

5


1

2

R RR R

= minus= +

1

1 2

1 1 1 1 2( 1)( 1)( ) ( 1)( ) 5mnC n nR R R R R







3pDin 2

2 1 21

4 20cmy y y yy


c


4pCum 2

2 1 11

4 4 5x x x d xx


1

2 1

16cm54 64cm

dx

x x

= minus = minus

= minus =1p

d


2 1 1 2

1 1 1 ( 16) 64 128 cm16 64

x xfx x f x x


minus minus minus2p


1 1 2( 1)( 1) ( )

Rfnn

R R


2p

2( 1) 128 cmR n f= minus = 1p



54


a


3p 3p

b

302



sinsin irn

rArr =

3sin8

r = 1p

c


5p

tg hOI

α = rArr 1038mtg 3

h hOI = = =α 1p


= rArr2

sintg 0809m1 sin

rAB H r Hr

= sdot = =minus

2p


d


4p



53

i n r i n i

i i i n

i

= rArr = minus


rArr =

2p



55

TESTUL 8

A MECANICAtilde

Subiectul I Punctaj

1 a 32 d 33 a 34 b 35 a 3



aG1 = m1g 1p


b




1 2

g m ma

m mminus

=+

1p


c


4p1 2

1 2

2m m gTm m

=+

2p


d

eF k l= sdot ∆ 2p

4p2Tlk

∆ = 1p




a





fFL = minus 1p

b

totalcE L∆ = 1p

4p

2

2cmvE∆ =




1p


56

c

util

consumat

LL



11 ctg

η =+ micro α

ctg xh

α = 1p


d

c totalE L∆ = 1p

4p

2

2cmvE∆ = minus

total fFL L= 1p

2

2mv mgd= micro

2

2vd

g=

micro 1p




57


Subiectul I Punctaj

1 d 32 d 33 d 34 a 35 c 3



a

1 1 1p V RT= ν 1p

3p

11

1

RTVp

ν=

1p


b


1 2

V VT T

= 2p

4p1 22

1

TVTV

=

1p


c


4p11

mV

ρ =

22

mV

ρ = 2p


d

( )12 1 2 1L p V R T T= ∆ = υ minus 1p


52VC R= 1p





58


a

11

NnV

= 1p

3p1 1 1p V RT= ν A

NN

υ = 1p

rezultat final 11

1

Ap NnRT


b

123 3 1U U U∆ = minus 1p

4p1 1VU C T= ν

3 3VU C T= ν 1p

( )123 3 1VU C T T∆ = ν minus 1p


c

43-4 3

3

ln VL RTV

= ν 1p


33-4 3

4

ln pL RTp

= ν 34 3 ln 2L RT= ν 1p


1p

d

cedat 4-1Q Q= 1p


1p

4p VC C R R= + = 1p




59


Subiectul I Punctaj

1 d 32 a 33 d 34 a 35 d 3



a

0

0p

R RRR R

=+

2p

4pe pR R R= + 1p


b


EIR r

=+

1p


( )0

A

E I r RI

Rminus +

= 1p


c

V eU R I= 1p

4pu r I= sdot 1p


d

A

EIr

= 1p

3p 0VU = 1p




a

21 1 1P R I= sdot


4p1

1

EIR r

=+

22

EIR r

=+

1p




60

b scEIr

=

2p3p


c

ext s

totala s

P RP R r

η = =+

2p

4p1 2sR R R= + 1p


η = = 1p

d rEP4

2

max =

2p




61


1 a 32 d 33 c 34 b 35 b 3



a1

1

1Cf

=

11

1fC

=

2p3p




c

112

111fxx

=minus

1p

4p2 2

1 1

y xy x

β = =

1p


d

1 2

1 1 1

sf f f= + 1p

4p2 1

1 1 1

sx x fminus = 1p







c0extL h= ν 2p

4prezultat final 14


dc sE e U= sdot 2p

4p




62

TESTUL 9

A MECANICAtilde

Subiectul I Punctaj

1 a 0

0

FllE s

∆ =sdot

3

2 c 2

m4 ms 22 s s

va at

∆= rArr = =


3

3 b


= sdot =

3

4d

355 10 2750 N20

PP F v Fv

F

= sdot rArr =

sdot= =

3


cos ( sin )cos sin


α + micro α 3



a


3px24 = 3 ∙ (4 minus 2) 1p

x24 = 6 m 1p

b

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

=

2p

4p

2

m(3 1) s(2 1) s

m2s

va at

a


∆ minus

= 2p

c

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

5p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1ptotal 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


1p

total 01 12 24

01 01

12 12

24 24

total

m1 1s 1 ms


3 2 6 m9 m

x x x x

x v t x

x x

x v t xx

= + +


+= rArr = =


d

total

total

9 m2254 s

m

m

xvA

v

=

= =

2p

3ptotal

total

9 m2254 s

m

m

xvA

v

=

= = 1p



63


a


3p2

0 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

=

1p20 002

08 J

A p A cA

A

m vE E E

E

rArr = + = +

= 1p

b


0

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus = 1p

4p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

20

20

cos2 sin

cos2 1sin

056 m

f ABB A F

mv hmgh mg

vhg

h

E E L


rArr =micro α + α

=

minus =

1p

c 056B

B

p

p

E mgh

E J

=

=

2p4p

056B

B

p

p

E mgh

E J

=

= 2p

d


4p024 J

f AB

f AB

pF B A

pF

L E E

L

= minus

= minus

2p

024 Jf AB

f AB

pF B A

pF

L E E

L

= minus

= minus 1p



64


Subiectul I Punctaj


2 d [ ]SI

J1mol K

C =sdot

3

3c 2

1

ln

1929 J

VL RTV

L

= ν

= minus

3


5 a A

NN

ν = 3



a

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

2p

4p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot

1p

23 16023 10 mol

A

A

A

m NN

mN N

N N minus

=micro

rArr =micro

= = sdot 1p

b0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

c

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

2p

4p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr =

1p

1 2 22 1

1 2 1

1

2

300 K600 K

p p pT TT T pT

T

= rArr = sdot

=rArr = 1p

d

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

4p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

=

1p

3

1 mol

016 m

A

pV RTN mN

RTVp

V

= ν


νrArr =

= 1p



65


a

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

4p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

=

1p

1 1 1

11

1

13

1

27 273 300 K

2493 m

p V RTRTVp

TV

= νν

=

= + =

= 1p

b

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

2p

4p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr =

1p

1 2

1 2

22 1

1

2 600 K

V VT T

VT TV

T

=

rArr = sdot

rArr = 1p

c

L12341 = Aria12341 1p

4p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot

1p

( )112341 1 1 1

112341 1

512341

22

212 5 J4 10

pL p V V

pL V

L

= minus minus

= sdot

= sdot 1p

d( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot

2p3p( )1 2 2 1

51 2 62325 10 J

VU C T T

Uminus

minus

∆ = ν minus

∆ = sdot 1p



66


Subiectul I Punctaj


2 a [ρ]SI = 1 Ω m 3

3 c q = I Δt 3

4 b 0 0 00

112 3 6eR R RR R= + + = 3

5 d 6 2003

UR RI

= rArr = = Ω 3



a2 3

12 3

11

e

e

R RR RR R

R

= ++

= Ω

2p3p

2 31

2 3

11

e

e

R RR RR R

R

= ++

= Ω 1p

b


4p


2

3

RR

1p

rArr 22

3

1 R

I RI =

+

1p

rArr 32

2 3

IRIR R

=+ rArr I = 04 A 1p

c

e

EIr R

= rArr+ 2p

4pe

Er RI

= minus 1p


d

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

2p

4p

2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =

1p2 2

32 2

2 3

04 A

04 6 24 V

AB

AB AB

U I RIRI I

R RU U

=

= rArr =+

= sdot rArr =1p



67


a

Din legea Joule

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

3p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr =

1p

4 A

W U I tWI

U tI

= sdot sdot

rArr =sdot

rArr = 1p

b

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr =

1p

4p

1

1

1

4

41 A

W U I t

WIU t

I

= sdot sdot

rArr =sdot

rArr = 1p

1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


1 2

2 1 2

Din 3 A

I I II I I I


c

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

2p

4p

2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω

1p2

2

275

e

e

e

W R I tWR I

I TR

=

rArr =sdot

= Ω 1p

d

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

2p

4p

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr =

1 1

11 1

2 2

22 2

110 W

330 W

W P tWP Pt

W P tWP Pt

= sdot

rArr = rArr =

= sdot

rArr = rArr = 2p



68


1 c 1

SI

m ms s

cminus

= = ν sdot 3


3 c Negativă 3


3

5 a sin 2 141sin

in nr

= rArr = = 3



a 3p 3p

b

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

2p

4p

2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p2 1

12

1

2

1 1 1

20 ( 50) 100 cm20 50 3

x x ffxx

f x

x

minus =

rArr =+sdot minus

rArr = =minus

1p

c

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


2p

4p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

2 2

1 1

22 1

1

2100 105 cm

3 ( 0) 3

Din

5

x yx y

xy yx

y

=

rArr =


1p

d

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

4p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1psistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr =

1p

sistem 1

sistem

sistem

sistem

1 1 12

1 2

210 cm

f f f

f fff

f

= +

=

=

rArr = 1p



69


a

00

814

0 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot

2p

3p0

08

140 9

3 10 574 10 Hz522 10

c

minus

ν = rArrλ

sdotν = = sdot

sdot1p

b 8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

2p

3p8

14

3 10 500 nm6 10

cλ =

νsdot

λ = =sdot

1p

c

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

4p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ


sdot= sdot = sdot

1p

0

034 8

9

17 19

66 10 3 10522 10

00379 10 J 379 10 J

extr

extr

extr

extr

L hcL h

L

L

minus

minus

minus minus

= ν

rArr =λ



d


34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

2p

5p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p

34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr =

1p34 14 19

19

66 10 6 10 379 1016 10

01 V

extr s

extrs

s

s

h L eUh LU

e

U

U

minus minus

minus

ν = +ν minus

rArr =


sdotrArr = 1p



70

TESTUL 10

A MECANICAtilde

Subiectul I Punctaj

1 a 32 a 33 b 34 c 35 d 3



a


2p4p


b


1 1 1 1

1

(2) ( ) ( )

G T m a G T m a


+ = rArr minus + =

+ + = + rArr + minus = +

1p

4p



1 2 1

( )2


minus += =

+ + +

1p


2 1 1 21 2

1 ( ) ( ) ( )2



+ +rArr 1 1

1

2 ( )2

m m m gTm m

+=

+ 1p

rezultat final 2

m02s

a = 25NT = 1p


1

4 ( )22

m m m gF Tm m

+= =

+ 2p

3p


d




2p

4p1

1

22

m mgNm m

=+

1p


1p


71



a




2p

4p2 2

90 m2 2

at F th gm

= = minus =

1p


1p

b


22

2

45 m2

2

FF g tv g t mm h hg

v gh


2p

4p



c



3pmax2v gh= 1p


d



4p2 maxL mgh= 1p





72


Subiectul I Punctaj

1 c 32 a 33 d 34 c 35 b 3



a


11

1

AA

A

pV RTNpV RTm NN

NN RTVpN


=

3p4p


b


11 A A

m N NmN N

microν = = rArr =

micro

1p2p


c


1A

NN

ν = 2 12 1

2 2A A

N NN N


3 3A A

N NN N

ν = = = ν 3p


1 2 3

2 36

amestecamestec

total


ν ν + ν + ν 1p


d



pν

= 1p


ν + ∆ + ∆= = 1p




73


a


3p 3p

b


4p


1

p Vp V RTRT

= ν rArr ν = 1p


2 11 2

3p p T TT T

= rArr = 1p


c


5p


2 22 2 3 3 3 2

3

32

p Vp V p V V Vp

= rArr = =

1p


2 3 2 1 12

3ln 3 ln 4800 J2



3

32

p TT Tp



1p


1p


1 2 2 3primit

L LQ Q Qminus minus

η = =+

1p3p




74


Subiectul I Punctaj

1 a 32 b 33 b 34 c 35 d 3




sc

E EI rr I

= rArr = 2p3p


b


4p


2 3 4

( )R R RRR R R

+=

+ + 1p



2 3 4

( )e

R R RR R RR R R

+= + +

+ +1p


c


EIR r

=+


eU IR= 1p

4p


1p

2 2 3 4 ( )U I R I R R= = +

1 2 5 1 5 2( ) U U U U I R R I R= + + = + + 1p

1 5

2

( ) eI R R RIR


d






75


a




2p

4p


W WIUt Ut

= = şi 22

23

W WIUt Ut

= =


2p

b


18URI

= Ω 22

9URI

= Ω 2p


1 2p

R RRR R

=+

1p



1 1 1WP R It

= = 2 21 2 2

WP R It

= = 2p4p



p

p

RR r

η =+ fie cu putil

consumat

R EPP EI

η = = 2p3p




76


1 d 32 c 33 a 34 d 35 a 3



a


1 1 1

0 4 4y x x x xy x x



2p4p


b



1 1 1 5Cf x x

= = minus = δ 1p

3p



c


= = 1p


2 1

1 1 1F x x

= minus 1p


1

FxxF x

=+

1p


d


FC

= = 1p


1

25 cm

F xXF x

= =+

1p





a



ν = = sdot 1p


1p


77

b



2extmvLε = + 1p


mε minus


c



LUe

ε minus= 1p


d



5p


1 4λ

λ = (1) 1p


exthcL h= ν =λ


hcL

λ = (2)


hcL

λ =

1p

1 1 11






78

TESTUL 11

A MECANICAtilde

Subiectul I Punctaj

1 b 2 2

2 22

m1kg kg m s2 sSI


2 b 3





m2 sin 10s



sdot sdot= = =

∆ ∆ 3



a


1p

4p



1 2 1 2



+ +

2p




c


1 2 0T Gminus + = 12 fG F= 2p


=micro

1p


d


1p

4p





79




Rezultă 42500 JAt

E = 1p

b


4p

2 2

2 2 f

B AG F


1p

Obţinem 2

2f

BF



c



sinhd =

α1p



1p


d




2p

4pObţinem 1 2

AtE

h sm g

=sdot sdot

1p




80


Subiectul I Punctaj

1d 1 1J kg K

SI

Q Jc cm T kg K


3


R Tp ρsdot sdot

=micro 3

kg8m

pR T

sdotmicroρ = =

sdot3



5 b 3



a


A

m NN

=micro

2p

4pRezultă 0A

mNmicro

= 1p



1p VT

Rsdot

=ν sdot

2p3p


c


4p


2

pR T

sdotmicroρ =

sdot

33

3

pR T

sdotmicroρ =

sdot 1p


3 1 1

2T TT T

ρ sdot= =

ρ1p

Rezultă 2

3

2ρ=

ρ 1p

d


Np V R TN

sdot = sdot sdot 1p

4p

Obţinem 1 12 4A

Np V R TN



1

24

Ap NnR T

sdot sdot=

sdot sdot1p

Rezultă 263

mol036 10m

n = sdot 1p



81


a


4pRezultă 3 11

22 R TV Vp



b


12 23tL L L= + 12 0L = 3

23 22

ln VL R TV






d







82


Subiectul I Punctaj


2 1J A sSI



sdot ∆= = = = Ω

∆

3

3 b 2 2

3 3eR R RR

Rsdot sdot sdot

= =sdot

3

2 3e

E EIR r R r

sdot= =

+ sdot + sdot3

4 d 3


WIR t

= =sdot ∆

50 mAI = 3



a





e e

EIR r

=+

= 5A 1p


b


2 32RI R I= sdot 1 2 3I I I= +2p

4p




1lR

Sρsdot

= 2

4dS πsdot

= 2

1 11 4


ρ sdotρ2p



4pJustificare

1

2 666Ate

EI I cresteR r

sdot= = rArr

+2p



83


a


4pRezultă 2bb

PRI

= 1p


b


1p

4pRezultă b

Rb

PU U sI

= minus 1p

RU R I= sdot

R

b

URI

= 1p


c




d

2R bP R I= sdot 2p

3p




84



= rArr = rArr =λ

3

2 d 3

3 c 22 1 1 1

1


2

i n in n n nr n r

= rArr = = sdot = 3

4 c 1 1 1 02 m = 20 cm5

C ff C

= rArr = = = 3

5 b 3




1 125 10

Cf mminus= =

sdot2p

3p

Rezultă 4C = δ 1p

b


2 1 2 1 1

1 1 1 1 1 1 fxxx x f x f x f x



4pobţinem 2

2 22

2fxx x ff xminus

= rArr =minus

1p



d


4pRezultă 1 2

1 1 100 10025 20

Cf f

= + = minus 2p




a


2p

4pRezultă 25 8

199

66 10 3 10 199 10450 10 4

minusminus

minus


sdot1p


1p


85


0

hcL h= ν =λ

2p3p


c

0eU L= ε minus 1p


eε minus

= 1p

Obţinem 19

19

(49 44) 1016 10

Uminus

minus

minus sdot=

sdot1P


d


2cmE eU ν

= =

1p


m mε minus

ν = = 2p




86

TESTUL 12

A MECANICAtilde

Subiectul I Punctaj

1 b 32 c 33 b 34 a 35 a 3






12

m3 s

F mg maFa gm

minus micro =

= minus micro =

2p

4p1

12

m3 s

F mg maFa gm

minus micro =

= minus micro = 2p

c


2 2

2

cos 0

sin 0f

f

F FN F mgF N

α minus =

+ α minus == micro

2p4p


mgF micro= =

α + micro α 2p


4pRezultă 3 kgFm

g∆ = =

micro2p




2

2MvMgh = 2p

4p




2

0 cos1802c


4p


2p


2 2


2p

4p


2p



87


Subiectul I Punctaj

1 c 32 c 33 a 34 d 35 b 3




1p3p


micro= = 2p

b

Din pV RT= ν 2p

3p

rezultă 4molpVRT

ν = = 4 moli 1p

c


5p

1 2 2

1 1

2871 1 004(3) 4(3)300

U U TfU Tminus



2

1 2 2 2

11 1 1

1 1 1 4(3)

RTp p p TVk RTp p T

V

νminus


scade cu k 1p

d


4p1 0


micro= = =

micro2p



88


a


1 1 1

8atm 300K 1200 1200 273 927 degC2atm

p T pT T K tp T p

= = = = = minus = 1p

4p


Graficul

2p

b

1 1 1p V RT= ν 1p

3p1 1

1

p VRT

ν =

1P

v cong 012 moli 1p

c

323 2

2

ln VL RTV

= ν

1p



2 1


= = = = 1p

223 2 1

1

V ln 11088 JpL pp

= = 1P

d

1abs ced ced

abs abs abs

Q Q QLQ Q Q


4p2

12 23 2 1 21




025 25η cong = 1p



89


Subiectul I Punctaj

1 b 3

2


2 3E E EI I I

R r R r R r= = =

+ + +




1 23

1 2

3 3 2727A3 4

E I IIR r I I

= = =+ minus

3

3 d 3

4


RR R r

η =+ +



R rη =

+



3

5


2

8 8( ) 9 9 4m

RE EPR r r

= = sdot+


5 34

2

rr rR r

plusmn = =


3



a

12 3

12

1 AEI R Rr RR R

= =+ +

+1p

4p2 3

12

2 VBAR RU I

R R= =

+ 1p

22

2 A3

BAUIR

= = 1p

33

1 A3

BAUIR

= = 1p

b

22 2 2qI q I t

t= = sdot ∆

∆2p

4p2

2 A 30 s 20 C3

q = sdot = 2p

c



scEI

R r=

+ 1p


d

p

EIR r

=+ 1p

4p1 2 3

1 1 1 1 142pp

RR R R R

= + + rArr = Ω 1p




90


a


2LL

U UI RR R R=

+ + 2p


Randamentul este 2

2

1 05 50( ) I 2L

RIR R

η = = = =+

1p

b


2 290400J3U tW I Rt

R= = =



290400 003 kWh36 10

W = congsdot

1p


kWhC = sdot = = 1p

c


xxrS


lRS

= ρ


=


UI xRl

=+

1p

4p


2(1 )

UP xRl

=+

1p



max0 605Ux P WR

= = =


min 20163Ux l P W

R= = cong

1p


1p





91


1 c 32 d 33 c 34 d 35 c 3



a


2 1

n nx x

=



= minus

2p

3p


nminus

δ = = 1p

b


1 1 1x x f

minus = 2p

4pobţinem 1

21

x fxx f

sdot=



c


1 1

x yx y

β = = 1p

5p

Rezultă 22 1

1

xy yx

= 1p

Rezultă 2 11

fy yx f

=+

1p








lλ

= 2p3p



4kk Dx

l+ λ

= 2p4p



92

c



4pSe obţine

2Di

lλ

= 2p


d


i i∆ minus

ε = = 1p


ε = minus 1p

Rezultă DD

δε = minus 1p





94

TESTUL 1

A MECANICAtilde




aDeoarece N = Gn 1p


b


0

v vat t

minus=

minus 2p



c



d





a




f g= sdot =

sdot1p



max 2 2cm v pE

msdot

= = 2p3p


c




d


4obţinem rF

r

Ld

F= minus 1p




95





a



obţinem 0

0

p L SR Tsdot sdot

ν =sdot

1p


b


1 2

m m=

micro micro1p

4Dar 1 2 ν = ν + ν sau 1 2 1 2

1 2

m m m m+= +


Deci 1 2

2micro + micro

micro = 1p


c


0

m mm R TV

p

ρ = =sdot

sdotmicro

1p

3pDeci 0

0

pR T

sdot microρ =

sdot1p


d


4Deci 0max

0

p TTpsdot

= 1p




96


a


31

1 11

2 10 Pa

= 4 dm

9627 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

4


32

2 22

4 10 Pa

= 4 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p


33

3 33

4 10 Pa

= 8 dm

38508 K

p

Vp VT

R

= sdot

sdot= cong

ν sdot

1p


34

4 44

2 10 Pa

= 8 dm

19254 K

pV

p VTR

= sdot

sdot= cong

ν sdot

1p

b


4

unde 34 V 4 3 4 35( ) ( )2


şi 41 P 1 4 V 1 4 1 47( ) ( ) ( ) ( )2



c




d


primit

LQ

η = 1p


Deci 12341

12341 cedat

LL Q

η =+ 1p




97





a


1 1 1

Pr r r= + 1p

3pObţinem 1 2

1 2P

r rrr r

sdot=

+1p


b


1 2P

E r E rEr r

sdot + sdot=

+1p


P

S P

EIR r

=+

1p


Rezultă 05 AI = 1p

c


2

P

P

EIR r

prime =+

3p4


d


4Deci 2 22

2

E I RIr

primeminus sdot= 1p




a






98

b


4

unde S

S

EIR r

=+

1p



RP




S

RR r

η =+

2p3p

Rezultă 50 η = 1p

d



P

EIR r

prime =+ 1p





99

DOPTICAtilde




a


1 1 1x x f

minus = 2p

4Obţinem 1 21

1 2

x xfx x

sdot=

minus1p



1

1Cf

= 2p3p


c


1 1 1 x x f

minus =prime 1p

4

cu 1 2

1 1 1 f f f

+ = 1p

obţinem 12

1

f fff f

sdot=

minus sau 2 2

22 2

x xfx x

primesdot=

primeminus 1p


d


1 1

y xy xprime prime

β = = 1p

4


1 1

y xy x

β = = 1p

Deci 2 2

2 2

y xy xprime prime

= 1p

Rezultă 2

2

12yyprime

= 1p



100


a


3Obţinem xiN

= 1p


b


Dil

λ sdot= 2p

4Obţinem

2l iD

sdotλ = 1p


c


2Dx

lλ sdot

=1p


42

Dxl

λ sdot= 1p



2Dx x x

lλ sdot

∆ = minus =1p


d


2Di

lλ sdot

= 1p

4unde 1

1 cn n

λλ = sdot =

ν1p

Deci 1 12

D inl i i

λ sdot= =

sdot1p




101

TESTUL 2

A MECANICAtilde

Subiectul I Punctaj


2gt hh t

g∆

∆ = rArr = = 3p

2 b pF F t pt

∆= rArr ∆ = ∆

∆

3p

3


L = 22 J 3p



α 3p



b



2p

3p

2 2

1 1

G T m aT G m a

minus =minus = 1 2

g mam m

∆rArr =

+ 2p

c



+ = = 1p


2 1

08 kg01 kg

m mm m

+ =minus =

2p



2p

4p

1 2( ) 126 kg ms= + = sdotp m m v 2p



102




2p3p


b

GL mg h= ∆

1p


2

2atx = 1p


45 JGL = 1p

c



4p

21 2 2c f

h xE mg F= minus

1p

21 (sin 03)

2cmgxE = α minus

1p

1 12 JcE = 1p

d

22 2

1 12c

cc

p p EEm p E

= rArr =

1p

4p



2 24 JcE =

1p

2

1

2pp

=

1p



103



1 2

1 2

m m+micro =

ν + ν

1 1 2 21 2

1 2

A A

N Nm mN N

N N

micro micro= =

=

1 2

2micro + micro

micro = b

3p


1

1 1 ii i f f f i

f

VTV T V T TV

γminus

γminus γminus

= rArr =

1

1iV i

f

VL C TV



c

3p

41 1 2 2

1 32 2 5

1 1

3 3

p V p V

V VV V

γ γ

γ

=

= rArr = b

3p

52 2

1 1

22

1

1 1

065 260 K

Q TQ T

T TT

η = minus = minus

rArr = rArr =

b3p



a

1 2

1 1 2 2( )

m m m

m p V p VRT

= +micro

= +3p

4p


b


4pRezultă 1 1 2 2

1 2

p V p VpV V

+=

+2p

5 28 10 Nm3

p = sdot

1p

c

mpV RT=micro 2p

4p1 21 2pV pVm m

RT RTmicro micro

= = 1p



104

d

2 22

2

m RTp V

=micro 2p

3p5 2

2 241 10 Nmp = sdot 1p



a


3p 3p

b

3 31

1


11

186 10 NmmRTpV

= = sdotmicro 1p

4p

3 312 2 12 4 10 m

2V V minusρ

ρ = rArr = = sdot

1p

5 21 12

2

93 10 Nmp VpV

= = sdot 1p

2 1

212 1

1

ln 25915 J

T Tm VL RT

V

=

= =micro

1p

c

13 450 K

2TT = = 1p

5p

11

3 323 2

3

224 10 mTV VT

γminusminus

= = sdot

1p

5 233

3

083 10 NmmRTpV

= = sdotmicro 1p


4 44 40 Kp VT

mRmicro

= = 1p

d

min

max

1CTT

η = minus 1p

3p4

1

1CTT

η = minus 1p

95Cη = 1p



105


Subiectul I Punctaj


21

2

1 A

AA

A

EI Rr R

EIR R r

= rArr = Ω+

= =+ + a

3


c3

4 22

225 W4

ER r Ir

EPr

= rArr =

= = c

3

5( 1) 10 ka V a

V

UR R n n RU

= minus = rArr = Ω


3



a

e

UIR

= 3 4 15sR R R= + = Ω 1p

4p1

1

5sp

s

R RRR R

= = Ω+ 1p

2 10e pR R R= + = Ω 1p

1I A= 1p

be

EIR r

=+ 2p

4p

2eEr RI

= minus = Ω 1p

c

2 3

2 3

25pR RR

R R= = Ω

+ 1p

4p

1 10s pR R R= + = Ω 1p

4

4

5se

s

R RR R R

= = Ω+ 1p

12 A7e

EIR r

= =+

857 VeU IR = = 1p


106

d


1p

4p1 2

2 1 4

1 26 A7

s

I I I I R I R

I I

= +=

rArr = =

2 3 4

3 2 4 3

3 46 A

14

I I I I R I R

I I

= +=

rArr = =1p

1 4AI I I = + 1p

9 A7AI = 1p



a


4p1 3 1 1 2 3( )P P r R R R R= rArr = + + 1p

( )21 2 1 1 2 3( )R R R R R R+ = + + 1p


b

2

max 84EP r

r= = Ω 1p

3pPmax = 450 W 1p

1 2

1 2

05R RR R r

+η = =

+ + 1p

c

2 Ann

n

PIU

= = 1p

4p1( ) 6 AnE I R r U I= + + rArr = 1p


4 15n

R

URI

= = Ω 1p

d

P UI= 1p


432 WP = 1p



107

DOPTICAtilde

Subiectul I Punctaj

1

2 22 1

1 1

12

1

2

53

23 3

y x y yy x

fxx ff x

fy

β = = rArr = β

= =+


3

22 5π

∆ϕ = δ = πλ c 3

3 375 nmaan

λλ = = b 3

4 206 nmexex

hc hcL eUL eU

= + rArr λ = =λ +

a 3

5 00


c 3



a


4

2( 1)C nR


( 1)2

CRnrArr minus = 1p

15n = 1p

b

2 1 2 1

1 1 1 1 1 Cx x f x x


3p12

1

15 m1

xxCx

= =+


2

1

5xx

β = = minus 1p

c



4

12( 1)

l

nCn R

= minus 1p

1 12

lnn C R=

+ 1p

163ln = 1p


108

d


4p

13 1 1 3 1

1 1 1 1 1 Cx x f x x


13

1 1

025 m1

xxC x

= = minus+ 1p

3

1

083xx

β = = 1p



a




2kk ix x iminus

= rArr = 1p


b

2Dil

λ= 1p

3p2liD =λ

1p

142 mD = 1p

c


4p

2 ( 1)klx e n kD




Pentru 33

1kx i en

λ= rArr =

minus 1p

42 me = micro 1p

d

1 2r vk kδ = λ = λ 1p

4p2 12k k= 1p

1 21 2k k= = 1p



109

TESTUL 3

A MECANICAtilde




a

1

2

AI dtgIB d

α = = 1p

4p1 0 =d v t

2

2 2atd = 2p

Rezultat final 2 tg

=sdot α

vta

15s=t 1p

b

2 0v v at at= + = 1p

3p1

21 1 135 kJ2

= =cm vE 1p

2

22 2 90 kJ2

= =cm vE 1p

c1 1= ∆ = minus

fF c cL E E 2p3p


d







a

0T F G+ + = 1p

4pcossin

T mgT F

α =α =

1p


α=

αF mg

3 N 09 N2

= congF 2p

b cosmgT =

α2p

3p



110

c


3p=AE mgh

2

2BmvE = 1p



d


4

p mv= 1p

final2mv mv= 1p


= =vv 1p



111





a

pV RT= ν 1p

4pν =

microm m

Vρ =

1p

p RTmicro = ρ 1p


b

1 2 3 1 2 3( ) ( )p V V V RT+ + = ν + ν + ν 1p

4p

1 1 2 2 3 3

1 2 3

p V p V p VpV V V

+ +=

+ + 1p

173pp = 1p


c

1 1 1 ∆ = minusU U U 1 1 1 15 5 2 2

= ν =U RT p V 1 15 2

U RT= ν 1p

4p1 1 1 1

5 2

= ν =pV RT U pV

1 1 1 15 2

= ν =pV RT U pV 1p

1 1 1 1 15 10( )2 3



d

1 1 2 2 3 3

1 2 3


ν + ν + ν p

3p1 2 34 9


micro = 1p




112


a


4p

2 1 1 1 2 13 3= =p V p V p p 2 1 1 1 2 13 3= =p V p V p p 1p

1 1 1 1 12 2 4 4L p V p V RT= = = ν

14LRT

ν =

1p


b


4p

12 1 13 2 32


12 34LU∆ = 1p


c


3p23 1 15 6 152



= sdot =Q L 1p

d

P

LQ

η = 1p



92PQ L= 1p


η = = 1p



113








c


2 2

23

U rU r

= = 1p

5pK icircnchis 1 11

1 1

=+p

r RRr R

2 22

2 2

=+p

r RRr R

11 2p p

U R RI

= = 2p


d

1

1G

p

U IR

= 2p

4p1 11

1 1

12 k= = Ω+p

r RRr R 1p




a

2

b

UPR

= 2p3p


b


5p

2 A= =bPIU

1p



UnE I r

1p


c4 (4 )= + +b b aE I r R R 2p


d

2

max 4

=EP

r cacircnd R r= 2p

4p




114

DOPTICAtilde



b 2 1 1

1 1 1x x f

minus = 2p3p


c


4p2 1 2

1 1 1x x f

minus = 1p


d



1 12

1 1

x fx x f

=+ 1p





2Dil

λ= 2p

3p


b


4p2 2 4d i i i= + = 2p


c


4p

2 π∆∆ϕ =

λr 2

3sdot λ

∆ = =y lr

D

Rezultă 2 3π




II

= 1p


115

d


21

15 cm= =+

x fxx f




1

22 sdot=

minusl xl

x

2p

4p


λ= 1p




116

TESTUL 4

A MECANICAtilde

Subiectul I Punctaj

1

b8

3

m 3 10 s

1 al 631 10 ua

1 an

c

dc dt

t


∆ ∆ =

3

2 d 3

3

a

m m

vF m a m

t∆

= sdot = sdot∆

m5s

v v∆ = =

= 25 NF

3

4

c

fR N F= +




θ = = micro

3

5


2

22 2

2 42

m v F dv v

m v F d


3



a

1

1

0const2m

d va vt

+= rArr = =

∆ 2p

4p11

1

2 100 m sdv vt


∆ 2p

b

22

2

50 sdv tt

= rArr ∆ =∆ 2p

4p

1 2 110 st t t t∆ = ∆ + ∆ rArr ∆ = 2p


d11 2 727 m sm m

d dv vt

minus+= rArr = sdot

∆ 4p 4p



117



b


0

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

2p

4p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p

1p

2 20

20

1 1

2p2 2

- 2 1p

= 2 15 m s 775 m s

R

m v m vEc L m g h

v v g h

v minus minus


= sdot sdot

sdot sdot = sdot

1p 1p

c


1 1 231 msin

hd d= rArr =α

1p


( )

2

2 2

2

2 2

0 sin - cos 2

237 m2 sin cos

ABAB R

m vEc L m g d m g d

vd dg



2p

1 2 868 mD L d d D= + + rArr = 1p

d


4p


2 2 1p

2 = 09 m 1p

pp

Ek xE xk

x


1p2

= 09 m 1p

pp

Ek xE xk

x

sdotsdot= rArr =

1p



118



11

1 1 2

2 2 12

2

1 8

A

A

m NN N

m N NN


micro

3

2

d

0 V

Q U LL U C T

Q= ∆ +



= 12465 JL

3

3

c

12 12

12 12

1213 13 13

13

12 13

31 2 32

5 61 3 80 1202 5

V

V

RC C RQ R T

Q C TQQ C T R T KQ

T T


3


3

5


2

1

1C

p

LQ

TT

η =

η = minus


3



119



3p 3p

b51

1 1 1 1 15 10 PaR Tp V R T p pV


c


1p

5p

0const 0 0 1i f

QV L U U U pQ U L

= = rArr = rArr ∆ = rArr == ∆ +

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p

1p

( )( )

( )

( )

1 2

1 2

1

2 2

1 2

1 2

5 2

2 1 1

3 5 2 25 3 5 2 1 2 2 2

5 5

i v v

f v v v

v v

U C T C R

U C T f C f C T p

C R C R

R T f R f R T

T f T

T



sdot = + sdot rArr

2 1

2

5 15

283 K

T pf

T

= sdot+

= 1p 1p

d ( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

2p

4p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p

1p( )

2 2 2 2

1 1 1 1

22 1

15

2

2

1 1

184 10 Pa


Tp f p pT

p


= + sdot sdot

= sdot 1p 1p



a

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

1p

4p


1

2 3 3 23

53 51 1 1 1

1

2 = 8 = 15

5 3

p

v

p V V pVp V V pVC

C

γγ

= rArr = rArr =

γ = =

1p

13 1 3

1

15 15 374 lR TV V Vp



120

b

2p

4p

( )12 2 1

12

37395 J

VQ C T TQ


=2p

c

1231 12 23 31 1pL L L L= + +

12 0 L =1p

4p

1 1 2 2 22 1

2 2 1 1 1

52 2 10 Pa


p


= sdot

( )

( )

23 23 3 2

23 1 1 13 315 22 4

vL U C T T

L R T T R T



1p

23 186975 J 1pL =


31 12465 J 1pL = minus

1231 62325 JL =1p

d


2 3 2 3

2 3 2 3

A A A

B B B

Q Q QQ Q Q

= += +

32 2

2

lnAVQ R TV

= ν sdot sdot sdot


32 2 2

2 3

ln ln BB

V VQ R T R TV V





( )2 3 23

ln 0BVR T R T TV


( )2 3 2 2 3 23

1

2 3 2 3

4ln ln 3

42 ln 15 2 0 3

B

B A

VR T R T T R T T TV

R T

Q Q



rArr gt

3p 3p



121


Subiectul I Punctaj

1

b

1

2

21

2

156

lRl LLR

l lR lR L

= ρ sdotsdot

= ρ sdotsdot

= =

3

2

a U R I= sdot

2

lRmS R

m m d Sd ll S d S


sdot sdot

2 mU Id S

= ρ sdot sdotsdot

= 3 VU

3

3


2 WRI t

=sdot ∆

1 R = Ω

3

4



02 AI =

3

5

b2

max 4

p

p

EP

r=

sdot

1 2

1 2

6 12 5p pr rr r

r rsdot

= rArr = Ω = Ω+ 1 2

1 2

26 Vpp

p

E E E Er r r

= + rArr =

max 14083 WP =

3



a

1

3 1 1

075 A

EI THORN IR R r

= =+ + 1p

3p1

1

1

45 C 1p

R

R

qI p

tq

=∆

=

1p1

1

1

45 C 1p

R

R

qI p

tq

=∆

= 1p

b

2

3 2 2

06 A

EI IR R r


4p2 2 1

84 1U E I r pU V p

prime= minus sdot=

1p2 2 1

84 1U E I r pU V p



122

c


( ) ( )( )

1 1 2 2

1 1 2 2

90 19p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1p

4p1 2

1 1 2 2

18 V 19

pp

p

E E E Er R r R r


3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= =

1p3

3

3

1

3 A 009 A 134

pR

p

R

EI p

R r

I p

=+

= = 1p

d



( )3

3

2 1

1 1 1 1

1

2

115 106

R

R

I I I

E I R r I

I AI A

+ =

= sdot + +

==

2p

4p

1 1 2 2 1 1883 1

AB

AB

U I R I R pU V p


1p1 1 2 2 1

1883 1AB

AB

U I R I R pU V p




a


1 2 1 2

085 A

E EI IR R r r

minus= rArr =

+ + +3p 3p

b


3 2 2

2 A 1p

EI IR R r


1p




123

c

[ ] 015 mint isin ( ) 1 1 1

685 V

AB

AB

U E I R rU

= minus sdot +

=1p

6p


1p


( ) ( )( )

1 1 2 2

1 1 2 2

079

p p

R r R rr r

R r R r+ sdot +

= rArr = Ω+ + +

1 2

1 1 2 2

7 V

pp

p

E E E Er R r R r

= + rArr =+ +

3

25 A 1p

p

p

EI I


+

1p


2p

d 3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


=

2p3p3

3

2 23 2 3 3 2

25650 J 1pR

R

W R I t R I t p

W


= 1p



124

DOPTICAtilde

Subiectul I Punctaj

1



02

2

sinsin n irnsdot

=

2sin 04r =


3

2


1 0

1 1ln gf n

= minus sdot

2

1 1l

a

n gf n

= minus sdot


( )2

1

14a l

l a

n nff n n

sdot minus= =

minus

2 1 4f f= sdot

3

3 c 3

4

d


1

22

2

2

2

extr

extr

m vL

m vL

sdotε = +

sdotε = +


3

5

c

2

11 2

2k

Dx kl

λ sdot= sdot sdot

sdot 2

2 2k

Dx kl

λ sdot= sdot

sdot

22 1 k k

d x x= minus

( )1 2 2 - 02k Dd

lsdot


3




b


4p



0 01 1

1

sin 2 cmsin

r rr d dd r

= rArr = rArr =

1p



125

c

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot

3p4p

0

0 3

82 2 cm = 11596 cm 1p

dvt

c n d pt

cnv


=

δ = sdot 1p

d


4p





2n = = 1p




1

13 1 3 3

2Dx ilλ


sdot 2p

3px3λ1

= 36 mm 1p

b

x4λ2

= 36 mm 1p

4p

x4λ2

=i2 1p


12

2

3 4

450 nm 1p

sdot λλ =

λ =1p

c

3

1

3 1

34

13 3

4 3

4 2

3 360 nm2

DxlDxl

x x

λ

λ

λ λ


sdot =

3p4p


d1 1

14 4 4 6 mm

08 2Dx x

lλ λ


4p 4p



126

TESTUL 5

A MECANICAtilde




a


0N Gn Gt Ff+ + + =

1p


2p

1p

b




1p




c


N T T= + 1p

4p( ) 306BT m a g N= + = 1p

( )( )2 22 1 cos 90N T= + minus α 1p

5911N N= 1p

d


3p( )2

2B B


072m sBv 1p



127


a




b




2B



c


2

cc

mvEc p mv= = 2p


2 2C B

Ffmv mv Lminus = 1p

Rezultă

6 m sCv = 12 m sCp = 36 JCEc = 1p

d


max

2 2Cmv kx

=


4p

max cmx vk

= 1p




128






HemNamicro

= 2p3p


b



c


1p

4p


1p

Obţinem

1 1 1

2 2 2 1 2

1 2 1 2


ν= ν =gt ν = ν =

ν + ν = ν + ν


d





a


1 2

p pT T

= 1p


3 4

p pT T

= 1p


T2 = T4 = 1 3 360KTT = 1p

b

LTOT = L12 + L23 + L34 + L41 1p



129

c

TOT

p

LQ

η = 1p



d

min

max

1 TT

η = minus 1p

3p1

3

1 TT

η = minus 1p

30η = 1p



130





a


1 2 3

1 2 3

1 2 3

36V1 1 1

E E Er r rEp

r r r

+ += =

+ +

1p

3p

1 2 3

1 1 1 1 1pp

rr r r r

= + + =gt = Ω 1p


p

EI

R r= =

+A 1p

b


1 2

1 1 1

2 2 2


= +


2p

5p


11 2 1 2

E r E E RI


=sdot + sdot + sdot

2 1 2 12

1 2 1 2


+ minus=

+ + 2p


c

Deoarece

ABU I R= sdot

1 2I I I= + 1p

4p1 2

1 2

1 2 3

1 1 1AB

E Er rU

r r r

+=

+ +2p

UAB = 40 V 1p

d

Icircn relaţia

1 2

1 2

1 2

1 1 1AB

E Er rU

r r R

+=

+ +1p

3p


vRrarr 1p




131


a


4p



sdot=

sdot


U tminus sdot

=sdot

2p

5AQIU t

= =sdot

1p

b


4pRezultă 2pQRI t

= 1p

Rp = 22 Ω 1p


3pE = 120 V 1p

d


x p

x p

R Rr

R R=

+

2p

4p

px

p

R rR

R r=

minus 1p



132

DOPTICAtilde




a



2 smv e U= sdot 1p

4p11

22

S ex

S ex

ch eU L

ch eU L

= + λ = + λ

2p


b


11

S exch eU L= +λ

2p

4p

11



c


4p0exLh

υ = 1p


d2

2

cW Nh Nh= υ =λ 2p

3p




a1 2

11 11e

m

fnn R R

=

minus minus

2p3p


b


2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2

21

11

2 1

1 1 1 xdx x f

d x xx f

= minus +minus

minus = =+

2p


minus x1 = 80 cm 1p


133

c


1 22 1

1 1 1d x xx x f

= minus + minus =

1 22 1

1 1 1d x xx x f



11

1y xy x

β = = = minusminus

2p


d


1 2

2

1

2 1

1 1 1

2

1 1 1

F f fxx

x x F

= +

minusβ = =

minus

minus =

2p4p



134

TESTUL 6

A MECANICAtilde




atg 1

3tg ϕ = micro = 2p

3p

30ϕ = 1p

b

t fG Fa

mminus

= 1p

4psin cosmg mga

mα minus micro α

= 1p


2310 =577ms3

a = 1p

c



1p

4pt fG F F= + 1p


310 3 10 100 N3

F = sdot = 1p

d



1p

4p

t fG F F+ = 1p

sintG mg= α 1p

3 1 1 100 3 100 3 200 N2 23




135


a

sin 5

1000m54 =15ms3600s

hpl

v

= = α =

= sdot1p


cosUcirc α


rArr sintg cos


α1p



4 352 sin 2 10 15 15 10 W=15kW100


b


4psincos


α1p

7500W=75kWP 1p


2p3p



3p25kJpE E∆ = ∆ = 1p



136





a

1

1 1 2 1 21

22 1 2 2 1

2

2 540 2 54 4 083 450 3 45 5

pVV T V TRT

pV T V V TRT


ν2p

3p

1

2

4 085

ν= =

ν 1p

b

1

1 1

22 2

p VVRT

p V VRT

ν= =

ν 2p

3p

1

2

08VV

= 1p

c

1 1 11 2 1 1 2 2

2 2 2

( ) ( )pV RT

p V V R T TpV RT

= ν rArr + = ν + ν= ν

1p

5p

1 1

1 2 1 22 2

( ) ( )

p V RTp V V RT

p V RT

= ν rArr + = ν + ν= ν

1p

1 1 2 2 2 2

1 2 2

5 08 450 540 ( ) 3 18

p T Tp T T


ν + ν sdot ν 2p


d

1 36lV =

12

2

2 3 36 3 18 54l3 2

V VV


1p

4p 1 2 1 2

1

1 22

9

08 08

V V V V lV V VV

+ = + =

= rArr =2p




137


a 4p 4p

b

212 1

1

41 1 4 1 4

ln 0

5( - ) ( ) 02V

VQ RTV

RQ C T T T T

= ν gt


4p

12 41 229356JprimitQ Q Q= + = 2p

c

23 4 1 4 1

134 4

2

5( ) ( ) 02

ln 0

VRQ C T T T T

VQ RTV


= ν = lt2p

4p


d1 cedat

primit

QQ

η = minus 2p3p

3043η = 1p



138





a

2AP R I= sdot 2p

3p25 WP = 1p

b

0

0 0

0

A

A

A

I I IxI R I Rl

l xE I R I Rl

= + = minus

= +

3p5p

05 mx = 2p

c

ee

E EI RR I

= rArr = 1p

4p0

00

05A

A

A

I I IR lI Ix R

= +sdot

= =sdot

2p

151 151eI A R= = = Ω 1p


3p2250Wh 225 kWhW = = 1p



139


a


3p200 4A50

totaltotal

PP U I IU


b



5p



1p

21

1 1 1 2

22

2 2 2 2

40 102 4

2

60 152 4

2

I PP R RI

I PP R RI

= rArr = = = Ω

= rArr = = = Ω

2p

10VABU = 1p



d

1

1 01 1 3

0

191 1800 C

5 10

RR Rt tR minus


α sdot

2p

4p

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot= t2 =

2

2 02 1 -3

0

11151 = 2300 C

5 10

RR Rt THORN tR

minus= + α = =

α sdot

2p



140

DOPTICAtilde




a

1 2


n ff R R


2p

4p

1 2


n ff n R R


2p

b

2 1

1 1 1

aerx x f= + 1p

4p2

24 cm 48cm5


2

1

xx

β = 1p

04β = 1p

c

2 1

1 1 1

apax x f= + 1p

4p

2

96 cm 192cm5


21

xx

β = 1p

04β = 1p

d2p

3p




141


a

412 10 m 12mm2Dil


3p


b

λapăapatilde nλ

λ = 1p

3piapă

12 3 09mm4apatilde

iin

= = sdot = 1p


c



2p




2p

10nouk = 1p

d






142

TESTUL 7

A MECANICAtilde


3


md dvt t t

= =+

Cum 112 6

d dtv v

= = şi 222 2

d dtv v

= =

6 154

6 2

md vv vd d

v v

rArr = = =+

a

3

4


∆= = minus

∆






Icircnlocuind 35

rArr micro =c

3

5


11 2

2mvmgh v gh= rArr =




∆ b

3



a


3p 3p

b


4p



Randamentul este 2

u

c

L mghL Fl

η = = 1p


l= =

η1p


143

c


0fG T N F+ + + =


2p


lα =

Rezultă 2fhF F mgl

= minus

2p

600 NfF = 1p

dcos

cosf

f f

FF N F mg


α 2p3p

065rArr micro = 1p



a


cE L∆ = 1p

4p



1p

Ff = μN = μmg

Rezultă 2 2

200 2 21 ms 458ms


2p

b



1p


1

2 2mv mv mgR= + 1p


1 11 ms 331msv = = 1p

c



max max2 2mv vmgh h

g= rArr =

2p3p

max 105msh = 1p

d


max 081 m2vh fh fg

= = = 1p



222 2


2p

2 204 v m s= 1p



144




1 2 2 1 1 21 2 2 3 2 3 2

2 3

( )( ) 150 J ( ) 200J 0752

p p V V LL L p V VL

rarrrarr rarr

rarr


c

3

5


max

1CTT

η = minus


1 max 1

1L T LQ T Q

η = rArr minus =


3 150 K8

TT = =d

3



a


1 1

0 1

V VT T

= şi

2 2

0 2

V VT T

= 1p

4p

dar 1 2V V V+ = şi 2 1 12


23VV = iar

1 2 2VV V= = 1p


32TT = şi 0

234TT = 1p

Rezultă 1 4095KT = şi 2 20475KT = 1p

b


4p

1 1 ( )

2 2l lV S V x S= = +


2 2 ( )2 2l lV S V x S= = minus 1p

Obţinem 50 00 1 1 2

500 2 2 0 2

2 N( ) 066 102 2 2 3 m

N( ) 2 2 102 2 2 m





1p

51 2

52 2

N066 10m

N2 10m

p

p

= sdot

= sdot1p


536 NF = 1p

d


mp V RT=micro

şi 2 2 2 20 2 2

1 1 1

m V m Tp V RTV m T

= rArr =micro 2p

5p

Dar 0 01 2

3 32 4T TT T= = şi 1 2

23 3V VV V= =

1p

2 2

1 1

2m Vm V

rArr = 1p

2

1

2mm

rArr = 1p



145


a


1p

3p


2p

b


1 2 1 2


= rArr = rArr = 1p


2 3

p pT T

=


31 13 3

p p ppT T

rArr = rArr =

2p

5 23 10 Nmp = 1p

c


4p1 1

32 32



1 2 2700JU rarr∆ = 1p

d

1

LQ

η =



1p



44 1 4 1 3

p p p p TTT T T T

= hArr = rArr =


2p


1 1

4 2 02218 9

p Vp V

η = = = 22rArr η = 1p



146



41 2sR R= 1 1 1 2

2 3pp

RRR R R

= + rArr = 22 53 3sR RR R= + =

1 1 23AB

RR R

= +


b

3



a

1 2R R R R= + = 1p

4p

22 1

1 1 1 3 22 3

RRR R R R

= + = rArr = 1p

3 22 53 3R RR R R R= + = + = 1p

3

1 1 1 3 1 85 5

5 6258

AB

AB

R R R R R RRR

= + = + =

= = Ω1p

b


UU IR IR

= rArr = = 2p


( ) 264ABAB

EI E I R r VR r

= rArr = + =+

2p

c




AB

EIR r

= =+

1p





d

2

max 4EP

r= 2p

3p

max 8712WP = 1p



147


a


2 4 1

3 4 5

00


minus minus =

2p


1 3 1 1 3 1 2 2

2 3 2 4 4 2 2

2 3 2 5 5



3p



4p19MNU V= minus 1p

c

24 4 4P I R= 2p

3p4 8P W= 1p

d2

1 1 1W I R t= 2p3p

1 10800 J 108 kJW = = 1p



148

DOPTICAtilde


680nma a aa

nnλ

λ = rArr λ = λ =

d 3

2 c 3

3



2MO PO AOMOAO PO

= rArr =

2ON PO O BONO B PO

= rArr =


=09 m2 2


b

3

4


k Dxl

λ=


2

1 1 mm222 2 mm2

Dk xl

Dk xl

λ= rArr = =

λ= rArr = =

c3

5


2 1 1 2

1 1 1 x xfx x f x x

minus = rArr =minus


1

x x xx

β = rArr = β 1 1

11 1

x xf βrArr = =

minus β minusβ


20 10 cm1 1

f minusrArr = =

minus minusb

3



a



1 2

1 1 1 1( 1)( ) n nC n RR R R C


R = 10 cm 1p

b

Din 1 1 20C f cmf C

= rArr = = cm 1p


2 1 1

1 1 1 fxxx x f f x

minus = rArr =+

2p


x minus= =

minus2p


149

c1 2d x x= minus +

2p3p


d


1 1

y xy x

β = = 1p

3p22 1

1

xy yx

= 1p

2 2 cmy = minus




a


1 11 1

2 12 2

( )(1 )

c ch L eU h L eU

c ch L eU h L eU f


2p

5p

11

1

2

08

ch LeU

c eUh L

minusλ

=minus

λ

1 2

1 2

0802


λ λ2p



LL hh


1410 HzL = 1p


1

ch hε = ν =λ

2p3p

191 011 10 J 006eVminusε = sdot = 1p

d


11

22

2

2

2

e

e

m v eU

m v eU

=

=

2p

4p1 1 1

2 2 108v U Uv U U

= = 1p

1

2

54

vv

rArr = 1p



150

TESTUL 8

A MECANICAtilde




a


4p2

1 2atx = 2 0x v t= 2p


= = 1p

b0v v at= + 2p

3p16msv = 1p

c


3prezultat final

-=fracircnare

fracircnare

va


d

1 2= +d d d 2p

5p1 0=d v t 1 1 2= = = 64d x x m 64 m 2p


2 2 fracircnare

vdaminus

= 2 = 32d m




aE mgh= 2p


b

c totalE L∆ = 1p

4p

2

2cmvE∆ = = +

ftotal G FL L L 1p



1p



151

c

c totalE L∆ =

1p

4p 0cE∆ = GL mgh=

f


rezultat final = -f

totalFL mgh = -4500

f

totalFL J 1p

d

= +f f f


4p2f


F oprire

F FL d

+= minus 1p







152





a

pV RT= ν 1p

4p1 1 1 2

2 2 2 1

p mp m

ν micro= = sdot

ν micro2p

rezultat final 1

2

74

pp

= 1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0


4pVf = 2 V 1p


d




4p 4p



a1 1

1 1 11

p Vp V RTRT

= ν rArr ν = 2p3p


b


4p

1 2

1 2 12 1

2 1

34

V

p pV ctT T U C T

T T


=2p




153

c

min

max

1 TT

η = minus 1p

4pmin 1=T T 1p

max 2 1= = 4T T T 1p


η = = 1p

d

P

LQ

η = 1p

4p


1


1-2 2-3 3-1L L L L= + + 1-2 0L = 32-3 2

1

ln VL RTV

= ν 3-1 1 3( )L R T T= ν minus 1p


minusη =

+ 25η 1p



154





a


( )1 1 1 3E I R R= + de unde 1 3AI =2p


b



2422

131 )()( 1p


c


2p4p


d


2p3p




a



bP1 + P2 = Rp I

2 2p3p


c

echivalent

echivalent

RR r

η =+ 2p



d


4p


EIR r

=+

1p




155

DOPTICAtilde




b

2 1

1 1 1x x f

minus = 1p

4p2

1

xx

β = 1p


c

21 CCCsistem += 1p


2 1

1 1sistemC

x xminus = 1p


d

1 2= -x d x 2p

4p

1 22

1 2

x fxx f

=+

1p




a


3p11 10

yi = 1p


b


2i 1p

4p1

12 22id i= + rArr d = 3i1

2p



156

c

22 29

2iy i= + 2

22 072 mm19yi = = 1p


1 2Dil

λ= 2

2 2Dil

λ=

1 22

1

ii

λλ = 1p


d


iin

= 2p


1 1

1

-025 i ii

= 1p


n = 1p



157

TESTUL 9

A MECANICAtilde


5

21 1 2 2

12 1 2

2

2 2m 1156s

m v m v

mv v vm

=

= =3



a 3p 3p

b

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

4p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

1 1 1

1 1 1 1

1

2 2 2

2 12 2 2 2

1 2

2

0 T

m2s



a


minus =

+ = sdotminus micro


=

1p

c2 2

08 NT m g m aT

= minus=

2p3p2 2

08 NT m g m aT

= minus= 1p


158

d

1 0 1 1 0

1 1 0

1 1 0

( ) 0

( )g 0 ( )


+ + + + + =

minus + == +

1p

5p


1p

2 2

2

2

0 0

m G TOy G Tm g T

+ =minus =

=

1p

2 1 0

2 10

( )m g m m gm mm


=micro

1p

m0 = 03 kg 1p



aEA = m1gh 2p

3pEA = 1 J 1p

b

ndashf AB

f AB

B A F

B A F

E E L

E E L

=

= +2p

4p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p1 1

1

cossin

(1 ctg )0654 J

B

B

B

hE m gh m g

E m ghE


= minus micro α=

1p

c


4p

( )( )

1 1 1 2 1

1 1 1 2

1 1

1 2

m v m m v

m v m m vm vv

m m

= +

= +

=+

1p

unde 1m25s

v = 1p

01 255 m08503 s

v sdot= = 1p

d


( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

2p

4p( ) 22

1 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

=

1p

( ) 221 2

1 2

2 2

0147 m

m m vkx

m mx vk

x

+=

+=

= 1p



159



VU VC T∆ = ∆ b 3


3pRT

microρ =

[ ] 3

kgmSi

ρ = c

3

41 1

1 1

33 22

L p VLU p VU

=

∆ = rArr =∆

a

3

5 36 gA A

m N Nm mN N


a 3



a15 mol

mν =

microν =

2p3p

15 mol

mν =

microν = 1p

b

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

2p

4p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ =

1p

2

2

1

1

3

Kg224m

pRTpT R

microρ =

microρ =

ρ = 1p

c

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

3p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot

1p

1 2

1 2

22 1

2

5 52 2 2

8 N N10 267 103 m m

p pT T

Tp pT

p

=

rArr = sdot rArr

= sdot = sdot 1p

d

2 2

3 3

mp V RT

m mp V RT

=micro

minus ∆=

micro

1p

5p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot

1p

( )

( )

( )

2 2

3 3

33

2 2

3 23 2

2 2

3 22

2

52

N114 10m

mp V RT

m mp V RT

m m Tpp mT

m m T mTp pp mT

m m T mTp p

mT

p

= =micro

minus ∆=

micro

minus ∆=



∆ = minus sdot 1p



160


a

L1231=Aria1231 1p

3p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot

1p( )( )2 1 2 1

4

12

3 10 J

L p p V V

L

= minus minus

rArr = sdot 1p

b

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

5p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot

1p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot05p

( )( )

( )( )

12 2 1

1 2 112

512

23 3 2

2 3 223

523

primit 12 23

5primit

075 10

28 10 J

355 10 J

v

v

p

p

Q VC T T

V p pQ C

RQ JQ VC T T

p V VQ C

RQQ Q Q

Q

= minus

minus=

= sdot

= minus

minus=

= sdot= +

= sdot 05p

c

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p

3p

( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot

1p( )( )

1 2 2 1

1 2 11 2

51 2 075 10 J

V

V

U C T T

V p pU C

RU

minus

minus

minus

∆ = ν minus

minus∆ =

∆ = sdot 1p

d

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= =

1p

4p

1 1min 1

2 3max 3

p VT TvRp VT TvR

= =

= = 1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η =

1p

1 1min 1

2 3max 3

min

max

1

91

C

C

p VT TVRp VT TVR

TT

= =

= =

η = minus

η = 1p



161




d 3

2 2W RI t= ∆ b 3

3 [ ] 2 1SI

V1 11 N m A sA


4

( )

( )

Aria trapez2

200 300 2500 500 mC

2

b B hQ

Q Q

+ sdot= =

+ sdot= = rArr = a

3

5

( )0

2

0

0

1

288

2110 C

R R t

UR RP

R Rt tR

= + α

= rArr = Ω

minus= rArr = deg

α


3



a


3p

1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω

1p1 4 2 3

2 34

1

4 45

R R R RR RR

RR

=

rArr =

rArr = Ω 1p

b

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

4p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω

1p

12 1 2 12

34 3 4 34

12 34

12 34

55

1

25

e

e

R R R RR R R R

R RRR R

R


sdotrArr = rArr

+= Ω 1p

c

Potrivit legii Ohm

22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

2p

4p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

=

1p22088 A

e

e e

e

EIr R

EIr R

I

=+

=+

= 1p


162

d

1p


( ) ( )2 4

2 1 2 4 3 4

2 4

I I II R R I R R

I I

= +

+ = +

rArr =

1p( ) ( )

2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr =

1p

( ) ( )2 4

2 1 2 4 3 4

2 4

4

4

2044 A

I I Ii R R i R R

I III

I

= +

+ = +

rArr =

rArr =

rArr = 1p



a

1 2

1 2

1 2

1 1 1

07 A

E Er rI

RR r r

I

+=

+ +

=

3p

4p

1

2

2E EE E

==

1

2

2r rr r

==

1p

bU = R I 3p

4pU = 7 V 1p

cP = RI2 3p

4pP = 49 W 1p

d88

e

RR r

η =+

η =

2p3p

88e

RR r

η =+

η = 1p



163

DOPTICAtilde

Subiectul I Punctaj

12

2 smv eU= rArr

2

2SI

2 ms

S

e

eUm

=

d 3

2 2

1

sinsin

i nr n

=

a 3


4 0 00

275 nmextext

hc hcLL


b 3




a

1

1 01 m10

fC

f

=

= =

2p

3p

1

1 01 m10

fC

f

=

= = 1p

b

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

4p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr =

1p

( )

( )

( )

11 2

1 1 11

1 11

104 01 004 m 4 cm

n Rf R R

nf R

R n fR R


= minus

rArr = minus sdot

= sdot = rArr = 1p

c

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

4p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


1p

2

1

12

1

1

10 10 210 15 5

xx

fxxf x

ff x

β =

=+

rArr β =+


d

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

2p

4p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus

1p

( )

( )

12 2

1

2 1 1

12 2

1

10 1530 cm

10 155 cm10 5

10 cm10 5

fxx xf x

x x d x

fxx xf x

minus= rArr = =



prime+ minus1p



164


a

Din grafic15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot

1p2p15

0115

02

15 10 Hz

3 10 Hz

ν = sdot

ν = sdot 1p

b

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

5p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot

1p

00

0101

8

01 15

02028

02 15

Din

3 10 200 nm15 10

3 10 100 nm3 10

c

c

c

λ =ν

rArr λ =λ

sdotλ = =

sdot

rArr λ =λ

sdotλ = =

sdot1p

c


4p

rArr

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p

( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot

1p( )1 01 1

011

34 15

1 19

66 10 15 10 618 V16 10

s

s

s

h h eUh v

Ue

Uminus

minus

ν = ν +

ν minusrArr =

sdot sdot sdot= =

sdot1p

d

( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p

4p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot=

1p( )

2 02 max 2

max 2 2 02

19max 2

max 2

132 10 J825 eV

c

c

c

c

h hv EE h v v

EE

minus

ν = +

rArr = minus

= sdot= 1p



165

TESTUL 10

A MECANICAtilde




a



2

7 msdvt



1

2 msvv a t at

= rArr = =

1p

4p


2 11 1 1 1

1

2 9m2vv a d da

= rArr = =


1 3 3 21 3 3

1 3 31 3 3

22

0

final

final

final

v v a dv a d

v v a tv a t

v

= + = minus= + rArr

= minus= 21

33

21

33

15ms

12m2

vat

vda


1p



b


4 1 3 41 3 4

4 1 4 113 3 3

3


t


3p4p


c



1p


1p


1p



166

d


2 20G N N G+ = rArr = 1p


2p


1p



a


2

2 20

45 m2

2 2

gth

v v g h v g h

∆ = =

= + ∆ rArr = ∆

2p

4p2

2cmvE mg h= = ∆

1p


b


( )c

p


∆ ∆= =

minus ∆ minus ∆

2p

3p

rezultat final 016c

p

EE

= 1p

c


2

22

c

c

L E Fd

mvE

v gh

= ∆ =

∆ =

=

2p

4p

2

2 2mv mghF

d= = 1p


d


2

( 1)

2

f i

i f i

f


mvE fmgh


2p

4p

( 1)( 1) ( 1)



1p




167





a


mpV RTm= ν


micro=

2p3p



A

N N pVpV RT NN RT

= rArr = 2p3p


c


1

p VmRT

micro= respectiv 2

22

p VmRT

micro= 2p


2 12 1

V p pm m mR T T


1p


d


2p



2 3 2






a


primit primit

QLQ Q

η = = minus 2p



1p


b


1 1

1 T TT T

∆η = minus = 1

TT ∆=

η1p


η1p


2 300 KT = p


168


2 22

m m RTp V RT Vp

= rArr =micro micro

2p3p


d


4p


2 3 2

p p pT TT T p

= rArr = 1p


2

1V Vm pQ C T

p

= minus micro

1p




169





a


= = Ω 1p


232 3

R RRR R

=+ 1p


2 3

R RR R R RR R

= + = ++

1p


b




s

EIR r

= =+

1p



c


2 2 3 3

I I IR I R I

= +=

2p3p


1p

d


1 3R R R= +1p


1 2

093As

e

EIR R r

= =+ +

1p


3 3U R I= 1p




170


a


1 1 1 11

33 3 3 3

3

125 A

05 A

PP U I IU

PP U I IU

= rArr = =

= rArr = =

2p


1p

Din legea lui Ohm R

R

URI

= 1p


b



Iminus

= + = + rArr = 1p




d




consumat

095PP

η = = 1p




171

DOPTICAtilde




a


1 20 cmfC

= =


1 1

30 cmf xxf x

= =+




1 25 cmfC



2 1

f xxf x

=+


1p


2p

b



= 1p


c


1

1346 cmFxxF x

= =+

1p


1

y xy x

β = = 1p


1

xy yx

= 1p


d


1

4xx

β = = minus 1p


2 1

1 1 1F x x

= minus 1p





172


a


1ext s

hc L eU= +λ

(1) 1p

4prespectiv 2

2ext s

hc L eU= +λ

(2) 1p


1 21 2 1 2

1 1 s ss s

e U Uhc e U U h

cminus λ λ


1p


b




se obţine 10

1

sc eUh

ν = minusλ

1p


c



4p

Respectiv 22


Se obţine 1 1

2 2

s

s

v Uv U

= 1p

rezultat final 1

2

073vv

= 1p

d


11

ext shc L eU= +λ

(1) respectiv 33

ext shc L eU= +λ

(2)


1p

3p

3 1

1 1s

hcUe

∆ = minus

λ λ 1p




173

TESTUL 11

A MECANICAtilde


c 2

2 2mkg sN s s=

kg m kg m


= adimensional 3

2 c 3

3


5h = 0cE =2

0

2m v m g hsdot

= sdot sdot 0m2 10s

v g h= sdot sdot =

2

2 05 kgc

o

Emvsdot

= =

3

4 c 3

5 d ( ) 60 kW


= = = 3



a


3p 3p



= =+ +

3p

4p

Rezultă 2

m75s

a = 1p

c


sdot=

sdot ∆


4dS πsdot

=

3p4p


d



4pRezultă 1 2





174


a


ti tfE E= 2

0

2 cf pfm v E Esdot

= +

20


= =

2p

3p


b


4pRezultă 1 1



c




h h= sdot =

1p


2 20


= + sdot sdot


2p

Rezultă m1038 kgs

p = sdot 1p

d



Rezultă 0GL = 1p



175



d 2 3

2 3


mol k

minussdot


sdot

3

2 c p vRc c= +micro

p vRc cminus =micro

3


2 8VV = 12

1

8TT

γminus= 53

p

v

cc

γ = =

2 232 3 3

2 11

8 (2 ) 4 4T T TT

= = = rArr = sdot 3

4 b 3





a



3 311

1

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro 1p

3pSe obţine 3 31

11

025 10 mm R TVp


sdotmicro1p


b


NN

ν = 1p


mν =

micro1p


mNmicro

= 1p


c


Np V R TN

sdot = sdot sdot 1p


V= 1p

Se obţine 1

1

Ap NnR T

sdot=

sdot1p



176

d


1 2

( )m R T m m R TV V



1 1 2

1 2

( )m R T m m R TV V



2p

4punde 1

11

m R TpV

sdot sdot=


1 22

2

( )m m R TpV

minus sdot sdot=


1p


1 2

( )m R T m m R TV V






a



5p



4 1

34 3 3 33 1

4ln ln 2 2 ln 2 4 ln 22


= = = =


2p


b






c


max

1 TT

η = minus 1p

4p1 1

min 1p VT TvR

= =

3 3max 3

p VT TvR

= =

1p

Obţinem 1 1

3 3

1 12 2

p V pVp V p V


Rezultă 75η = 1p

d


3p 3p



177



c SI

V RA


SI

WR RI t


3

2 d 11

lRS

ρsdot= 2

2

lRSρsdot

= 1

2

14

RR

= 3

3 b 34 d 3




a


1p

4p





c


3

13 AABUIR

= = 2 24 AI = 2p




1 2 1 2 1 2( )I R R r r E Esdot + + + = + 3p4p




a


4extPP =

22

16ER I

rsdot =

sdot1p


2

P

EIR r

= +

1p


Rezultă 12 2784 016R = Ω Ω 1p


178

b



r= =

sdot


Rezultă 15 kJW =

c



Rezultă 32 WP = 1p

d


1 1 1 11

8E r IE R I r I RI


1

1

RR r

η =+

2p3p

Rezultă 80η = 1p



179

DOPTICAtilde


a 22


2 c 3


3

i rir

= rArr = = 3

4 c 15

2 1214

1 2

1

5 10 252 10

c

cλ ν sdotν

= = = =λ ν sdot

ν

3

5 c 9

43

600 10 2 6 10 06 mm2 2 10Dil

minusminus

minus


sdot3



a


1 1 1x x f

minus = 1p

4Rezultă 1

21

20 ( 30)20 30

fxxf x

sdot minus= =

+ minus2p


b

Distanţa focală 2

1 1 2

1 1 11nf n R R

= minus minus

unde 1 1n =

( )1 1 11nf R R

= minus minus minus

2p

4p

obţinem ( )2 11 1

2n Rn

f R fminus

= rArr minus = 1p

Rezultă 20 1 152 20

n = + =sdot

1p

c


3p 3p

d


1 2 1

2( 1)

nCf n R

C nR

= = minus

= minus

2p


1 15 1151 112

C nnCn

minus minus= =

minus minus1p

Rezultă 2CC

= 1p



180



8

1 91

3 10600 10

cminus

sdotν = =

λ sdot2p

3p


b


16 152 9

2

3 10 1 10 055 10 Hz54 10 18

cminus


λ sdot 1p



2 1

( )( )

h e UhU

e


∆ = unde 152

2


λ1p


c


cL h h= ν =λ

1p

4pObţinem 17


Dar 17

190 19

0044 101 16 1016 10

eV J Lminus

minusminus

sdot= sdot rArr =

sdot1p


d


11 0 ch h Eν = ν +1p


1p





181

TESTUL 12

A MECANICAtilde




a


1 1

1 1

0

f

F T m aN m gF N


2p


2 2

2 2

0f

f

T F m aN m gF N

minus =

minus == micro

2p


1 2

( ) m125s

F g m mam m


+1p


c


1 1 2

m2s


m1s


4p






2 21 cos180


5p

Rezultă 2 2

1 0482

v vgdminus

micro = = 2p


3pp = 8000 Ns 1p

c


21

112 2

MvW =

2p3p

1 8000JW = 1p

d


4 2 2Mv kx

= 2p

4p

Rezultă 212

kN3204 mMvk

x= = 2p



182





a


1 2


1p

3p1 1 2 21 2p V p Vm m

RT RTmicro micro

= = 1p

1 1 1

2 2 2

35m pm p

micro= =

micro1p

b


1 2


1p

4p

1 2

21 2

2 2

( )

2

m m RTm RTp V

V p

+micro micro

= =micro

1p


12

2 2 1 12 1 21 2

2 1 2 2 2

2 2

1 1( )1 1 1( ) (p p )

2 22

mmp pmp m RT p

p



micro

1p


p = + =


1p

c


1 2

1 2


+micro= =

+micro micro

2p


12

2

12

2 1 2

( 1)

1 1( )

mmm

mmm

+micro =

sdot +micro micro

1p


1 2

kg12kmol

p pp p


+1p

d

0 2 rmp V RT=micro

1p

4p02r

p VmRTmicro

= 1p

0 0

1 2 1 1 2 2 1 2

2 2 2 0(6)3

rm p pm m p p p p

micro= = = =

micro + micro +2p



183


a

1 (273 27)K 300KT = + =

1p

3p


2 1

2 2 600Kp p T TT T

= = = 1p


3 2

V VT T


3 1

p pV V

=


1p

b



1p

4p


13 3 3 3 1 1 3 1 1 3 11 1( ) ( )2 2





sdot1p

c

123 12 23L L L= +



123 1 1 13 3 300JL RT p V= ν = = 1p

d

1 ced

abs

QQ

η = minus 1p

4p31 3 1( )( )



1 1

1 1 1 1

2 (4 ) 11 773 5 13(2 ) (4 2 )2 2

R T TR RT T T T



1p



184



2

b


=+

+


2

2

( )( )

E Rx R xRx rR rx

+=

+ +


=

Rezultă RrxR r

=minus

3

3

c


=





3

4

c


2EI

R r=

+


22

EI rR=

+


2

22 2122 7

rRI EI R r E

+= sdot =

+

3

5


nal 1

2nn n

E EIR r R r

= =+ +

Rezultă 174

E E=3





1

171AEIr R

= cong+

2p

b




22 ( ) 22ABx R xR x x

R Rminus

= = minus

2p

4p

22 ( ) 22ABx R xR x x

R Rminus

= = minus


2p


185

c


max 2ABRR =

2p

4pRezultă min

1 2

EI Rr R=

+ + 1p


dsc

EIr

= 2p3p

12AscI = 1p



a


R r=

+1p

4p


22( )

REP RIR r

= =+

1p


2 2

2 4

( ) 2 ( )0 0( ) ( )



1p


4mEP

r= De aici

2

14 m

ErP

= = Ω 1p

b


2p3p


c


02

0( ) 4R E Ef

R r r=

+ 2p


2

02 4(1 )

4 m

f fERP f



2p


186

d

1

2

22

2

22

RR r

RrR

η =+

η =+

2p

4p1

2

1223

η =

η =1p

2

1

43

η=

η 1p



187

DOPTICAtilde




a



1p

5p

2 1 2 1

1 1

n n n nx x R

minusminus = 1p

3 2 3 2

2 2

n n n nx x a R

minusminus =

minus 1p


2 1 1 2

n n n n n nx x R R


b


1 2 1

y n xy n x

= 1p


1 3 1

y n xy n x

= 1p



3 1 1

n x yn x y

β = = 1p

c


2 1 3 2

1 2

imnx f n n n n

R R


+

15p

3p


2 1 3 2

1 2

obnx f n n n n

R R

minusrarr infin =

minus minus+ 15p

d


3pRezultă

22( 1)Rf

n=

minus 2 2

1 1

y xy x

β = = 2 1

1 1 1x x f

minus = 2p



188


a




b

Putem scrie 2

kx rD l










2Dx a l b a l bl


c


4pRezultă

2lDi

lλ

= 1p


Rezultă 2icircDi

lλ

= 1p

d


D D fi il f l f

λ λ∆ = minus =

minus minus2p

4p


2 2D g Di gi

l l+ λ λ

∆ = minus = 2p




190

TESTUL 1

A MECANICAtilde




a


4punde 1 2 1 2( ) sin ( )t

HG m m g m m gL



= + sdot sdot minus

1p


b


5p

unde 2 2

1 2 1 2( ) cos ( )nL HG m m g m m g

Lminus




1 2( )fF L

m m g L H

sdotmicro =

+ sdot sdot minus1p


c


4p




Lminus





d


3pDeci 2 2


L+ micro sdot minus

= + = sdot sdot 1p

Rezultă 28 NF = 1p



191


a


4p

Unde 2 2

0 2 2

M MM cM pM



Obţinem 2

max 2Mvhg

= 1p


b



4 52

QQ cQ pQ cQ cQ cQ

m vE E E E E E

sdot= + = + = =

1p

4pDar M QE E= 1p


5Q Mv v= 1p


c



Deci 2

2M

rm vF m g

dsdot

= sdot + 1p


d


2p3p




192





a


3pObţinem 1 11

1

p VR T

sdotν =

sdot1p


b


4pObţinem 2 2 2

22 A

p V NR T N

sdotν = =

sdot1p

Dar 22

A

NN

ν = 1p


c




1 2


ν + ν1p


d


4p



Rezultă 2 2

1 1

05VV

prime ν= =

prime ν1p



193


a



Dar 12341 0U∆ = 1p


b






c




d


4p






194





a


PR R R R R= + + + 1p

3p



b



P S

EIR r

=+

1p


Rezultă 02 AI = 1p

c




Rezultă 4 Vu = 1p

d


S

EIr

= 2p

4pObţinem SC

EIr

= 1p




195


a


EIR r

=+

1p

4pUnde 12 34

12 34BD

R RRR R

sdot=

+ 1p

Iar 12 1 2R R R= + 34 3 4R R R= + 1p

Rezultă 1 AI = 1p

b


5p









EIr

= 2p3p




196

DOPTICAtilde




a



= 1p


b


1 1 1x x f

minus = 1p

3pObţinem 1

21

x fxx f

sdot=

+1p


c


1 1 1x x f

minus =prime prime

1p


1

12

xxprime


1p


β1p


d


1 1 1x x f


1p


1

xxprimeprime


1p

Obţinem 1

fx f


1p




a


Dil

λ sdot= 1p

3pObţinem

2l iD

sdotλ = 1p



197

b


1p




∆α = = 1p


c


D iil

λ sdot= = 1p

4pDeci 1 2

DD = 1p


Rezultă 1 md = 1p

d


Dil

λ sdot= 1p

4pUnde 2

1 cn n

λλ = sdot =

ν1p

Deci 12 2 2

D iil n n

λ sdot= =

sdot1p




198

TESTUL 2

A MECANICAtilde



∆ SIpF Ft 3

2 c 15 mgiT G F= + = 3

3 a 2

0max2 2 10 m 0

2vd h r

g= = = ∆ = 3

4

b 2 2

5 45 4

4 5

452 2

4 s 5 s

at atx x x a

t t


= =22 msarArr =

2 210 9

10 9

9 10

19 m2 2

9 s 10 s

at atx x x x

t t


= =

3

5


3



a


1 1

2

0

0f

f

F T FT F

minus minus =

minus =

2p

4p

1 1 1 2( )F m m g= micro + 1p

2 2 kgm = 1p

b


1p


2 025micro = 1p


199

c


3 1

2

0

0e f

e f

F F FF F

minus minus =

minus =

2p

4p

3 2 1 2( )F m m g= micro + 1p

3 15 NF = 1p

d


4 1 1

2 2

e f

e f

F F F m aF F m a

minus minus =

minus =

1p

4p24 2 1 2

1 2


minus micro += =

+ 1p

eF k l= ∆ 1p

2 2( ) 1333 Nmm a gkl

+ micro= =

∆1p



a

00 50 mspv

m= =

1p

3p0u

vtg

= 1p

5 sut = 1p

b


4p0

32 2pA pA

cA c

E EE E= rArr = 1p

20

3vh

g= 1p

833 mh = 1p

c

0 3c cAE E= 1p

4p0

3vvrArr = 1p

29 msv = 1p


d

20

max 125 m2vh

g= = 1p


0 104 mc

f

Ehmg F

= =+ 1p

max 12hh

= 1p



200



1

2


= ν ∆ = ν ∆

∆rArr = γ

∆

3

2c

11

1 1 ii i f f f i

f

TTV T V V VT


= rArr =

44

if f i

VV = rArr ρ = ρ

3

3


2 2

V pV p

=


2 1 2

V p TV p T

=


2 2

3p Tp T

= =

3

4c

11

1 21 1 2 2

2 1

03V pp V p VV p

γγ γ γ

= rArr = =

2 1

1 2

N n VnV n V

= rArr =

3

5b 2 2

1 1

22 1

1

1 1

12 kJ

Q TQ T

TQ QT

η = minus = minus

rArr = =

3



a

1 2

1 2

m m+micro =

ν + ν 1p

4p

1 21 2

1 2

pV pVm mRT RT

micro micro= = 1p

1 2 2 1

1 2

T TT T


+ 1p

165 gmolmicro = 1p

b

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν2p

4p1 2

2 1

83 5

TTTT T

=+ 1p

2844 KT = 1p

c

pRT

microρ = 2p

3p30699 kgmρ = 1p


201

d

1 21 2

22 fpp pT T T

ν + ν = ν rArr + =

1 1 1 2 2 2 1 1 2 2( )i f

V V V V


ν + ν = ν + ν

2p

4p1 2

2 1

116 5

TTTT T

=+ 1p

5 22 1

2 1

11 (2 ) 153 10 Nm2(6 5 )f

p T TpT T

+= = sdot

+ 1p



a

2 3T T= 1p

4p

11 1 1

1 1 2 2 2 12

VTV T V T TV

γminus


1p

1 11 300 Kp VT

R=

ν 1p

2 14 1200 KT T= = 1p

b

323 2

2

ln VL RTV

= ν 1p


2 3

V pV p

= 1p

22 1

2

32RTp pV

ν= = 1p

23 2441 JL = 1p

c

31 31 31Q L U= + ∆ 1p

4p

1 3 1 331

( )( )2


83

V V= 1p

31 1 3( )VU C T T∆ = ν minus 1p

31 1314 JQ = minus 1p

d

min

max

1CTT

η = minus 1p

3p1

2

1CTT

η = minus 1p

75Cη = 1p



202





4 a 2

1 2max

1 2

( ) 4 W4( )

+= =

+E EP

r r 3

5 c 00

0 0

90 mA(1 ) (1 )

U U II IR R t t

= = = =+ α + α

3



a


e e

EIR r

=+

1p

4p1 3 1 3e eE E E r r r= + = + 1p

2 31

2 3e

R RR RR R

= ++

1p

1 AI = 1p

b

1 1 1U I R= 1p

3p1

1

e

e

EIR r

=+ 1p

1 154 VU = 1p

c

2V pU I R=

1p

4p

2 1

e

p e

EIR R r

=+ + 1p

2 3

1 1 1 1

p VR R R R= + + 1p

11 VVU = 1p

d


e e

E IR r

=+

1p

4p

1 23 3 3

1 2


r r= + = + = +

+ 1p

1 2

1 2

1 2

8 V1 1 3p

E Er rE

r r

minus= =

+1p

061 AI = 1p



203


a

22 2

2

PRI

=

2p3p

2 10R = Ω 1p

b


5p

1 21 3

1 2e

R RR RR R

= ++

pentru K deschis 1p

1 22

1 2e

R RRR R

=+


1 2er r r= + 1p

2 45r Ω 1p

c

22 int 2P r I= 1p

4p1 2I I I= + 1p

1 1 2 2I R I R= 1p

22 6 AI I= = 2 int 162 WP = 1p

d2

2

e

e e

RR r

η =+ 2p

3p

037η = = 37 1p



204

DOPTICAtilde


b 2

1

ff

β = 3

2 a 2 2 22

1 1 1

1 45 cma

x n x xx n x n

= rArr = rArr = 3

3 c 15 mmaa

iin

= = 3

4


c

1 2

1 1 11

l l

l

f n nnn R R

= =

minus minus

lfrArr rarr infin 3

5a 1 1 2 2

192 61 10 J

ex c ex c

ex

L E L E

L minus

ε = + = +

rArr = sdot

3



a2p

3p


b

2p

4p


2 25 cmf = 1p

c

1 12

2 1 1 1 1



4p1 2 3 3 35 cmd x x x= + rArr =

1p

4 3 2

1 1 1x x f

minus = 1p

4 875 cmx = 1p


205

d

3

1

yy

β = 1p

4p1 2β = β β 1p

21

1

15xx

β = = minus 42

3

25xx

β = = minus

1p

3 1 5625 cmy y= β = 1p



a

W N= ε 1p

4p

hcε =

λ1p

W Pt= 1p


= = sdot 1p

b

ex cL Eε = + 1p

4p

c sE eU= 1p

ex shcL eU= minusλ

1p


c

1s

N eIt

=

2p

3p18


= = sdot 1p

d1p

4p




206

TESTUL 3

A MECANICAtilde




a

2 0yG F N+ + =

1p



b

1 2 0x x fF F F+ + =


4p


1 2

2

cossinm

F Fmg F

+ θmicro =

minus θ 1p


c

1 2x fF F F ma+ + =

1p



F F F mgam

1p

2412 msa 1p

d



vv = 1p

2 cos2matP F= θ 1p

1854 W mP 1p



207


a

2

2PkxE = 2p

3p


b


4p2

2

= =i pkxE E sistem

1 2= + =f c c cE E E E

2p


c

2 2 21 1 2 2


1p


1p

21

1 1 2

( )

=+

kmv xm m m

12

2 1 2( )kmv x

m m m=

+1p


d

fc FE L∆ = 1p

4p

21 1


21

2vd

g=

micro1p




208





a1 0 2

lp p g= + ρ 2p3p

51 15 atm 15∙10 Pa p = 1p

b

1 1 2 2p V p V= 1p

4p

1 0 2

= + ρlp p g 1

2=


2 00( ) 0


1p

10 cm x 1p

c

1 1p V RT= ν 1p

4p1p ShRT

ν = 1p


d

2iU RT= ν 1p


2

5 078 2

= νNU RT2

5 02 2


aer 138 J U 1p



b

1 1 2 2 amestec

1 2

V VV

C CC ν + ν=

ν + ν 2p

4p2

=ViC R 1

3 2

=VC R 252VC R= 1p



209

c

min

max

1 ndashCTT

η = 1p



d

cedat

primit

1 ndashQQ

η =

1p

5p



3ndash1 1 13


= ν = ν 1p

p VC C R= + 1p


η = =V

p

C RC




210





a2

12

Ve A

V

R RR R RR R

= + ++ 2p

3p


b


2 2 3

1 2 3

( )A

V


= + + + = = +

111 16

=I A

25 8

=I A

31

16I A=

2p

4p


069 AAI 6875 V VU 1p

c 1

AA

EI R r R

=+ +

0VU = 2p4p


d

er R = 2p

4p1e AR R R= + 1p



a


=E

2p

4p2

max 4EP

r= 1p


b

2middotP R I= 2p

4pEIR r

=+

1p


c

ext

totala

p

p

RPP R r

η = =+

2p

4p2pRR = 1p



211

d

lRSρ

= 1p

3p2

4dS π

= 1p




212

DOPTICAtilde




a2 1 2

1 1 1 minus =x x f

2 22

1 1

y xy x

β = =

2p



b2 1 1

1 1 1ndash x x f

=

2 2

1 1

y x y x

β = = 2p



c

2 1 1

1 1 1 minus =x x f 2 120 cm=x 1p


2 1 2

1 1 1ndash x x f

=

1p


d 1 2

1 1Cf f

= + 2p3p




a

11

2λ

=Dil

2

2Dil

λ=

2p

4p1

1

cλ = rArr

ν1 2 1

12id

cλ ν

=

1p


b

1 1 2 2k i k i= rArr 1 1 2 2k kλ = λ 2p


1p



213



d


δ =l yD

2p


k λδ = = 1p



ν= = 1p



214

TESTUL 4

A MECANICAtilde


1

1

01 kgpm mv

= rArr =

m m

vF m a m

t∆

= sdot = sdot∆

m 2 2 2 s


2 NF =

3

2

a 0N F G+ + =


cosm gN sdot

=α

3




12 2 1 v v v= minus 12km 216 h


5





a




3p 3p


215

c


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α


( )

20

1 1

1 1

20 1

0 cos 2 sin

2 sin cos

m v m g h m g d

h d

v g d


= sdot α



20

1 1 307 m 2 sin cos

vd dg


2p

5p



u

d vv tt

+= = rArr ∆ =

∆


1p


( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot

( )

22

1 1

22 1

12

0 cos 2 2 sin - cos

45 m s

m v m g h m g d

v g d

v minus



= sdot



c

d vv tt

+= = rArr ∆ =

∆


1p


1p

d

Notăm cu R



1p

4p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




2p( )2 2 2 2 2

2 2

2

1

1 cos 1

cos 1 087 87

f fR N F R N F N

R N m gRG




1p




b


( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot

3p4p

( )2

1 01 0

10

2 2 1 cos 2

316 m s

m vm g h v g h g l

v minus


rArr = sdot1p


216

c




( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) ( )

2 21 1 1 0 1 2

1 1

2 21 0 1 2 1

2 2 2

1

c R

m v m vE L m g d

v v g d



( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

( ) 2 21 2

1 2

2 2

m m v k x

kv xm m


prime = sdot+

2p

4p


2p

d


11 1 1 1

1

= 14 m s


minus



1p

4p

( )1 1 1 2

11 1 1 1

1

= 14 m s


minus



( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr =

1p( ) ( )2 20 1

11 2

1

1

08 m

v vdg

d

minusrArr = rArr

micro + micro sdot

rArr = 1p



217



2 2CO C Og = + 44


5

29 10 Pa = 29 atm

m R Tp V R T pV

p


micro sdot

rArr = sdot

3

2

a

12

1213

13

12 13

cu 22

5 cu 082

V

p pp

RQ C T C C R

Q CQ C T C RQ C

T T T


∆ = ∆ = ∆

3

3

d-1 -1


1 1 1

2 2 2 1 2

2

V S hV S h S h


2 75

p

v

C iC i

+γ = = =

1

12 1

2

VT TV

γminus

= sdot


3

4



3





b


1

12 1

2

VT TV

γminus

= sdot

(1) 1p

5p

p

v

CC

γ =






2p


γ = = 1p



218

c1 1

111

1

1344 g

cmm


3p 3p

d

2 2

222

2

4368 g

cmm


1p

4p

3 3

333

3

08 g

cmm


1p

1 2 3 5792 gm m m m m= + + rArr = 1p

32

kg 29 m

mV

ρ = rArr ρ = 1p



a


micro ν 1p

3p1 2 3 4m m m m m= + + +

1 2 3 4ν = ν + ν + ν + ν1p


molmicro = 1p

b

4p 4p

c


5p


1 2 3 41 2 3 4v v v vv

C C C CC


ν

1p


1 1 11 1

2121 2

22

2

p V R TT TTV Tp R T


sdot = ν sdot sdot

1p

112 1

1 1 15

31 2

- 155 - 155

155 10 J

ced

ced

TQ Q R T

R T p VQ



= minus sdot

1p


219

d


1 1 1 1 123 2

2 1

2ln ln ln 22 2

V T V p VQ R T RV V


1p

3p5

23 035 10 JQ = sdot

( ) 131 1 2 1 1 121 105


1p

531 105 10 JQ = sdot


1p



220



1 1 2 2 3 31 2 3

1 1 2 2 3 31 1 2 2 3 3

1 2 3


S S S S


1

2

2775 n m222 n m


3

2

b

( ) ( )1

1 10 1

1 0 1

1 1

UI UR IR

R R


(1)

Analog ( )2

0 2

1

UIR

=sdot + α sdotθ

(2)


3



4


1 2 3

1 1 1 1 pR R R R




3

5

a2

U N SW tl

sdot sdot= sdot ∆


3



a

1K rarr

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

2p

4p1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω

1p

1

1

2 2 3

3 8

3

EI R R r

ER rI

R

=sdot

+ sdot +

rArr = sdot minus

rArr = Ω 1p

b

2K rarr

2

2

2

154 A

EIR r

I

=sdot +

rArr =

1p

4p2

2

2

154 A

EIR r

I

=sdot +

rArr =1p



221

c2 2

3e eRR Rsdot

= rArr = Ω 3p 3p

d3 3 36 A2

3

EI IR r= rArr =

sdot+ 2p

4p




a


400 = 3

b b b b

bb b

b

P U I IUR RI

= sdot rArr

= rArr Ω2p

4p

bb c

EIR R r

=+ + 1p

485 1616 3cRrArr = Ω = Ω 1p


c

icircnchisK rarr b

t

PP



100 VR bU U= = 1 2R R R+ =


2 b RI I I= +

2 2b RU I R= sdot

2p

4p



η = = 1p

d


4p


1 1 1

e sR R R= +

prime

1p


r= rArr =

sdot


2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =

2 337 2

2 033 A

s s

b

s s

b

s ss

E I r I RR RRR RR R

E I rI IR

= sdot + sdot

sdot= + rArr = Ω

+

minus sdot= rArr =



632 Wbb b

b

UP PR

= rArr =

2p



222

DOPTICAtilde


( )( )1


3

2 a 3

3

a1

2

50 cm1 80 cm

fC

ff=



1 2

1 2

1 22

1

2

16 cm

h hf f

h fhf

h

= rArr

sdot= rArr

=

3

4c

39 Vextrextr s s s

LL e U U Ue


5

a

3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3 3

2tg tg 20 10 20 10 rad

236 72 23

iD

Dil

minus minus



3



a 3p 3p

b




i i = rArr =

2p

c

2 5 2sin cos tg = 3 3 5

r r r= rArr = rArr

2p

4p1

1

1

tg i = = 50 cmtg = = 775 cmtg

tg

x hh i H H

r hxrH

rArr rArr=

2p


223

d

2p

4p

0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p0

2


tg

2 = 907 cm

an n l

l l

rlh

D r

sdot deg = sdot

rArr rArr =

=

rArr = sdot

1p



a

19min min

max

= 262 10 J = 163 eVR


λ λ 2p



1p

b


λ2p

4p max

min c

extrextr c

EL im

L E=




4p252 m = 66 10

2 sc

cm v EE v v

msdot sdot




224

TESTUL 5

A MECANICAtilde




a


1p



b






d





a


0


4p2 20

2G gFr v v df f


V = 9 ms 1p

b


2p


1 1 1 2( )m v m m vprime= + 1p

Rezultă 1 1

1 1

3 msm vvm v

prime = =+ 1p


225

c


4p

1 2( )0 m m g Fr Df

+minus = sdot 1p

1 2( )m m gFrf

+= 1p


d

21 2

21 1

( )cs

cl

E m m vxE m v



3p

Deci 103

x vprime= 1p



226





a

ν1 = 2Omicro

m 1p

3p1

2

116

νν

= 1p

2

2Hmicrom

ν = 1p

b


4pp 2

ν= ν1RT1

p 2ν

= ν2RT2

2p

Rezultă 21

12

16TT

νν

= = 1p

c


pV2prime= ν2RT1

1p

4p

1

22

1VV

=νν


1p


V2prime= 2

L x S +

1p


d


4p1 2

2m m m+ =


Obţinem 1 2

1 2

2micro micromicro =

micro + micro1p

Rezultă 1 2

1 2

2micro micromicro =




227


a


4p 4p

b


1p

4p


12

1

lnc T VQ RV



1 1 3

13 1

3

4

1 44

PV RTPV RT

T T TT

= ν= ν

= rArr =

1p


c


p

QQ

η = minus 1p


215η = 1p

d

minCarnot

max

1 TT


3pCarnot

3 754

η = = 1p



228





a


V

ERR r+

1p


2vR (2) 2 2 2 2

v v

v

R ERU IR r

= =+

2p


b


EIR r

=+ 1p


EIR r

=+

2p


c2 1

1 1VrR

EU U

=

minus

1p

4p

2 1

1 1AR E

I I

= minus

2p


d

RV rarr infin 1p

3p1V

VV

V

ER EU ErR rR

= = =+ + 1p



a


4p

1 1 3

2 2 4

S

S

R R RR R R

= += +

1p

1 3 2 4

1 2 1 2 3 4

1 1 1 ( )( )tot

tot S S


+ += + =gt =

+ + + 1p


1 2 3 4

( )( ) 4


+ += + rArr = Ω + + +

1p


3pW = 504 J 1p


229

c


4p1 2 3 4

1 2 3 4

EI R R R RrR R R R

=+ +

+ +2p


d


tot

RR r

η = =+

2p

4p


tot

RR r

η = =+ 2p



230

DOPTICAtilde




a


1 2

11 1( 1)

fn

R R

=

minus minus

2p

4p1

2 0RR

rarr infinlt

1p

R = 15 cm 1p

b

2


4p2

1 2

1 1 1( 1)l

fn

R R

=

minus minus

1p

|R1| = |R2| = 15 cmRezultă nl=137 1p

c

2 2

1 1

1y xy x

β = = = minus 1p

3p

2 1

1 1 1x x f

minus = 1p


d


21

1

xx

β = 2 1 1

1 1 1x x f

minus =

1p

4p


2 1 2

1 1 1x x fminus =

2

21

xx

β =


1 21 1 2

1 1

f xx f ff x

= minus minus+

1p


11 1

1 1

f fff x

f x


+

1p

2

1

ff

β = minus


1p



231



2kk Dx

lλ

=k = 5

2p3p



2lDi

l= 2p


c


1p

5p



2knx eD

lminus

=

xk = 04 m

1p


d



2 2k D k

l lλ λ

= 1p





232

TESTUL 6

A MECANICAtilde




a

2

2f

BG F cB cA



4 msBv = 1p

b

0cE∆ = 1p

3p20J

20JpE mgh

E

∆ = =

∆ =2p

c

2 0

2f

BF cB


4p2 2

2 2B Bv vgd d

gmicro = rArr =

micro1p

4md = 1p

d

( ) 0 0


4p


cos ( )sin


1p

40JL = 1p



233


a

1 1F t psdot ∆ = ∆

1p


1p

12mvF

t=

∆1p

1 2kNF = 1p

b


1p

4p

2

2 2

22

p p mv

F t pp mvFt t

∆ = =

sdot ∆ = ∆

∆= =

∆ ∆

2p

2 1 kNF = 1p

c

H F t= sdot ∆

H F t= sdot ∆1p

3p

1 1

2 2

2 N s1 N s

H F tH F t


2p

d

2p

4p

3 2 2 2N sp p mv∆ = = = sdot

2p



234





a 4p 4p

b

A BU U U∆ = ∆ = ∆ 1p

4p2 1 2 1 2 2 1 1


250 JU∆ = 1p

c


5p


650 JAQ = minus 1p


50 JBQ = minus 1p





a

12 34 0Q Q= =

123 3 2 3 2 3 2 2 2 2 3 2 3 2

5 5 5 5( ) ( ) ( ) ( ) ( )2 2 2 2V


2p

5p1 1 2 2 2 1

4 1 3 2 3 4

1 2

3 4


γ γ γ

γ γ γ



-11

23 4 1 4 1 15 5( ) ( )2 2primit


3p


235

b41 1 4


4p


c

1 cedat

primit

QQ

η = minus 1p

3p4 1 1

1-1

4 1 1

5 ( ) 121 15 ( )2

p p V

p p Vγminus

γ


εminus ε sdot2p

d

pL Q= ηsdot = 1p

3p-1 -1

4 1 1 4 1 1-1

1 5 5(1 ) ( ) ( 1) ( )2 2


ε2p



236





a

11

1 1 2 1 1 2 1 48 16 8 8 8 4 p

p

RR

= + = + = = = Ω 1p

5p

1 4 20 24sR = + = Ω 1p

22

1 1 1 3 1 69 18 18 6 p

p

RR

= + = = = Ω 1p

2 6 6 12sR = + = Ω 1p

1 1 1 3 1 824 12 24 8 AB

AB

RR

= + = = = Ω 1p

b

1 1 3

1 1 05 8 4 VABU I R IR

I R= += sdot =

2p

4p

1 11 1 2 2 2

2

4 1 025 A16 4

I RI R I R IR

= rArr = = = =

1 22 1 A4 1 20 4 20 24 VAB

I I IU

= + == + sdot = + =

2p

c

ABe

AB

UIR

= 2p3p

3AeI = 1p

d

ABAB e e

e

E UE U I R rI

minus= + rArr = 2p

3p

1r = Ω 1p



237


a

1 2

( )

( )


I

= + ++

= + +

3p

5p

2

6024 24

2 20 50 05A

II

I II

= +

minus + ==

2p

b( )e e e



43eR = Ω 1p

c


4p1

1

21

24 2A1236 3A12

b

b

PIUPIU

= = =

= = =2p

d

11

12 62

bURI

= = = Ω

22

12 43

bURI

= = = Ω2p

3p




238

DOPTICAtilde




a1 1 1 1

22 1 1 2 1 1 1 1


+= + hArr = hArr =

+ 2p3p


b



4p 2 1 2 12

2 1 2 2 2 1 2 1


+= + hArr = hArr =



c

2 2

1 2 1 1

sistemx xx x



d 5p 5p



239




b


2p3p


c



λ= 2p


relaţia 2

xD l∆ δ

= care devine ( 1) 45 ( 1) 45 ( 1)2 2 2 2



en

λ=

minus

2p


d


4p

( 1)nouke

nλ

=minus

1p



240

TESTUL 7

A MECANICAtilde


3



EE mgh m

gh= rArr = =


892 ms2

poco po c p p o

EmvE E E E E vm

+ = + rArr = rArr = =

3

4 c 3

5

d

0

0 sin 0 cos

f

x

f y f


+ + + =


cos 20 10 3 027

sin 10f

f

F mg FF NN F


sdot α

3



b


4p


1 1

2 1 2

l ml m m

∆=

∆ + 1p

1

2

25

ll

∆=

∆ 1p

c


4p2 2 1eF m g m g= + 3 3G m g= 1p


d


1m gm g kx k

x= rArr = 2p

3p

100 Nkm

= 1p



241


a


2p

4p


2 1 2 2 1 2

2

0 ( )80 N

N N G N m m gN


b





1p

5p


1 22 1 2

2 1 2 2 1 2


Oy N N G N m m g

minus minus minus =




1p


33 2 3

3 3 3 3

sin


Oy N m g N m g

α minus minus =




2p

Rezultă 3 2 1 1 2 1 2 1 2 3

3 2 1 1 2 1 22

1 2 3

(sin cos ) 2 ( ) ( )(sin cos ) 2 ( ) 342


m m m s


= =+ +

1p

c

1 1 1 1( ) 1626 NT m a g T= + micro = 1p

3p2 1 2 1 1 2 1 2

2

( ) 2 ( )4736



d

1

1 1

22 ( )

R TR m a g

== + micro 2p

3p3252 NR = 1p



242



4



υ ∆ = υ ∆ + 2 1


1 2 1 1 2 2 2 12 1 2 1

3( ) ( )2 2



2 2

p aVp aV

= =

1 11 2 2 1

2 2

p V p V p Vp V

rArr = rArr =

Obţinem 2 2 1 12 1 2 1

3( ) ( )2 2



2 2 2

p V RTp V RT

= υ= υ


3

5



2pV V p= rArr =

adiabat 11 1 2 2 2 2


2

2pp

γminus= 3



a


4p


1 1 1 1

2 22 2 2 2

1 21 2

( )( )


p V Vp V V RTRT

= ν rArr ν =

= ν rArr ν =

++ = ν rArr ν =

1p

Rezultă 51 2 1 11 2 1 1 2 2 2 2

2



b


2 1 1

p Vp V

ν=

ν 2p3p

1

2

5ν=

ν 1p

c


1 2

m m m= +

micro micro micro1p

4p1 2

1 21 2

1 2

2m m

m m m

=micro micro


2p

7gmolmicro = 1p

d


4p


1 2 1 2

1 2 1 2

1 21 2

1 2

2 1

1 2 1 2

( )22

V V V V V V

V V V VVV


C C C CC C




1p

1 22 1

1 2

134

7716

V VV

V

C CC R

mU C T KJ

micro + micro= =

micro + micro

∆ = ∆ =micro

1p



243


a


3p 3p

b



4p

1 1 1 33 1 2 1

3 3 1


= rArr = = rArr = =

1p



3 3 4 4p V p V= 4 1p p= 1 1 1 49 p V p VrArr =

4 19 45lV VrArr = =1p

c


p Vp V RT TR

= υ rArr =υ

1 12 2 2 1 1 2 2 1



1 13 3 3 1 1 3 3 1



4 3 19T T T= =

2p

5p

32vC R= rArr 1 1

1 3 3 1 1 13 8( ) 122v


υ2p

1 3 6000 J = 6 kJU rarr∆ = 1p

dmin

max

1cTT

η = minus

1

3

1cTT


88cη = 1p



244



3


rArr = = Ω


= rArr = Ω


η = = rArr η =+

3

4


182 31 1 1 1 11

2 3

k

ke

k

E E E Er Er r rE

r r r r

sum + += = =

sum + +

1 1 61 1 1 1 11

2 3

e

k

rr

r r r r

= = =sum + +

3

5

d Puterea maximă 2

max 4EP


2 22 2

2

2 6 04 ( )E E R R Rr r


+


3



a


1

1 1 1 22 3e

e

RRR R R

= + rArr = 1p

4p1 18V 2AU I= = 11 1 eU I R= sdot 1p

11 1

1

2 33 2R UU I R

I= sdot rArr = 1p

6R = Ω 1p

b


2

1 1 1 32e

e

RRR R R

= + rArr = = Ω 1p

5p

1 1 1 1

2 2 2 2



1 1 2 2U I r U I rrArr + = + 1p

1 2

2 1

1U UrI I

minus= = Ω

minus E = 10 V 2p

csc

EIr

= 2p3p

10AscI = 1p

d

2

max 4EP

r= 2p

3p

max 25 WPrArr = 1p



245


a

1R şi


lRS

= ρ

22

lRS

= ρ 1p

3p


1 2 2 1 1 2 1 1 2 2l lR R R R R R R l R l

S Sρ ρ


1p



1 2 1 2

1 1 1 ( )e

e

R R RRR R R R R R R

+ sdot= + rArr =

+ + + 3p4p

5eR = Ω 1p

c


3e e

nE EIR nr R r

= = =+ +

2p4p




e

EIr

= = 2p



246

DOPTICAtilde


2

a


2 2 2 2

2 2 2 1 2

1 1 1

62 4V

(155 )155055





ν minus

140 0 118 10 HzLL h


3

3

b



2

sintg tg1 sin

PB rr PB h r hh r

= rArr = =minus


2

sintg tg 3 07m1 sin

x rx PB hPB r


3

4


2r r

v v

D iil i

λ λ= rArr = =

λ


v

ii i fi i f fi


f = 30

3

5


4l

n= =

2

tg tg


rl r h lh

l lr h h ml l

= rArr =

= = =minus

3



a



1 2

1 20 cm1 1 1( 1)( )

Rfnn

R R


1p


2 1 1

1 1 1 ( 20)( 30) 12cm20 30

fxxx x f f x


+ minus minus 2p


1 1 1

( 12) 2 08 cm30

x y x yyx y x


minus


b


1 2

11 1( 1)( )

a

a

a

n Rf n n nn R R

minus= =

minusminus minus2p

3p

80 cmf = minus 1p


247

c


R ffnn

R R

minus= = = = minus


2p


1

2 2 1 1

1 1 1 ( 10)( 30) 15 cm10 10

f xxx x f f x


+ minus minus 1p


2 2

1 1

x yx y

β = = rezultă

2 12

1

( 15) 2 1cm30

x yyx

minus sdot= = =

minus 1p

d


a

a a

f fFF f f f f

sdot= + rArr =

+1p

4p

1 15 cm1 1 2( 1)( 1)( )( )

aa

a

Rfnn

R R


minus1p

( 20) 15 30 cm30 20


minus1p

12

2 1 1

1 1 1 ( 30)( 30) 15 cm30 30

F xxx x F F x


+ minus minus 1p




2Dil

λ= = 2p


i= rArr = = 1p

b


lλ

=


Dxl

λ= =

1p


λ= + sdot 1p


Dx x x xl


c


Dδ = minus =



∆ = minus minus

2p

5p


λ = minus minus



= minus rArr =minus

2p

0 33 45 mm2

Dx xl


dDin relaţia 1

2KK Dx K

lλ

= = rArr 22l x

Dsdot

λ = 2p3p

2 480nmλ = 1p



248

TESTUL 8

A MECANICAtilde




a

2

2cpEm

= 2p3p


b

pFt

∆=

∆ 1p

4pF ma= 1p

pam t∆

=∆

1p


c

amFFF fx

=++ 21 1p





d

2

2atd = 1p

4p2 1d d∆ = minus

21

1 2atd =

22

2 2atd = 1 1 st = 2 2st = 2p




249


a


4p1iE m gh= 2

1

2fm vE = 2p


b


2gth = 1p


2

2atd vt= + 1p


1p


c


1 1 112 2


1 1 1 2i fp p m v m m v= hArr = +

2p

4p( )1

1 2

2m g h dv

m mminus micro

=+

1p


v = 1p

d


1 2

2sistem

m m vE

+=

2

2sistemk lE ∆

=

1 2m ml vk+

∆ =

2p

3p


l∆ =

1p



250





a

pV RT= ν 1p

4p1

2

2

1

2

1

2

1

micromicro

νν

sdot==mm

pp

2p

rezultat final 47

2

1 =pp

1p

b

1 2

1 2

1 2

amestecm mm m

+micro =

+micro micro

2p3p


c

0

2p V RT= ν

1p

4p0 finalp V RT= ν

1p


d


2p




b1 1L p V p V= ∆ sdot ∆ = 2p


c

min

max

1CTT

η = minus 1p




251

d

efectuat

primit

LQ

η = 1-2 2-3primitQ Q Q= + 2p

5p


p

V

CC

γ = p VC C R= + 1p

RC γ=

γ minus 1V

RC =γ minus

1p

12 1γ minus

η =γ +

1p

rezultat final 2 11

19η = = 1p



252





a

4 52

4 5s

R RR RR R

= ++ 2p

4p3

13 3

se

R RR RR R

= ++ 1p


b

1e

EIR

= 1


3 3

sp

R RRR R

=+

2p


c 1

EIR

=

2p3p


d3

1 3

EIR R

=+ 3p

4p




alR

Sρ

= 2p3p




c

becuri b

sursatilde

P UP U

η = =

3p4p


η = 1p

d

middotx

xRS

ρ= 1p

4p2 x

UIR

= 2p




253

DOPTICAtilde




b

2 1 2

1 1 1x x f

minus =

2 2

1 1

12

y xy x

β = = =

2p

4p012

21

=+ff 1p


c

( )1

1 11nf R

= minus 2p



d


4p2 1 1

1 1 1x x f


2 1 2



1 2

16 cm

x fx xx f

= = minus+

1p






b

2 1r rδ = minus 1p



enδ minus δ

=minus

1p


c 2Dil

λ= 2p

3p



254

d


δ = 1p

4p



δ = δ + 1p


= minus 1p

rezultat final( )12

ed nh

lminus

= 75 mh = micro 1p



255

TESTUL 9

A MECANICAtilde



5 20 205


= rArr = = ms 3

4

d [ ]

( )

21

2

0 0

m1s

2

sC

atx t x v t

C a

=

= + +

rarr

3

5

a ( )

1 1 2 2

1 2

1 2

1 2

1 2

1 2

N12m

s

s

s

s

F k F kF k

F Fk F Fk k

k kkk k

k

= ∆ = ∆

= ∆ + ∆

rArr = =∆ + ∆ +

rArr =+

=

3



a

1p

4p

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1p1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

=

1 1 1

1 2

2 2

1 2

1 2

2

m2s


m m ma gm m m

a



+ +

= 1p

b2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+

2p4p

2

1

2 2

1 1( )(

16 N N) 1 2

T m g a TT m g a T

= minus == =

rArrrArr+ 2p


256

c


3p( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot =

1p

( )2

1 1

2 2

1

0

02 10 2 N 02 10 2 N

T m g TT m m g T+

= = sdot =

= = sdot = 1p

d( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

2p

4p( )

1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

=

1p

( )1 1

1

1 23

3

3 2

2 2

0

0

0

04 kg

m g TT m m g TT g

m mm m

m

m

=

minus micro + =

minus

minus

minus

=minus

rArr = minusmicro

= 1p



a 05 JA

A

E mghE

==

2p3p

1p

b

2p

4p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=

1p

( )

2

0

sin cosm327s

t f nG F ma N G

a g

a

minus = minus =

= α minus micro α

=1p

c


( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

4p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p

( )1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

=

1p( )

1

1

cossin

1 ctg0327 J

f ABB A F

B A

B

B

E E L

hE E mg

E mghE

minus =


= minus micro α

= 1p

d


2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p

4p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

=

1p2

2

0

1308 m

f BDD B F

B

B

E E L

E mgXEXmg

X

minus =

minus = minusmicro

rArr =micro

= 1p



257



b [ ] Jmol KSi

C =sdot

3

2 a 2100 kJ

Q m c tQ

= sdot sdot ∆=

3

3d 1 1 2 2

1 2


N NN Nmicro + micro

micro =+

micro = =

3

4

c 2

1

11 1

2

ln

ln

280 J

VL RTVpL p Vp

L

= ν

=

= minus

3



a0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot

2p3p

0

230 531 10 g

A

mN

m minus

micro=

rArr = sdot 1p

b

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

4p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot

1p

26 36023 10 m

AA

A

NnVN N NNNnV

minus

=

ν = rArr = ν sdot

ν=

η = sdot 1p

c

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

4p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

1p

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot

33

13

31

33 3

22

g64 10cm

mV

m m

VV

V

minus

ρ =


=

νmicrorArr ρ =

rArr ρ = sdot 1p

d

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

4p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= =

1p

13 2 1 3 2

2 12 1

2 1

3 1

3 1

1

22

2

43 754

Vp p V p p

p p p pT T

p pp p

p

= rArr =

= rArr =

rArr =minus

= = 1p



258


a

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

3p

3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p3 3

1 3 33 1

1 3 1

3 1 3

3

3 900 K18700 KJ 187 MJ

VU C TV V VT TT T V

T T TU

= ν

= rArr = sdot

rArr = rArr == =

1p

b

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

5p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

( )

1 2 1 2 12

0 01 2 1 2 0 0 1

0 22 0

0 0

1 2 2 1

0 0 2

0 02 1

1 2 1

1 2 1

1 2 1 2

2 2Aria 2 2 5 MJ2

33

3 39 9

5 82

2050 MJ 55 MJ

V

Q U Lp VL p V RT

p p p pV V

U C T Tp V RT

p VT TR

RU T

U RTU Q

rarr minus

minus minus

minus

minus

minus

minus rarr

= ∆ +sdot

= = = = ν =

= rArr =

∆ = ν minus

sdot = ν

rArr = =ν

∆ = ν sdot sdot

∆ = ν∆ = rArr =

1p

c( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

=

2p3p( )3 1 1 3

3 1 1745 MJPQ C T T

Qminus

minus

= ν minus

= 1p

d

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p

4p

( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p( )( )1 2 3 1 1 2 3 1

0 0 0 01 2 3 1

1 2 3 1 0 0

1 2 3 1

Aria3 3

225 MJ

Lp p V V

L

L p VL


minus minus minus

minus minus minus

minus minus minus

=

minus minus=

=

1p



259



d 2 WRtU

= [ ]SI1st = 3

2 b nEIne r

=+

3

3a

2

32 2 323 7221320

AB

ABAB

AB

RR RR R RR RR

R R RRR R

sdot sdot= + = +

+

sdot= =

+

3

4

c ( )0

00

02 1

1

10 grad

R R tR RR R A t

Rminus minus

= + α


rArr α =

3



a

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω

2p3p

2 31

2 3

20 301020 30

22

e

e

e

R RR RR R

R

R

=+

sdot= +

+= Ω 1p

b

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

2p3p

1

1

12 05 A2 22

e

EIr R

I

=+

= =+

1p

c

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

5p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1p

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus

1 1

1 2

1 2 1

1 1

1 1

04 A

112 V

AB B A

B A

B A

AB B A

AB

V V VV E I r V

E EI Ir r R





rArr = minus 1p


260

d


( )( )

1 2 23

1 1 1 1 23 23

2 2 2 2 23 23

I I IE I R r I R

E I R r I R

+ =

= + + sdot

= + + sdot

1p

4p

cu 2 323

2 3

23

12

8 A3

R RRR R

I

= = Ω+

rArr =

1p

Din 2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr =

1p2 2 3 3

2 3 23

2 16 A

I R I RI I I

I


primerArr = 1p



a


EIr R

=+

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus

unde 2 3

220

e

e

e

R RR R

REr RI

= + +

= Ω

rArr = minus1p

4p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p2Din 2

2 01 A

20

BC

BC

RP i

PI IR

r

= sdot

rArr = rArr =

rArr = Ω

1p

rArr r = 20 Ω 1p

b

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

3p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

=

1p

2

2

middot

middot

4 W30

=CD CD

CD

CD

P

I

P

RR I

P =

= 1p

c

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

2p

4p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr =

1p

2

232

108 J

AC AC

AC

AC

W R I tRW I t

W

= sdot sdot

= sdot sdot

rArr = 1p

d

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

4p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

1p

2max

220 92240

4

e

e

RR r

EPr

η =+

η =

=

şi eR = r 2p



261

DOPTICAtilde



2

2c

mvE = 3



3

4

d 2 2 22 1

1 1 1

x y yx xx y y

β = = rarr =

Din grafic 1 2

2

1 2

1 2

40 cm 10 cm40 cm

20 cm

x yx

x xf fx x

rArr = =

rArr =

rArr = rArr =minus

3

5 b 9

63

550 10 1 275 10 m2 2 10

275igrave m

Di ie

i

minusminus

minus


sdot=

i = 275 μm 3



a


3pcorespunde cu 1 2

05


1p1 2

05


b

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

4p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr =

1p

2

1

2

1 2

1 2

05

375 cm

25 cm

xx

xx xf

x xf

β = = minus

rArr =

=minus

rArr = 1p

c

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

4p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = =

1p

3 1 2

1

1

1

21

2

2 8 m025

s

s

s

C C CC C

Cf

Cf

C minus

= +=

=

rArr =

rArr = = 1p

d

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

4p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p

2sistem

1

12

1

2

sistem

1 125 cm

15 cm15 175 5

s

s

s ss

xx

f xxf x

f fC

x

primeβ =

sdotprime =+

= rArr =

primerArr =


1p



262


a 4p 4p

b0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


2p3p0

34 14 1566 10 92 10 6 10 Jextr

extr

L h

L minus minus minus

= ν


c

00

86

0 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot

2p

3p0

08

60 14

3 10 0326 10 326 nm92 10

C

minus

λ =ν

sdotλ = = sdot =

sdot1p

d

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

2p

5p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot

1p

( )

2max

0

0max

34 14

max 31

5max

22

2 66 10 48 1091 10

m836 10s

e

e

vh h m

h v vv

m

v

v

minus

minus

ν = ν +

minus=


sdot

= sdot 1p



263

TESTUL 10

A MECANICAtilde




a



4p


1 12

2

m v m vm vvm

=

=

1p


b


112 2

v gtda

micro= =

2p3p


c


222

2

0375m s2vad

= = 1p

3p


aa gg



d


5p

21


22


1p

( )2 2 2

1 21 2 1 2 1 2( )





264


a






b


3p20

0(1 cos ) 2 (1 cos )2



c


m= = 1p

4p




m M= =

+


ecuaţia vitezei 2

2vd

g=

micro

1p


d



m M m M= minus =

+ + 2p

4p




265





a


pV RT

mpV RTm= ν


1p

4pm pV RT

microρ = =

1p


3128 kg mρ = 1p

b


1 1 11

p V RTmp V RTm


1p

3p1 1

1

p VmRT

micro=

1p


c


2 1 2 1

22

1

p V RTp VRT

= ν

ν =

1p



1




1



d


1p

5p

1 11

1 1 1 2 2

2 2 12

1



1p

1 2 1 1 1 2 2( )p V V RT p V p Vν+ = = + 1p

1 1 2 2

1 2final

p V p VpV V

+=

+ 1p




266


a


3p 3p

b



2 22 1 2 1 2 1 2 12 1

( )( ) ( )( ) ( )2 2 2


1p

3p

2 22 1

2LkV V

=minus

1p


c


21 1 1


R R= ν rArr = =

ν ν

1p

5p2

2 2 22 2 2 2

p V kVp V RT TR R

= ν rArr = =ν ν

1p

2 22 1 2 1( )kT T T V V


ν 1p

2 22 1

3 ( )2VU C T k V V∆ = ν ∆ = minus

1p


d


1 2

1 2

1 2 1 2 1 2

100 J300 J

400 J

LU

Q U L

minus

minus

minus minus minus

=∆ =

= ∆ + =

1p

4p


23 12 3 2 2

2 2

2 3 2 3

0

ln ln 245 J

245 J

UV VL RT kVV V

L Q

minus

minus

minus minus

∆ =

= ν = = minus

= = minus

1p


1p




267





3p 3p

b


1p

4p



4 4s

ee s

r rrr r

= rArr = =1p


015er = Ω 1p

c


1 23 ( )4

E i r I R RI i

= sdot + + = sdot


142A34

neIr R R

= =+ +

1p

4p




= 1p


d2R Sl =ρ

3p4p




268


a


1 2

1 2 1

( )

1

E E I r r RE EI A

r r R

+ = + ++

= =+ +

1p

3p


1

1 1

2 2

b

b

U IRU E IrU E Ir

== minus= minus

1p

rezultat final

1

2

8 V55V25V

b

b

UUU

===

1p

b


1 2

1 2

035Ass

E EIr r R

+= =

+ +

1p


1 2

266pR RR

R R= = Ω

+ respectiv 1 2

1 2

214 App

E EIr r R

+= =

+ +1p


1p


c



1

ext s

PfP

=

1p


d


2ext p pW R I t=

1p


1 2 1 2

057p pext

total p

R I RWW E E r r R

= = =+ + + 1p


total

WW

= 1p



269

DOPTICAtilde




a


2 2

1 1

12

y xy x

β = = = 1p

3p


1 1 1 125Cf x x

= = minus = δ 1p


b


1 1

10 cmf xxf x

= =+


4p




1 40 cmfC



2 1

f xxf x

=+



c


1

025xx

β = = respectiv 22

1

133

xx

β = = 1p




d


= = 1p


1 1 1F x x

= minus 1p


1

FxxF x

=+

1p




270


a


1 0108 mmxik

= =


Did

λ= se obţine 1

1i d x dD k D

λ = =

1p

3p

Din 22 2 2

D di id D

λ= rArr λ = 1p


b


3 1 13 3 Dx id

= = λ

1p


d= = λ 1p




c


1p


1 1 2 2 11


λ 1p


1p


d


1

nλ

λ = respectiv 2

2

nλ

λ =



2 12

k Dxn d

+ λ=

1p



n dλ

= = 1p






271

TESTUL 11

A MECANICAtilde


c 2 2 2 2

2 2 2 cp m v m v E

m msdot sdot

= = =sdot sdot


2 c 2


= + sdot + 2

55 2

g tx sdot=

24

4 2g tx sdot

= 5 4 45 mx x x= minus = 3


αmicro =

α3

4 b 1N100m

Fkl

= =∆

2N200m

k = 1

2

12

kk

= 3

5 b p F t∆ = ∆ F G= m100 kgs




a




2 2 1

1 2 1 2


= =+ +

1p

Rezultă 2

m375s

a = 1p

b




23 3F T T= sdot =

2p


c


4pRezultă


sin cosmM = =



d


V = a Δt2p

3p

Obţinem m5s

v = 1p



272


a


i fp p=

21 1 2 0

1 2

m v m vu

m msdot + sdot

=+

2p

3p

Rezultăm60s

u = 1p

b


4p




Obţinem Bp =m56kgs

1p

c



4pObtinem

21 2

1 2( ) ( )

2B

Bt


Rezultă 1700 JBt

E = 1p

d


2C f

Bc G F


4pDeci 2

2sin cos2C

t Bc t t




2p




273



b 12


2 b 201204 10A

A




4 b 1 2 1 2

1 2

m m m m+= +

micro micro micro 3

1 2


+micro micro

3


3 2 11mp R T m m mV


3



a


Np V R TN


11

025 10 mm R TVp


sdotmicro 3 31

11

025 10 mm R TVp


sdotmicro

1p

4pSe obţine 1

11

AN P NnV R T

sdot= =

sdot2p

Rezultă 25

1 3

molec12 10m

n = sdot 1p

b


22

p VR T

sdotν =

sdot2p

3p


c


11

m R TpV

sdot= sdot

micro 1p

4p


11

pR T

sdotmicroρ =

sdot

22

2

pR T

sdotmicroρ =

sdot1p

Se obţine 1 1 2

2 2 1

p Tp T

ρ sdot=

ρ sdot1p

Rezultă 1

2

113

ρ=

ρ 1p


274

d


2 2 2 2

p V R Tp V R T


SF 1 1 1

2 2 2

p V R T

p V R T



1p



1 1 2 2

1 2

1 2

1 2

p V p VT Tp V V

T T

sdot sdot+

=+

1p




a


1

ln ln 2vL R T R Tv


34 3 4 3 3

1( )2


141 1 1

4

1ln ln2

vL R T R Tv


4p





3p

34 12465 JU∆ = minus 1p

c


4pObţinem

1

4 3 14





d


p

LQ

η = 1p


2p

Obţinem 505η = 1p



275



c 2

2 UW R I t tR


= 3

2 c 3


4 c eSC

e

EIr

= 65er = Ω 24 V

5eE = 4AeSC

e

EIr

= = 0 VU = 3

5b

2 2U UP RR P

= rArr =

2

10 mS UlP

sdot= =

ρ sdot3



a


e e

EIR r

rArr =+

1p


1 2

1 2

1 2

30 V1 1e

E Er rE

r r

+= =

+ 1

2err = = Ω 1p


1 2

4s se

s s

R RRR R

sdot= = Ω

+ 1p

Obţinem 6 AI = 1p

b



Obţinem 1

2

1II

rArr = 1p

c



10 A3

e

e

EIR r

prime = =+

2p






276


a




b




c

Randamentul va fi e

e

RR r

η =+

1p

4p




2p

Rezultă 886η = 1p

d


4p


12

21

8022 3

2

e

RR RR RR

sdot= + =

+

30 A43e

EIR r

= =+

2p




277

DOPTICAtilde




a


1 1 1 1 1 12x x f x x f


2p

4pRezultă 2

1 32

cf x

minus= =

minus1p

Obţinem 2

3 2015 10

c minus= = δsdot

1p

b

2 1

1 1 1x x f

minus = 1p

4pobţinem 2

1 21

1 22

x x xx



c


2 1

1 1 1x x f

minus = 1p



1

x fxx f

primeprime =

prime +1p

Rezultă 275 5 15 cm75 5


1p




D a ii Da

λ sdot= rArr =

λ2p

3p

Rezultă 3 3

9

2 10 1 10 5 m400 10

Dminus minus

minus

sdot sdot sdot= =

sdot1p

b


4pRezultă

axD

δ = 2p

Obţinem 3 3

72 10 16 10 8 10 m4


δ = = sdot 1p


278

c



4pObţinem

9

max 3

400 10 442 10

xminus

minus

sdot sdot= sdot

sdot1p


d


naλprime =

1p

4pDin Dia

λ= rezultă ii

n= 2p

Rezultă 3

3 31 10 3 10 075 10 m4 43

iminus




279

TESTUL 12

A MECANICAtilde




a


00

( )

05

kt M m m akta t

M m m

= + +

= =+ +

2p


2

2 2

m8 02s

Mg ma

a

micro =

= micro =

2p

b


0 1( )Mg ma

kt Mg m m amicro =

minus micro = +

2p

3p


m( 2)s

kt Mga tm m


+ 1p

c


4p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + =

2p

( )

1 2

0

0

4

t t a akt Mg g

m mgt m m M sk

= =minusmicro

= micro+micro

= + + = 1p

d


4p


( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus =

1p( ) ( ) ( )

( )

( ) ( ) ( )

0

2

0

2 2 2m2 1s

2 2 2 195 N

F T fm a

a

T F fm a

minus =

=

= minus = 1p



a


30 5 10 J



Deci 0cE∆ = 1p


280

b




3p4p


c



4p


d


1p


2p




281



5

b

1 11

1

abs ced

abs

Q QQ

minusη = respectiv

2 22

2

abs ced

abs

Q QQ

minusη =


1 2

abs abs

L LQ Q

η = η = iar

2 1abs cedQ Q=


3



a


1p

3p0p VmRT

micro=

1p


b


4p


A

Np V RTN

= 1p


A

N RTpVN+ α

= 1p

Rezultă 0

1 14pp

= + α = 1p

c


kmolmicro = şi


kmolmicro =

1p

4p

012i ii

A i

N m iN

ν = = =micro

1p

1 2

1 2

m m m= +

micro micro micro

1p

0

1 2 1 2

1 2


A

A A

Nm N

m m N NN N

micromicro


micro micro

1p


282

d


4p


= ν ν = 1p


1 2Q Q Q= + 1 1 1 2 2V VQ C T C T= ν ∆ + ν ∆1p

( 5) 450 J2

pV TQT

α + ∆= sdot =

1p



a


2 1

p pT T T nTT T

= = = 1p

3p


3 2

V V T nkTT T

= = 1p


1 4

V V T kTT T

= =

1p

b


1p

4p


= minus minus


= + minus 1p



1p


283

c


Ccald

T T nkT nkT nk


Cη gt η adică 1 2 2 2 25 2 3

nk nk n knk nk n



adevărat

1p

4p



minusη = =



minusη = =

1p




minus minus


5n n DA q e d

nminus

lt gt minus




minus minus minus

1p



krarrinfinrarrinfin

η = lt

1p

d


cedQL

ϕ =

1p

4p


1 5 2 32 2 2 2

cedQ nk nL nk n k

minus minusϕ = = =

η minus minus +


absQL

ε =

1p


1 3 2 512 2 2 2

nk knk n k

+ minusε = minus =

η minus minus +



1p







284



2




ERIR r r x xr

=+ + +


ERIr R r

= =+

3

3 c 3

4




2 2

U UI R R R= =


R UUab I= sdot =


abR

U UIR R


2

4RUP RI

R= =

3

5


0 i D

i D

U U EI U U E

r R

le += minus minus +


1 5 V2

i

e ii

UU U U

le= +

gt


3



a


n

ki

k k k

I I

U E I rU IR

=

=

minus = minus =

sum


k k

E RI Ir r

= minus

2p

4p


1

11

nk

i kn

i k

ErI

Rr

=

=

=+

sum

sum și respectiv 1

1

11

nk

i kn

i k

ERrU IR

Rr

=

=

= =+

sum

sum2p

b


1 1

lim1 11

n nk k

i ik kgol n nR

i ik k

E ERr rU

Rr r

= =

rarrinfin

= =

= =+

sum sum

sum sum 1p

4prespectiv 1

1

lim 011

nk

i kgol nR

i k

ErI

Rr

=

rarrinfin

=

= =+

sum

sum 1p


0 0n n

ksc sc k sc

i ik

ER I I Ur= =



285

c


1

1

1

1

11

nk

i kn

i k echivalent

echivalentn

i k

Er

r EIR rR

r

=

=

=

= =++

sum

sum

sum

2p

4p

adică 1

1

1

nk

i kechivalent n

i k

ErE

r

=

=

=sum

sum 1p

respectiv

1

11echivalent n

i k

r

r=

=

sum

1p

d

echivalentE E= 1p

3pechivalent

rrn

= 1p

EI rRn

=+

1p



a


0 2

A B

A

B A

I I IE Ir RI

RI RI

= + = + = minus

1p

3pRezultă

3 22

3 23

3 2

B

A

EIr R

EIr R

EIr R

= + = +

= +

2p


286

b

( 2 )3 2CD B A

E R xU RI xIr R

minus= minus =


5p

Pentru max

min

03 2

3 2

2

CD

CD

CD

ERx Ur R

ERx R Ur R

RU o x

= = + = = minus +

= =

1p

Graficul

1p


CDdU Edx r R

α = = minus+

1p


tg CDU

R = minus = Ωα


2 mrE Uα

= minus + = 1p

c


2

1 2



1p

3p

Rezultă

2 2

(2 )E x RIRx R x

minus=

+ minus1p

Pentru

0 2A

2A

02

Ex IR

Ex R IR

Rx I


= = 2

max max 20 WP RI= =

1p



=


xEI cu xy nx rRn

= =+

1p

4p


= =+

1p



R= 1p




287

DOPTICAtilde




a


2 1

1 1n nx x R

minusminus = 1p

5p



1 1n nx x R

minusminus =

minus 1p

Se deduce

12

1( 1)Rxx

nR n x=

+ minus 1p


b


1 1

y nxy x

= 15p

4pLa a doua trecere

2 2 1 1

y xy x

= 15p

Dar 1 2y y= deci

2 03 cmy = minus 1p

c



1 1 2x x R

+ = Rezultă 12

1

2 214 cm2

xxx R

= =minus

1p


21

007 cmxyx

minus= = 1p


d


12 1

1

5 cm 7cm2Rxx R xx R


3p


2 x

Ryy yx R


1p




288


a



2 0 2 2 2 2( ) ( ) ( )

nn a l b n r e n e


minus 1p


b


D l∆

= 1p


Rezultă ( )2 2 2 20 2 2 1 1( ) ( ) ( ) ( )



c


4pRezultă

2Dinlλ

= 2p

d



0 1 1n n= = 1p



289

Cuprins




Documents

TESTE DE FIZICÃ PENTRU BACALAUREAT