JEE MAIN EXAMINATION - 2015 QUESTION WITH SOLUTION PAPER CODE - A€¦ · (Page # 2) JEE MAIN Examination(2015) (Code - A) Rank Booster Test Series [JEE Advanced] 12th & 13th Students

JEE MAIN EXAMINATION - 2015

QUESTION WITH SOLUTION

PAPER CODE - A

[HINDI VERSION]

Fastest Growing Institute of Kota (Raj.)

FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | CBSE | SAT | NTSE | OLYMPIADS

JEE MAIN Examination(2015) (Code - A)(Page # 2)

Rank Booster Test Series [JEE Advanced]12th & 13th Students Start from 6 April. 2015

[PHYSICS]1. nks i RFkj 240 m Å¡ph , d pV~Vku ds fdukj s l s , d l kFk

Åi j dh vksj Øe' k% 10 m/s o 40 m/s dh i zkjafHkd pkyl s Qsads t kr s gSaaA i zFke i RFkj ds l ki s{k f} r h; i RFkj dhl ki sf{kd fLFkfr dk l e; i fj or Zu dks fuEu esa l s fdl xzkQesa mfpr cr k; k x; k gS \(ekuk i RFkj /kj kr y i j Vdj kus ds ckn mNyr k ugha gS r Fkkok; q i zfr j ks/k ux.; ] g = 10 m/s2 gSA)(fp=k O; ofLFkr gS r Fkk i Sekuk ughas fy; k gSA)

(1)

8t 12t(s)

(y –y ) m2 1

240

(2)

8 12t(s)

(y –y ) m2 1

240

(3)

8 12t(s)

(y –y ) m2 1

240

(4)

12 t(s)

(y –y ) m2 1240

Sol. 3i zFke i RFkj

0 t 8secvr = 40 – 10= 30 m/sar = 0

t=2sec

t=6sec

sr = vr × t = 30 × 8 = 240 m

(y –y )m2 1

240

8t(sec)

`

t=8sec

t=1sec

8 sec < t 12 secvr i fj ek.k esa c<+r k gS r Fkk l ki sf{kd Roj .k g uhps dh vksj

(y –y )m2 1

240

8 12t(sec)

2. , d l j y yksyd dk nksyudky LT 2g gSA L dk

eki k x; k eku 20.0 cm gS] 1 mm ' kq) r k gS r Fkk yksyd

ds 100 nksyuksa ds fy, 1s fo; kst u {ker k dh , d ?kM+h l s

90 s eas i zkIr gksr s gSaA g dh ' kq) r k Kkr gksxh :(1) 1% (2) 3%(3) 2% (4) 5%

Sol. 2

dT 1 dL 1 dgT 2 L 2 g

90 1100 100

1 dg 1 dL dT2 g 2 L T

1 0.1 1/100 1 12 20 90 /100 400 90

1 dg 1 12 g 400 90

dg 490 2g 400 90

490200 90 = 0.02722

3%2.72%100g

dg

JEE MAIN Examination(2015) (Code - A) (Page # 3)


3.

A BF

fn; s x; s fp=k esa nks xqVds A vkSj B ft uds Hkkj Øe' k%

20 N o 100 N gSA ; s fp=kkuql kj , d cy F ds } kj k nhokj

ds fo: ) nck; s x; s gSaA ; fn xqVds ds chp ?k"kZ.k xq.kakd 0.1

gS r Fkk xqVds B o nhokj ds chp 0.15 gS] xqVds B i j nhokj

ds vkj ksfi r ?k"kZ.k cy gksxk :(1) 100 N (2) 150 N(3) 120 N (4) 80 N

Sol. 3

A BF

ekuk fd fudk; l kE; voLFkk esa gS] dqy xq: Rokd"kZ.k cynhokj l s ?k"kZ.k cy ds } kj k l Ur qfyr gksxkA?k"kZ.k cy = 120 N

4. nzO; eku m dk , d d.k pky 2l s x fn' kk esa xfr ' khy gSt ks nwl j s d.k ft l dk nzO; eku 2m dh pky l s y fn' kk esa

xfr ' khy gS l s Vdj kr k gSA ; fn VDdj i w.kZr % vi zR; kLFk gS]VDdj ds nkSj ku Åt kZ esa i zfr ' kr gkfu yxHkx gS %

(1) 44% (2) 62%(3) 56% (4) 50%

Sol. 3

m 2v

v

2m

VDdj ds i gys

xP

= 2mv i

yP

= 2mv j

V

3m

VDdj ds ckn

xP

= 3mv'cos

yP

= 3 v' sin

l aosx l aj {k.k l s %

{kSfr t esa 2mv = 3mv' cos ...(i)

m/okZ/kj esa 2mv = 3mv' sin ...(ii)

(i) vkSj (ii) l s tan = 1; = 45°

vfUr e pky v' = 3

v22

i zkj fEHkd K.E. ; 1/2 (m) (2v)2 + 1/2 (2m) (v)2 = 3mv2

vfUr e K.E. ; 1/2 (3m)

2

3v22

= 4/3 mv2

% deh i

fi

)KE()KE()KE(

× 100%

= 55.55 ~ 56%

5. , d Bksl , d l eku ' kadq ds ' kh"kZ ds nzO; eku dsUnz dh nwj h

z0 gSA ; fn bl ds vk/kkj dh f=kT; k R gS r Fkk bl dh Å¡pkbZ

h gS] r c z0 cj kcj gS &

(1) 5h8 (2)

3h4

(3) 5h8

(4) 23h

8R

Sol. 2

03hz4 z0

h

(theory l s)



Sol. 2Bksl xksyk ft l dk nzO; eku M, f=kT; k R gSAxksyh; Hkkx gVk; k t kr k gS ft l dh f=kT; k R/2, bl fy,bl dk nzO; eku M/8

xqgk ds dsUnz i j foHko= Vsolid sphere + Vremoved part

CR/2

=

3

GM2R

22 R3R

2 + )2/R(2)8/M(G3

= RGM

8. , d yksyd , d l eku r kj ft l dk vuqi zLFk dkV {ks=kQy A

dk cuk gS ft l dk vkor Zdky T gSA t c , d vfr fj Dr

nzO; eku M bl xsan esa feyk; k t kr k gSA r ks vkor Zdky

cnydj TM gSA ; fn r kj ds i nkFkZ dk ; ax xq.kkad Y gS] r c

1/Y cj kcj gS & (g = xq: Roh; Roj .k)

(1)

2MT A1T Mg (2)

2

M

T A1T Mg

(3)

2MT A1T Mg (4)

2MT Mg1T A

Sol. 1

T = 2 g

; TM = 2 g'

=

Mg/ A/

' =

MgA =

' = 1 +

MgA

Also:

TTM =

' TM =

1/2MgT 1A

2

2M

TT

= 1 + AMg

1

TT

2

2M

= MgA

1

= A

Mg

1TT 2

M

6. nzO; eku M o f=kT; k R ds , d Bksl xksys l s vf/kdr e l EHko

vk; r u dk ?ku dkVk x; k gSA ?ku ds dsUnz o bl ds , d

Qyd ds yEcor ~ xqt j us okyh v{k ds i fj r % ?ku dk t M+Ro

vk?kw.kZ gksxk %

(1)

2MR32 2

(2)

24MR3 3

(3)

24MR9 3 (4)

2MR16 2

Sol. 3

I = 6

Mx2

fdukj s dh yEckbZ : (x)

2R = 3 x (fod.kZ)

x = 3R2

vc]?ku dk nzO; eku :

m =

3R34

M

3

3R2

R

R

3R4M3

33R8 3

m = 3

M2

I = 31

3M2

3R4 2

= 39

MR4 2

7. fp=kkuql kj nzO; eku M o f=kT; k R dk , d Bksl xksys l sf=kT; k R/2 dk xksyh; Hkkx gVk; k x; k gSA xq#Roh; foHkoV = 0, r = i j ekus] cuh xqgk ds dsUnz i j foHko gksxk(G = xq#Roh; fu; r kad)

(1) R3GM2

(2) RGM

(3) R2

GM(4)

RGM2



9. r ki T i j f=kT; k R ds , d xksyh; dks' k i j fopkj dhft , Abl ds vUnj d f".kdk fofdj .k QksVksuksa dh vkn' kZ xSl dsvuql kj eku l dr s gSa] ft l dh vkUr fj d Åt kZ i zfr , dkad

vk; r u 4Uu TV

r Fkk nkc

1 Up3 V gSA ; fn

dks' k vc , d : ) ks"e i zl kj ds vUrxZr gS r ks T o R ds chpl EcU/k gksxk :

(1) RT e (2) 3

1TR

(3) 1TR

(4) 3RT eSol. 3

4Uu TV

1 UP3 V

: ) ks"e i zl kjTV–1 = K

4TV C

14

3 14

43

(r = )

4TV C

13TV C

1334T R C

3 1T

R

10. fu; e Å"ek {ker k 1 J/°C dh , d Bksl oLr q nks r j g ds

fj t oZj l s l Ei dZ esa j [ kus ds } kj k Åf"er gSA

(i) 2 fj t oZj ds l kFk mÙkj ksÙkj l Ei dZ esa j [ kus i j r kfd

i zR; sd i j Å"ek dh l eku ek=kk l s l IykbZ dj r k gSA

(ii) 8 fj t oZj ds l kFk mÙkj ksÙkj l Ei dZ esa j [ kus i j r kfd

i zR; sd fj t oZj Å"ek dh l eku ek=kk l IykbZ dj r k gSA

nksukas fLFkfr ; ksa esa oLr q i zkj afHkd r ki 100°C l s vfUr e r ki

200°C r d yk; k t kr k gSA nksuksa fLFkfr ; ksa esa oLr q dh

, UVªkWi h i fj or Zu Øe' k% gS %(1) ln2, 2ln2 (2) ln2, ln2(3) ln2, 4ln2 (4) 2ln2, 8ln2

Sol. 2

(i) 150 200

1100 150

dQ dT dTs ms msT T T

150 200ln ln100 150

3 4ln ln2 3

1s ln2

(ii) 112.5 125

2100 112.5

dQ dQ dQs ......T T T

112.5 125ln ln .....100 112.5

9 10 16ln ln ln8 9 15

16ln8 = ln2

11. , d foyfxr cUn psEcj esa , d vkn' kZ xSl i j fopkj dhft , AxSl , d : ) ks"e i zl kj ds vUr xZr ] v.kqvksa ds chp VDdj dkvkSl r pky Vq ds vuql kj c<+r k gS] t gk¡ V xSl dk vk; r u

gSA q dk eku gksxk

V

P

CC

:

(1) 6

53 (2)

21

(3) 2

1(4)

653

Sol. 3ek/; eqDr i Fk

2

12 d n

n v.kqvk sadh l a[ ; k

vk; r u

avg.v T T.V-1 = C

avg.

Vtv T v vk; r u gSA

1

2

r 1

V VC

v

1

q 2v v 1q2



12. , d l j y yksyd ds fy, ] , d xzkQ bl ds foLFkki u ds l ki s{kbl dh xfr t Åt kZ (KE) o fLFkfr t Åt kZ ds (PE) ds chpvkj sf[ kr gSA fuEu esa l s dkSul k l gh gS\ (xzkQ O; ofLFkr gSr Fkk i Sekus l s ugha cus)

(1) d

KEE

PE

(2) d

PEE

KE

(3) d

KEE

PE

(4)

E

PE

KE

Sol. 2 KEmax ek/; fLFkfr i j

U=0KE=0

pj e fLFkfr pj e fLFkfrMPV=max=0

U=0KE=0

PEmin ek/; fLFkfr i j

13. , d j syxkM+h , d l h/ks i Fk i j 20 ms–1 dh pky l s xfr ' khygSA ; g 1000 Hz dh vkofÙk i j l hVh ct k j gh gSA i Fk dsfudV [ kM+s , d O; fDr ds } kj k ¼t c j syxkM+h ml l s xqt j r hgSA½ l quh vkofÙk esa i zfr ' kr i fj or Zu gksxk &(/ofu dh pky = 320 ms–1)(1) 18% (2) 12%(3) 6% (4) 24%

Sol. 2

f1 = 1000 320

300 20

= 1066 Hz

f2 = 1000 320

300 20

= 941 Hz

Change is ~ 12%

14. , d yEch csyukdkj dks' k esa Åi j h v) Z esa /kukRed i "Bvkos' k o fupyh v) Z esa _ .kkRed i "B vkos' k – gSAcsyu ds pkj ksa vksj fo| qr {ks=k j s[ kk, ¡ gSA fn; s x; s fp=k esafn[ kkbZ gS : (fp=k esa O; ofLFkr gS r Fkk i Sekuk ugha fy; k gSA)

(1) ++

+++++++

–––––––––

– (2) ++

+++++++

––––––––––

(3)

++

+++++++

––––––––––

(4)

++

+++++++

–––––––––

–

Sol. 1Tangent to the electrical field lines will giveus the direction at a given point.

15. f=kT; k R dk , d l eku : i l s vkosf' kr Bksl xksys dk foHko

V0 (ds l ki s{k eki k x; k gSA) bl dh l r g i j gSA bl xksys

ds fy, foHko 2V3 0 ,

4V5 0 ,

4V3 0 r Fkk

4V0 ds l kFk

l efoHko l r gas gS dh f=kT; k Øe' k% R1, R2, R3 r Fkk R4 gS]

r c %

(1) R1 = 0 r Fkk R2 > (R4 – R3)

(2) R1 0 r Fkk (R2 – R1) > (R4 – R3)

(3) R1 = 0 r Fkk R2 < (R4 – R3)

(4) 2R < R4



Sol. 3 & 4

3V /20

V0

r=R

0 0 0 01 2 3 4

3V 5V 3V VR ; R ; R ; R

2 4 4 4

2 23

KQr R V 3R r2R

01

3Vv , R 0

2

2 2023

5V KQ 3R R4 2R

2RR2

r > R

0

3

3V KQ4 R

30

4KQ KQ R RR3V 3 KQ 3

0

4

V KQ4 R

40

4KQ 4KQR R 4RV KQ

r qyuk dj us i j i zkIr dj r s gS %(1) ; k (2)

16. fn; s x; s i fj i Fk esa] 2µF l a/kkfj =k i j vkos' k Q2 ,C dsvuql kj cnyr k gSA 1µF l s 3µF r d i fj ofr Zr gSA 'C' dsQyu ds : i esa Q2 mfpr fn; k x; k gS % (fp=k O; ofLFkrgS r Fkk i Sekuk ugha fy; k gSA)

C

1µF

2µFE

(1) Q2

1µF 3µFC

Charge

(2) Q2

1µF 3µFC

Charge

(3) Q2

1µF 3µFC

Charge

(4) Q2

1µF 3µFC

Charge

Sol. 2

3Cq EC 3 q = CV

q C

23C 2q E

C 3 3

22Cq E

C 3

22Cq E

31C

q = CV

C q2 ; fn C , q = fu; r eku

17. t c 5V foHkokUr j 0.1 m yEckbZ ds r kj i j vkj ksfi r fd; kx; k gS] r c bysDVªkWuksa dh vuqxeu pky 2.5×10–4 ms–1

gSA ; fn r kj esa bysDVªkWu ?kuRo 8×1028 m–3 gS] r ks i nkFkZdh i zfr j ks/kdr k yxHkx gS &(1) 1.6 × 10–6 m (2) 1.6 × 10–7 m(3) 1.6 × 10–8 m (4) 1.6 × 10–5 m

Sol. 4i = neAVd

RV

= neAVd { lRA

}

V A = neAVd

50.1 = 8 × 1028 × 1.6 × 10–19 × 2.5 × 10–4

= 1.56 × 10–5 m ~ 1.6 × 10–5 m



18. i znf' kZr i fj i Fk esa] 1i zfr j ks/kd esa /kkj k gksxh &

(1) 1.3 A, P l s Q(2) 0.13 A, P l s Q(3) 0.13 A, Q l s P(4) 0A

Sol. 36V P0

9V

Qx

59x

+ 3

6x +

1x

= 0

15

x1530x527x3 = 0

x = 233

A

from Q to P

19. fHkUu&fHkUu f=kT; kvksa dh nks l ek{kh; i fj ukfydkvksa esa /kkj k I

l eku fn' kk esa i zokfgr gSA ekuk 1F

ckgjh ds dkj .k vkUr fj d

i fj ukfydk i j pqEcdh; cy gS rFkk 2F

vkUr fj d ds dkj .k

ckgj h i fj ukfydk i j pqEcdh; cy gks] r c %

(1) 1F

= 2F

= 0

(2) 1F

f=kT; h; : i l s ckgj dh vksj gS r Fkk 2F

= 0

(3) 1F

f=kT; h; : i l s vUnj dh vksj gS r Fkk 2F

= 0

(4) 1F

f=kT; h; : i l s vUnj dh vksj rFkk 2F

f=kT; h; : i

l s ckgj dh vksjSol. 1

vuqi zLFk dkV n' ; %

i2i1

(nksuksa i fj ukfydk, ¡ i zd fr esa vkn' kZ gSA )nksukas r kj i j Li j vkdf"kZr gksaxs] fdUr q dqy cy i zR; sd r kji j ' kwU; gksxkA

fl ) kUr %nks /kkj kokgh ?kVd i j Li j vkdf"kZr gSA ; fn /kkj k dh fn' kkl eku gSA

F.B.D

0F1

0F2

20. nks yEcs /kkj kokgh i r ys r kj ] nksuksa esa /kkj k I gS] t ks yEckbZ L

ds vpkyd /kkxsa ds } kj k LFkkfi r gS rFkk fp=kkuql kj l kE; koLFkk

esa gS] /kkxs Å/okZ/kj l s '' dks.k cukr s gSA ; fn r kj ksa dk

nzO; eku i zfr , dkad yEckbZ gS] r c I dk eku gksxk &

(g = xq: Roh; Roj .k)

(1)

0

gL2 tan

(2)

0

gL2sincos

L

II

(3)

0

gLsincos

(4)

0

gL tan

Sol. 2

xid

LL

d2i20

cy i zfr , dakd yEckbZ fy; k x; k gS

tan =

.g.d2i20



i2 =

cos.sin.g.

0 (2) d [d = 2L sin]

i = 2 sin

cosLg

0

21. Hkqt k, ¡ 10 cm o 5 cm dk , d vk; r h; ywi esa 12 A dh

/kkj k I i zokfgr gSA fp=kkuql kj fofHkUu foU; kl ksa esa fLFkr gSA

(a)

x

I

I

I

I

z

y

B

(b)

x II

II

z

y

B

(c)

x

II

I

I

z

y

B

(d)

x II

II

z

y

B

; fn ; gk¡ i j , d l eku pqEcdh; {ks=k 0.3 T /kukRed zfn' kk esa] foU; kl ksa esa ywi (i) LFkk; h l kE; koLFkk r Fkk (ii)vLFkk; h l kE; koLFkk esa gksxk \(1) Øe' k% (a) r Fkk (b)(2) Øe' k% (b) r Fkk (c)(3) Øe' k% (b) r Fkk (d)(4) Øe' k% (a) r Fkk (c)

Sol. 3l kE; voLFkk ds fy,

= 0

= MB sin n; fn, sin = 0;

= 0

; fn M

o B

ds chp dks.k ' kwU; gS r c LFkk; h l kE; voLFkk

; fn M

o B ds chp dks.k gS\ r c vLFkk; h l kE; voLFkk

22. , d i zsj d Ro (L=0.03H) r Fkk , d i zfr j ks/kd(R=0.15 k) fp=kkuql kj i fj i Fk esa 15V EMF dh cSVªhdh Js.khØe esa l a; ksft r gSA dqat h K1 yEcs l e; ds fy,yxk; h t kr h gSA r c t = 0 i j , K1 [ kqyh gS r Fkk dqat h K2

l kFk&l kFk yxh gSA t=1 ms i j ] i fj i Fk esa /kkj k gksxh &

5(e 150)

(1) 6.7 mA (2) 67 mA(3) 100 mA (4) 0.67 mA

Sol. 4nh xbZ fLFkfr ds vuql kj :

i0 = RV

= 31015.015

= 0.1A

i = i0 LRt

e

= 0.1 × 03.0101015.0 33

e

= 0.1 × e–5 = 150

1.0 = 0.67 mA

23. , d yky LED bl ds pkj ksa vksj , d l eku : i l s 0.1watt dk i zdk' k mRl ft Zr dj r h gSA Mk; ksM l s 1 m dh nwjhi j i zdk' k ds fo| qr {ks=k dk vk; ke gksxk &(1) 5.48 V/m (2) 2.45 V/m(3) 1.73 V/m (4) 7.75 V/m

Sol. 2

r hozr k 20

P 1 E CA 2

202

P 1 E C24 R

20

2PE4 R C

9

8

2 0.1 9 10E1 1 3 10

9

8

1.8 10 18E33 10

E 6 2.45 V /m

24. , d o.khZ; i zdk' k dks.k A dh , d dkap fi zTe i j vki fr rgSA ; fn fi zTe ds i nkFkZ dk vi or Zukad gS] , d fdj .kQyd AB i j dks.k i j vki fr r gS t ks i z; qDr fi zTe dsQyd AC l s i kj xfer gksxh %

A

B C

(1) > sin–1 1 1sin A sin

(2) < cos–1

1sinAsin 1

(3) > cos–1

1sinAsin 1

(4) > sin–1 1 1sin A sin



Sol. 1r2 < cr2 < sin–1 (1/)

A

r1 r2

sin r2 < 1/sin = sin r1r1 = sin–1 (sin/)sin (A – r1) < 1/

sin

sinsinA 1

< 1

A – sin–1

sin

< sin–1

1

A – sin–1

1

< sin–1

sin

m1sinAsin 1

< sin

sin–1

m1sinAsin 1

<

25. xehZ dh j kr ] ok; q dk vi or Zukad /kj kr y ds fudV U; wur e

gS r Fkk /kj kr y l s Å¡pkbZ ds l kFk c<+r k gSA t c , d i zdk' k

i qUt {kSfr t : i l s gS] gkbxsu ds fl ) kUr l fEefyr gSA bl ds

xeu dj us i j ] i zdk' k i qUt %

(1) uhps dh vksj eqM+r h gSA

(2) fcuk fdl h fo{ksi .k ds {kSfr t t kr h gSA

(3) l dM+h gks t kr h gSA

(4) Åi j dh vksj eqM+r h gSASol. 4

Åi j dh vksj eqM+r h gSA

26. ekuk O; fDr dh i qr yh dh f=kT; k 0.25 cm r Fkk 25 cm

nwj ns[ kus ds fy, vko' ; d gS] nks oLr qvksa ds chp U; wur e

vyxko nwj h D; k gksxh] r kfd O; fDr dh vka[ k 500 nm

r j axnS/; Z fo; ksft r gS &

(1) 100 µm (2) 30 µm(3) 1 µm (4) 300 µm

Sol. 2 Dy 1.22d

9 2

2

500 10 25 102 0.25 10

y 30 m

27. , d bysDVªkWu , d gkbMªkst u t Sl s i j ek.kq@vk; u dh mÙksft rvoLFkk l s /kj kr y voLFkk r d l aØe.k dj r k gS %(1) bl dh xfr t Åt kZ c<+r h gS fdUrq fLFkfr t Åt kZ o dqyÅt kZ ?kVr h gSA(2) xfr t Åt kZ o dqy Åt kZ ?kVr h gS fdUr q fLFkfr t Åt kZc<+r h gSA(3) xfr t Åt kZ ?kVrh gS] fLFkfr t Åt kZ c<+rh gS fdUrq dqyÅt kZ l eku j gr h gSA(4) xfr t Åt kZ] fLFkfr t Åt kZ r Fkk dqy Åt kZ ?kVr h gSA

Sol. 1

ETotal = –13.6 eV 2

2

nZ

KE = |ETotal|

PE = 2 EtotalD; kasfd n ?kVr k gS] dqy Åt kZ ?kVr h gS] fLFkfr t Åt kZ ?kVr hgS r Fkk xfr t Åt kZ c<+r h gSA

28. l wph-I (ewyHkwr i z; ksx) dks l wph –II (bl ds l UnHkZ esa) l sfeykb, r Fkk nh xbZ l wph l s j [ kh fodYi pqfu, &List–I List–II

(A) Ýasad–gVZt + (i) i zdk' k dh d.kh; i zd frdk i z; ksx

(B) i zdk' k-fo| qr i z; ksx (ii) i j ek.kq ds for fj r Åt kZ Lr j

(C) Msfol u-t eZj i z; ksx (iii) bysDVªkWu dh r j ax i zd fr(iv) i j ek.kq dh l aj puk

(1) A i ; B iv ; C iii(2) A iv ; B iii ; C ii(3) A ii ; B i ; C iii(4) A ii ; B iv ; C iii

Sol. 3i zdk' k fo| qr i z; ksx i zdk' k dh d.kh; i zdfr l s l EcfU/kr gSA

29. 5 kHZ vkofÙk dk , d fl Xuy] 2 MHz vkofÙk dh okgd

r j ax i j vk; ke eksMwfyr gSA i fj .kkeh fl Xuy dh vkofÙk; k¡

gS@gSa %

(1) dsoy 2 MHz(2) 2000 kHz r Fkk 1995 kHz(3) 2005 kHz, 2000 kHz r Fkk 1995 kHz(4) 2005 kHz r Fkk 1995 kHz



Sol. 3vkofÙk; k¡ gSFC, FC ± FS

30. , d LCR i fj i Fk esa voefUnr yksyd ds r qY; gSA , d LCRi fj i Fk esa l a/kkfj =k Q0 l s vkosf' kr gSA r c fp=kkuql kj L oR l s l a; ksft r gS :

R L

C; fn , d Nk=k L ds L1 o L2 (L1 > L2) ds nks fHkUu&fHkUuekukas ds fy, l e; t1 ds l kFk l a/kkfj =k i j vf/kdr e vkos' k(Q2

max) ds oxZ ds xzkQ vkj sf[ kr dj r k gS] r c fuEu eas l sdkSul k xzkQ l gh gS \ (vkj s[ k O; ofLFkr gS r Fkk i Sekuk ughafy; k gSA)

(1)

Q2Max

t

L1

L2 (2)

Q2Max

t

(For both L and L )1 2Q0

(3)

Q2Max

t

L1

L2 (4)

Q2Max

t

L1

L2

Sol. 1

KVL

R L

C

dI qIR L 0dt C

2

2

d q dq qL Rdt Cdt

voefUnr nksyu dh l ehdj .k ds l kFk r qyuk dj us i j

2

2

d y dyd kydtdt

vk; ke dh l ehdj .k bty Ae

t gk¡ Rb

2m 2L

Rt2L

max 0q q e

Rt2 2 Lmax 0q q e

l e; dky = RL

pwafd L1 > L2

1 < 2 vr % l gh mÙkj 3 gSAoSdfYi d gy %

Qmax dk eku ?kVr k gS] D; ksafd i zfr j ks/kd esa Åt kZ mRl ft Zrgksr h gSA D; ksasfd i zsjdRo dk eku c<+rk gS fuj kos' ku dh {kerki j ; k vkos' k c<+us i j A bl fy, Å"ek mRl t Zu l e; ?kVr kgSA vr %l gh xzkQ 3 gSA



[CHEMISTRY]31. O; kol kf; d j sft au dk vkf.od l w=k C8H7SO3Na (vkf.od

Hkkj 206) gSa t ks fd i kuh dh l kf¶Vfuax ¼i kuh dh dBksj r k

gVkuk½ esa vk; uksa ds i fj or Zu ds fy, mi ; ksx esa vkr k gSaA

jsft ua ds } kj k Ca2+ vk; uksa dk vf/kdre eku D; k gksxk t c

j sft ua dks eksy@xzke esa cr k; k x; k gSa\

(1) 1031

(2) 2061

(3) 3092

(4) 4121

Sol. (4)2C8H7SO3Na + Ca7+ (C8H7SO3)2Ca2 eksy 1 eksy2 × 206 xzke Ca2+ dk 1 eksy ysr k gSA

1 xzke ysxk = 4121

Ca2+ ds eksy

32. dk; dsfUnzr ?kuh; t kyd esa l ksfM; e /kkrq fØLVyhdr gksrhgSa ft l dh , dd l sy dh yEckbZ 4.29 Å gSaA l ksfM; ei j ek.kq dh f=kT; k yxHkx gSa%(1) 1.86 Å (2) 3.22 Å(3) 5.72 Å (4) 0.93 Å

Sol. (1)

a3 = 4r

r =4

29.4732.1 = 1.86Å

33. gkbMªkst u dh l EHkkfor mÙksft r voLFkk dh Åt kZ fuEu esa l sdkSul h gSa\(1) +13.6 eV (2) –6.8 eV(3) –3.4 eV (4) +6.8 eV

Sol. (3)

2

2

nz6.13– gkbMªkst u ds fy, : z = 1

2n6.13–

l EHko gSA –13.6, –3.4, –1.5 etc.34. var j vkf.od vkd"kZ.k] i zfrd"kZ.k t ks fd O; qRde ?ku dh nwj h

i j fuHkZj dj r k gSa v.kqvksa ds e/; gSa&

(1) vk; u&vk; u cy

(2) vk; u&f} /kqzo vkd"kZ.k

(3) yanu cy

(4) gkbMªkst u ca/kSol. (4)

gkbMªkst u cU/k , d f} /kzqo & f} /kzqo i zHkko gSA

35. fuEufyf[ kr vfHkfØ; k 298 K i j gksr h gSa\2NO(g) + O2(g) 2NO2(g)NO(g) ds fuekZ.k dh ekud eqDr Åt kZ 86.6 kJ/mol gS]298 K i j A NO2(g) ds fuekZ.k dh ekud eqDr Åt kZ298 K i j D; k gksxh\ (Kp = 1.6 × 1012)(1) R(298 ln (1.6 ×1012) – 86600(2) 86600 + R(298) ln (1.6 × 1012)

(3) 86600 – )298(R)106.1ln( 12

(4) 0.5[2×86,600–R(298 ln (1.6 × 1012)]Sol. (4)

2106.1n298R–

12

= Gºr = 0NO

0NO G2–G2

2

0NO2

G =86.6×103 – 2

106.1nK298 12

36. , l hVksu dk ok"i nkc 20ºC i j 185 Vksj gSaA t c vok"i ' khyi nkFkZ dk 1.2 g, , fl Vksu ds 100 g esa ?kqyr k gS] 20ºCi j A bl dk ok"i nkc183 Vksj gSaA i nkFkZ dk eksyj nzO; eku(g mol–1) gSa%(1) 32 (2) 64(3) 128 (4) 488

Sol. (2)Pº = 185

PP–P0

= Nn

183183185

= 58/100M/2.1

M = 64

37. 2A B + C vfHkfØ; k ds fy, ekud fxCt Åt kZ

esa i fj or Zu 300K i j 2494.2J gSaA fn; s x; s l e; i j

vfHkfØ; k feJ.k dk l a?kVu [A] = 21

, [B] = 2 r Fkk [C]

= 21

gSaA v fHkfØ; k fd l fn' kk esa t k; sxhA

[R = 8.314 J/K/mol, e = 2.718](1) vxz fn' kk D; ksafd Q > KC

(2) i ' p fn' kk D; ksafd Q > KC

(3) vxz fn' kk D; ksafd Q < KC

(4) i ' p fn' kk D; ksafd Q < KC



Sol. (2)Gº i j 300K = 2494.2 J

2A B + C

Gº = –RT n K–2494.2 = –8.314 × 300 n KK = 10

Q = 2]A[]C][B[ = 4

21

212

2

.

Q > KC reverse direction.38. CuSO4 ds foy; u esa 2 QSj kMs fo| qr xqt kjh xbZ gSaA dSFkksM

i j t ek gksus okys Cu dk Hkkj gksxk&(Cu dk i j ek.kq nzO; eku = 63.5 amu)(1) 0 g (2) 63.5 g(3) 2g (4) 127g

Sol. (2)Cu2+ + Ze Cu2 eksy bDdBk dj r k gS = 1 eksy Cu2F 2 eksy 1 eksy Cu 63.5 xzke

39. mPpr e Øe (>3) dh vfHkfØ; k nqyZHk gSa] dh ot g l s%(1) l Hkh vfHkfØ; k ; qDr Li h' kht dh , d l kFk VDdj ksa dhde l EHkkouk(2) , UVªksi h esa c<+ksr j h vkSj l fØ; .k Åt kZ t Sl s fd T; knkv.kq l fEefyr gSa(3) i zR; kLFk VDdj ksa ds dkj .k l kE; vfHkdeZd dh vksjfoLFkkfi r gksrk gSa(4) VDdj i j l fØ; Li h' kht dh gkfu

Sol. (1)vk.kfodr k vkSj dksfV > 3 l EHko ugh gSA D; ksfd l HkhvfHkfØ; k djus okys vo; oks dk , d l kFk Vdjko dh l EHkkoukde gSA

40. i k=k esa l fØ; pkj dksy dk 3 g, 50 mL ds , l hfVd vEyfoy; u (0.06N) esa feyk; k x; kA , d ?kaVs ckn bl dksNfu=k fd; k x; k vkSj Nfu=k dk l keF; Z 0.042N i k; k Xk; kA, fl fVd vEy dh vo' kksf"kr ek=kk ¼pkj dksy dk i zfr xzke½ gSa&(1) 18 mg (2) 36 mg(3) 42 mg (4) 54 mg

Sol. (1)CH3COOH (0.06M)50 mlfeyh eksy = 50 × 0.06 = 3cps gq, feyh eksy = 50 × 0.042 = 2.1vo' kksf"kr feyh eksy = 0.9

vo' kksf"kr Hkkj = 3

60109.0 3– × 103

= 354

= 18 feyh xzke

41. N3–, O2– vkSj F– (Å esa) dh vk; fud f=kT; k Øe' k% gSa&(1) 1.36, 1.40 vkSj 1.71(2) 1.36, 1.71 vkSj 1.40(3) 1.71, 1.40 vkSj 1.36(4) 1.71, 1.36 vkSj 1.40

Sol. (3)l Hkh vo; oks esa bysDVªksuksa dh l a[ ; k l eku gSA ; fn i zksVksuksadh l a[ ; k T; knk gksxh r ks vkdkj NksVk gksxkA

42. Al ds fu"d"kZ.k ds fy, gkWy&gsj ksYV i zØe esa] fuEu esa l sdkSul k dFku vl R; gSa\(1) CO vkSj CO2 bl i zØe esa cur s gSa(2) Al2O3, CaF2 ds l kFk fefJr gksr s gS t ks fd feJ.k dsxyukad fcUnq dks de dj r k gSa vkSj pkydr k ykr k gSa(3) Al3+ dSFkksM i j vi pf; r gksr k gSa] Al cukus ds fy,(4) Na3AlF6 fo| qr vi ?kVuh dh r j g gS

Sol. (4)

43. H2O2 ds l anHkZ esa fuEufyf[ kr dFku l s] vl R; dFku dkspqfu; s\(1) ; g dsoy vkWDl hdkj d , t saV dh r j g O; ogkj djl dr k gS(2) ; g i zdk' k esa , Dl i kst j i j fo?kfVr gksr k gS(3) bl s IykfLVd esa j [ kuk i M+r k gS vkSj oSDl ykbu Xyklcksr y va/ksj s esa(4) bl s /kqy l s nwj j [ kk t kr k gS

Sol. (1); g vkWDl hdkj d o vi pk; d nksuks dh r j g dke dj r k gSA

44. fuEu esa l s fdl esa] {kkj h; enk /kkr q l yQsV dh gkbMªs' kuÅt kZ vf/kdr e gSa ml dh t kyd Åt kZ l s\(1) CaSO4 (2) BeSO4(3) BaSO4 (4) SrSO4

Sol. (2)dsoy BeSO4 ?kqyu' khy l YQsV gSA D; ks fd bl dh t kydÅt kZ bl dh gkbMªs' ku Åt kZ l s de gksr h gSA buds vykokckdh l c vo{ksi h; gksr s gSA

45. fuEu esa l s dkSul k l cl s T; knk vfØ; gSa\(1) Cl2 (2) Br2(3) I2 (4) ICl

Sol. (4)bl ds fy, f} /kqzo vk/kq.kZ ' kqU; ugh gS] vr% ; g /kqzoh; gSA ckfdl c v/kqzoh; gSA r Fkk = 0.



46. mRi szj d dk feyku dj ks] l gh i zØe ds l kFk%mRi szj d i zØe

(A) TiCl3 (i) osdj i zØe

(B) PdCl2 (ii) ft xy j &ukVk

cgqydhdj .k

(C) CuCl2 (iii) l Ei dZ i zØe

(D) V2O5 (iv) Mhdku i zØe(1) (A) - (iii), (B) - (ii), (C) - (iv), (D) (i)(2) (A) - (ii), (B) - (i), (C) - (iv), (D) (iii)(3) (A) - (ii), (B) - (iii), (C) - (iv), (D) (i)(4) (A) - (iii), (B) - (i), (C) - (ii), (D) (iv)

Sol. (2)

47. fuEu esa l s fdl dk DoFkukad fcUnq mPp gSa\(1) He (2) Ne(3) Kr (4) Xe

Sol. (4)i j ek.kq Hkkj ft r uk T; knk gksxk DoFkukad fcUnq mr uk ghT; knk gksxkA

48. T; kfefr l eko; fo; ksa dh l a[ ; k t ks fd oxkZdkj l er yh;[Pt(Cl)(py) (NH3) (NH2OH)]+ ds fy, fo| eku gSa(py = i sj hfMu) :(1) 2 (2) 3(3) 4 (4) 6

Sol (2)dsp2 Mabcdbl fy, T; kfer h; l eko; oh; ks dh l a[ ; k = 3

49. KMnO4 ds j ax dk dkj .k gSa&(1) M L vkos' k LFkkukUr j .k l aØe.k(2) d - d l aØe.k(3) L M vkos' k LFkkukUr j .k l aØe.k(4) * l aØe.k

Sol. (3)vkos' k dk LFkkukUr j .k yhxsUMl ~ l s /kkr q i j gksr k gSA bl fy,KMnO4 dk j ax cSxuh gksr k gSA

50. dFku : ok; qe.My esa ukbVªkst u vkSj vkWDl ht u eq[ ; ?kVdgS ysfdu ; s vfHkfØ; k dj ds ukbVªkst u dk vkWDl kbM ughacukr s gSaAdkj .k: ukbVªkst u vkSj vkWDl ht u ds e/; vfHkfØ; k ds fy,mPp r ki dh t : j r gSaA(1) nksuksa dFku vkSj dkj .k l gh gSa ysfdu dkj .k] dFku dsfy, l gh O; k[ ; k ugha dj r k gSaA(2) nksuksa dFku vkSj dkj .k l gh ugha gSaA(3) dFku l gh ugha gSa ysfdu dkj .k l gh gSaA(4) nksuksa dFku vkSj dkj .k l gh gSa vkSj dkj .k] dFku ds fy,l gh O; k[ ; k dj r k gSaA

Sol. (1)

51. gSykst uksa dh x.kuk dh D; wfj ; l fof/k esa] dkcZfud ; kSfxd ds250 mg, AgBr ds 141 mg nsr k gSaA ; kSfxd esa czksfeudk i zfr ' kr gSa% (Ag dk i j ek.kq nzO; eku = 108, Br = 80)(1) 24 (2) 36(3) 48 (4) 60

Sol. (1) Br ds eksy = 1 × AgBr ds eksy

= 1 × 188

10141 3–

Br dk Hkkj = 80188

10141 3–

Br dh % = 10010250

80188

101413–

3–

= 24%

52. fuEu esa l s dkSul k ; kSfxd T; kfefr ; l eko; ork esa fo| ekugSa\(1) 1-Qsfuy-2-C; wfVu(2) 3-Qsfuy-1-C; wfVu(3) 2-Qsfuy-1-C; wfVu(4) 1, 1-MkbZ Qsfuy-1-i zksi su

Sol. (1)H3C – HC = CH–CH2 – Phnksuks f} cU/k dkcZu i j ek.kq vyx&vyx f} i zfr LFkkfi r gSA

53. dkSul k ; kSfxd vkst ksfudj .k dj us i j 5-dhVks-2-esfFkygsDl susy nsxk\

(1)

CH3

CH3 (2)

CH3

CH3

(3)

CH3

CH3

(4) H3C

CH3

Sol. (2)

CH3 CH3

CH3 CH3

vkst ksuhdj .k OCHO

4

3 2

1

5

6CH3

CH3



54. , Ydkby ¶yksj kbM dk l a' ys"k.k l cl s vPNk gksr k gSa] fdl ds}kj k&(1) eqDr ewyd ¶yksj huhdj .k(2) l sUMes; j vfHkfØ; k(3) ÝsUdyfLaVu vfHkfØ; k(4) LokZVl vfHkfØ; k

Sol. (4)R–Cl or

R – BrR–F + AgCl/AgBrAgF/dmF

Swart reaction

55. fuEufyf[ kr vfHkfØ; k ds Øe esa

VkWyqbZu A B

C

KMnO4 SOCl2

H /Pd2 BaSO4

mRi kn C gSa%(1) C6H5COOH (2) C6H5CH3(3) C6H5CH2OH (4) C6H5CHO

Sol. (4)COOH

(A)

(B)(C)

CHO

CH3

COCl

SOCl2

KMnO4

H /Pd2

BaSO4

56. vfHkfØ; k esaNH2

CH3

NaNO /HCl2

CuCn/KCN

D

E + N2

0 - 5°C

mRi kn E gSa%

(1)

COOH

CH3

(2) H3C CH3

(3)

CN

CH3

(4)

CH3

Sol. (3)NH2 N Cl2 CN

CH3 CH3 CH3

NaNO /HCl2 CuCN/KCN0-5ºC + N2

(E)(D)57. fuEu esa l s dkSul k cgqyd i saV vkSj ysDoj ds fuEkkZ.k esa

mi ; ksxh gSa\

(1) csdsykbV (2) fXyi Vsy

(3) i ksyhi zksfi u (4) i ksyhfoukby Dyksj kbMSol. (2)

fXyIVy cgqyd i saUV o ysdj cukus esa mi ; ksx gksr k gSA

58. fuEu esa l s dkSul k foVkfeu i kuh esa ?kqyu' khy gSa\

(1) foVkfeu C (2) foVkfeu D

(3) foVkfeu E (4) foVkfeu KSol. (1)

dsoy foVkfeu B o C i kuh esa ?kqyu' khy gSA buds vykok

foVkfeu ol k ?kqyu' khy gksr s gSA

59. fuEu esa l s dkSul k ; kSfxd i zfr vEy ugha gSa?

(1) , sY; qfefu; e gkbMªksDl kbM

(2) l hesVhMhu

(3) QSuyt hu

(4) j sfuVhMhuSol. (3)

fQusyt kbu i zfr vEy ugh gSA

60. fuEufyf[ kr esa l s fdl ; kSfxd dk j ax i hyk ugha gSa\(1) Zn2[Fe(CN)6](2) K3[Co(NO2)6](3) (NH4)3 [As(Mo3O10)4](4) BaCrO4

Sol. (1)(1) (NH4)3 [As(Mo3O10)4] = i hyk

(2) BaCrO4 = i hyk

(3) Zn2[Fe(CN)6] = l Qsn

(4) K3[Co(NO2)6] = i hyk



[MATHEMATICS]

61. ekuk A r Fkk B nks l eqPp; gS ft l esa Øe' k% pkj r Fkk nks

vo; o gS r c mi l eqPp; A × B dh l a[ ; k Kkr dhft ; s]

ft l esa çR; sd esa de l s de r hu vo; o gksA(1) 219 (2) 256(3) 275 (4) 510

Sol. 1n(A × B) = 8Total subsets = 28

8C0 + 8C1 + 8C2

= 37No. of Req. Subsets = 256 – 37 = 219.

62. , d l fEeJ l a[ ; k z dks eki kad , d okyh l fEeJ l a[ ; k dgk

t kr k gS ; fn |z| = 1 ekuk z1 r Fkk z2 bl çdkj l fEeJ

l a[ ; k, gSA fd 1 2

21

z –2z

2 – z z eki kad , d okyh l fEeJ l a[ ; k

r Fkk z2 eki kad , d okyh l fEeJ l a[ ; k ugha gSA r c fcUnq z1

fLFkr gksxkA

(1) f=kT; k 2 dk oÙk

(2) f=kT; k 2 dk oÙk

(3) x-v{k ds l ekukUr j l j y j s[ kk

(4) y-v{k ds l ekukUr j l j y j s[ kkSol. 3

1 2

1 2

z –2z1

2 – z z

(z1 – 2z2) 1 2 1 2 1 2z – 2z 2 – z z 2 – z z

221 1 2 2 1 2 1 2 1 2| z | –2z z – 2z z 4 z 4 – 2z z – 2z z

2 21 2z z

2 2 2 21 2 1 2z z – z – 4 z 4 0

2 21 2z – 4 z –1 0

|z1| = 2

63. ekuk r Fkk l ehdj .k x2 – 6x – 2 = 0 ds ewy gS ; fn

n 1 ds fy , an = n – n r c 10 8

9

a – 2a2a dk eku

cj kcj gksxkA(1) 6 (2) –6(3) 3 (4) –3

Sol. 3

x2 – 6x – 2 = 0 2 – 6 – 2 = 0

2–6– 2 = 0 .....(1)an = n – n n 1a40 – 298 = 10 – 10 – 28 +28

= 8 (a2 – 2) – b8 (b2 – 2)= 8(6)– 8(6) (using (1)= 69 – 69

= 6a9now

10 8 9

9 9

a 2a 6a3

2a 2a

64. ; fn A =

1 2 22 1 –2a 2 b

, d vkO; wg gS t ks fd l ehdj .k

AAT = 9I dks l r qa"V dj r k gS t gk¡ I , 3 × 3 Øe dk

bdkbZ vkO; wg gS r c Øfer ; qXe (a, b) cj kcj gksxkA(1) (2, – 1) (2) (– 2, 1)(3) (2, 1) (4) (– 2, – 1)

Sol. 4AAT = 9 I

1 2 2 1 2 a 9 0 02 1 –2 2 1 2 0 9 0a 2 b 2–2b 0 0 9

a + 4 + 2b = 0 a + 2b = – 4 ....(i)2a + 2 – 2b = 0 a – b = – 1 ....(ii)From i and ii3b = – 3 b = – 1a = – 2

65. ds l Hkh ekuks dk l eqPp; ft l ds fyl j s[ kh; l ehdj .kksa dk

fudk;

2x1 – 2x2 + x3 = x1

2x1 – 3x2 + 2x3 = x2

– x1 + 2x2 = x3

vl axkeh gy j [ kr k gS &

(1) fj Dr l eqPp; j [ kr k gSA

(2) , dy l eqPp; j [ kr k gSA(3) nks vo; o j [ kr k gSA

(4) nks l s vf/kd vo; o j [ kr k gSASol. 3

= (2 – ) (2 + 3– 4) + 2 (– 2 + 2) + 1 (4– 3 – ) = 0– 3 – 2 + 6 + 8 – 3 – – 8 = 0– 3 – 2 + 5 – 3 = 0



3 + 2 – 5 + 3 = 0( – 1) (2 + 2 – 3) = 0( – 1) ( + 3) ( – 1) = 0

= 1, 1, –3

66. 6,000 l s cM+s i w.kkZdksa dh l a[ ; k t ks fd vadks 3, 5, 6, 7r Fkk 8 ds mi ; ksx l s cukbZ t k l dr h gSA t cfd vadksa dh

i quj kofr u gksA(1) 216 (2) 192(3) 120 (4) 72

Sol. 26/7/8– – – –3 × 4C3 × 3! = 72– – – – – = 120Total = 192

67. f} i n i zl kj 501–2 x es x ds i w.kkZadh; ?kkr ks ds xq.kkdks

dk ; ksx gksxk &

(1) 501 3 12

(2) 501 32

(3) 501 3 –12

(4) 501 2 12

Sol. 1for sum of integral power of xput x = 1 in

50 501 2 x 1 2 x

2

503 12

.

68. ; fn m nks fofHkUu okLr fod l a[ ; kvksa l r Fkk n(l, n > 1)dk l ekUr j ek è; gS r Fkk G1, G2 r Fkk G3, l r Fkk n, ds

eè; r hu xq.kksÙkj ekè; gSA r c 4 4 41 2 3G 2G G cj kcj

gksxkA(1) 4 l2mn (2) 4 lm2n(3) 4 lmn2 (4) 4 l2m2n2

Sol. 2

m = n

2

2m = + r4

G1 G2 G3 n r r2 r3 r4 = n4r4 + 24r8 + 4r12

4r4(1 + 2r4 + r8)4r4 (1 + r4)2

4r4 22m

n · 3 2

24m

4m2n

69. Js<+h 311

+ 3 31 21 3

+ 3 3 31 2 31 3 5

+ ..... 9 ds

çFke 9 i nksa dks ; ksx gksxkA(1) 71 (2) 96(3) 142 (4) 192

Sol. 2

Tn = 3n

(2n 1)

=

2 2

2n (n 1)

4 n

Tn = 14 2n 2 n 1

= 14

n(n 1)(2n 1) 2n(n 1) n

6 2

= 14

9 10 19 90 96

= 14

{285 + 99} = 96

70. x 0Lim

(1– cos2x)(3 cos x)x tan4x

cj kcj gSA

(1) 4 (2) 3

(3) 2 (4) 12

Sol. 3

x 0Lim

(1 cos2x) (3 cos x)x tan4x

x 0Lim

(1 cos2x)· (3 cosx)tan4xx .4x

4x



x 0Lim

2(1 cos2x) (3 cosx).

tan4x(2x)4x

12

·(3 + 1) = 2

71. ; fn Qyu g(x) = k x 1, 0 x 3mx 2, 3 x 5

vodyuh;

gS] r c k + m dk eku gksxkA

(1) 2 (2) 165

(3) 103 (4) 4

Sol. 1

g(x) = k x 1 x 0,3

mx 2 x 3,5

g(x) diff g(x) continuous g(3–) = g (3+)

k 4 = 3m + 2 2k = 3m + 2 ......(1)

Againg’(3+) = g’ (3–)

m = x 3

k k42 x 1

4m = k.....(2)from (1) & (2)2k = 3m + 2 8m = 3m + 2

5 m = 2

m = 25

& k = 4m = 85

k + m = 105

= 2

72. oØ x2 + 2xy – 3y2 = 0 i j fLFkr fcUnq (1, 1) i j

vfHkyEc

(1) oØ dks i qu% ugha feyr k gSA

(2) oØ dks i qu% f} r h; pr qZFkka' k es feyr k gSA

(3) oØ dks i qu% r r h; pr qZFkka' k esa feyr k gSA

(4) oØ dks i qu% pr qZFk pr qFkkZa' k esa feyr k gSASol. 4

x2 + 2xy – 3y2 = 0diff. w.r.t. x2x + 2x (y') + 2y – 6yy' = 02 + 2y' + 2 – 6y' = 0

4y' = 4y' = 1

slope of normal = – 1So equation becomesy – 1 = –1 (x – 1)x + y = 2

solving it with curvex2 + 2xy – 3y2 = 0x2 + 2x(2 – x) – 3(2 – x)2 = 0x2 + 4x – 2x2 – 3(x2 – 4x + 4) = 0–4x2 + 16x – 12 = 0x2 – 4x + 3 = 0(x – 1) (x – 3) = 0x = 1, 3

y = 1, –1thus second point of intersection is (3, –1)is in 4th qud.

73. ekuk f(x) pkj ?kkr dk , d cgqi n gS] ft l ds pj e eku

x = 1 r Fkk x = 2 i j gSA ; fn x 0Lim 2

f(x)1

x

= 3,

r c f(2) cj kcj gksxkA(1) –8 (2) –4(3) 0 (4) 4

Sol. 3f(x) =

x 0Lim 2

f(x)1x

= 3

f(x) must not contain degree 0 &degree 1 term f(x) = ax4 + bx3 + cx2

now f'(x) = 4ax3 + 3bx2 + 2cxf'(1) = 4a + 3b + 2c = 0 ......(1)f'(2) = 32a + 12b + 4c = 0 ......(2)

and x 0Lim 2

f(x)1x

= 1 + c = 3 ......(3)



c = 2 12b 24

b 21a2

(1) 4a + 3b = – 4(2) 32a + 12b = – 8(1) 32a + 24b = – 32

74. l ekdyu 2 4 3/4dx

x (x 1) cj kcj gSA

(1)

1/44

4x 1 c

x

(2) 1/44x 1 c

(3) – 1/44x 1 c (4) –

1/44

4x 1 c

x

Sol. 4

2 4 3 /4dx

x (x 1)

= 5 4 3 /4dx

x (1 x ) = 5

4 3 / 4x dx

(1 x )

...(1)

put 1 + x–4 = T4

– 4x–5 d x= 4T3 dT (1) become

– 3

3T dTT = – T + C

= – (1 + x–4)1/4 + C

= –

1/44

41 x C

x

75. l ekdyu

4 2

2 22

logxlogx log(36 –12x x ) dx cj kcj

gSA(1) 2 (2) 4(3) 1 (4) 6

Sol. 3

4 2

222

log xI dxlogx log x 6

.....(1)

using b b

a a

f x dx = f a b x dx

I =

24

2 22

log 6-xdx

log 6 x logx ....(2)

(1) + (2) gives

2I = 4

2

1dx 2I = 1

76. nh?kZoÙk 2x9

+ 2y5

= 1 dh ukfHk; t hok ds vafr e fcUnqvksa

i j Li ' kZ j s[ kk } kj k cus pr qHkZqt dk {ks=kQy ¼oxZ bZdkbZ esa ½gksxkA

(1) 274

(2) 18

(3) 272

(4) 27

Sol. 4

P

P1

35

,2

2 2x y 19 5

a = 3 b = 5

e2 = 1 –

2

2ba

= 1 – 5 49 9

e = 23

now the quadrilateral formed will be a rhombus

with area = 22a

e

= 2.9 32

= 27

77. ekuk y(x) vody l ehdj .k (x log x) dydx + y = 2x

log x, (x 1) dk gy gS] r c y(e) cj kcj gksxkA(1) e (2) 0(3) 2 (4) 2e

Sol. 3



dydx

+ 1

x log x . y = 2

I.F. = 1

log log xx logxe e log x

y· log x = 2·log x dx

y log x = 2 (x log x – x ) + cx = 1 c = 2x = e y = 2(e – e) + 2 = 2

78. mu fcUnqvksa dh l a[ ; k ft uds nksuksa funsZ' kakd i w.kkZad gSA t ks

' kh"kZ (0, 0), (0, 41) r Fkk (41, 0) l s cus f=kHkqt ds

vUnj fLFkr gS] gksxhA(1) 901 (2) 861(3) 820 (4) 780

Sol. 4

x + y < 411 to eachx + y < 39 x + y 38 x + y + z = 38

38 + 3 –1C3–1 = 40C2 = 40 39 780

2

.

79. fcUnq (2, 3) ds çfr fcEc dk j s[ kk (2x – 3y + 4) + k (x– 2y + 3) = 0, k R esa fcUnqi Fk gksxkA

(1) x-v{k ds l ekUr j l j y j s[ kk

(2) y-v{k ds l ekUr j l j y j s[ kk

(3) 2 f=kT; k dk oÙk

(4) 3 f=kT; k dk oÙk

Sol. 3

PR = RQ(x – 1)2 + (y – 2)2 = (2 – 1)2 + (3 – 2)2

(x – 1)2 + (y – 2)2 = 2

80. oÙkksa x2 + y2 – 4x – 6y – 12 = 0 r Fkk x2 + y2 +6x + 18y + 26 = 0, i j mHk; fu"B Li ' kZ j s[ kkvksa dh

l a[ ; k gSA(1) 1 (2) 2(3) 3 (4) 4

Sol. 3

x2 + y2 – 4x – 6y – 12 = 0

C1 (2, 3) r1 = 2 22 3 12 = 5

x2 + y2 + 6x + 18y + 26 = 0

C2 (–3, –9), r2 = 2 23 9 26 = 8

C1C2 = (52 + 122) = 13

81. {(x, y) : y2 2x r Fkk y 4x – 1} ds } kj k i fj Hkkf"kr

{ks=k dk {ks=kQy ¼oxZ bdkbZ esa½ gksxkA

(1) 732 (2)

564

(3) 1564 (4)

932

Sol. 4

y2 = 2x

2y y 12 4

2y2 – y – 1 = 02y2 – 2y + y – 1 = 0(2y + 1) (y – 1)



21–12

y 1 yA – d4 2

1213

1–21–

2

y y y2A –4 6

1 12 3

1 1– –2 2

y 2y yA –8 6

1 –13 1 14A – –8 8 6 48

3 3 8 1 12 3 9A – –

8 32 48 32 48

15 9 3 5 3A – –

32 48 16 2 3

= 3 15 – 6 3 9 9

16 6 16 6 32

82. ekuk i j oy; x2 = 8y i j O ' kh"kZ r Fkk Q dksbZ fcUnq gS]

; fn fcUnq P j s[ kk[ k.M OQ dks 1 : 3 esa vUr % foHkkft rdj r k gSA r c P dk fcUnqi Fk gksxkA(1) x2 = y (2) y2 = x(3) y2 = 2x (4) x2 = 2y

Sol. 4

Let P : (h, k)

1. .0h 4h4

1. 3.0k 4k4

(,) on Parabola 2 = 8 (4h2) = 8.4 k

16h2 = 32kx2 = 2y

83. fcUnq (1, 0, 2) dh j s[ kk x – 2

3 = y 1

4

= z – 212

r Fkk

l er y x – y + z = 16 ds çfr PNsnh fcUnq l s nwj h Kkr

dhft ; sA

(1) 2 14 (2) 8

(3) 3 21 (4) 13

Sol. 4P(3 + 2, 4 – 1, 12 + 2)3 + 2 – 4 + 1 + 12 + 2 = 1611 = 11 = 1Point of intersection (5, 3, 14)

Distance = 2 2 24 3 12

= 169= 13

84. ml l er y dh l ehdj .k Kkr dhft ; s ft l esa j s[ kk

2x – 5y + z = 3; x + y + 4z = 5 gS r Fkk l er y

x + 3y + 6z = 1 ds l ekukUr j gSA(1) 2x + 6y + 12z = 13(2) x + 3y + 6z = – 7(3) x + 3y + 6z = 7(4) 2x + 6y + 12z = –13

Sol. 32x – 5y + z – 3 + (x + y + 4z – 5) = 0x( + 2) + y( – 5) + z(4 + 1) – 3 – 5 = 0

2 5 4 11 3 6

3 + 6 = – 5 6 – 30 = 12 + 32 = – 11 6 = – 33 = – 11/2 = – 11/24x – 10y + 2z – 6–11x – 11y – 44z + 55 = 0– 7x – 21y – 42z + 49 = 0x + 3y + 6z – 7 = 0



85. ekuk a,b

r Fkk c

r hu v' kqU; l fn' k bl i zdkj gS fd mues

l s dksbZ nks l j s[ kk, sa r Fkk (a b

) × c

=13 b

c

a.

; fn l fn' k b r Fkk c

ds e/; dk dks.k gS] r c sin dk

eku gksxk &

(1) 2 2

3(2)

– 23

(3) 23 (4)

–2 33

Sol. 11(a b) c bc a3

1(a.c) b – (b.c) a bc a 0.b3

1–b.c bc3

, a.c 0

– bc cos = 1 bc3

cos = 1–3

sin = 2 23

86. ; fn 12 , d l eku xsans 3 , d l eku fMCcks esa Mkyh t kr hgS] r c i zkf; dr k Kkr dhft , t cfd bu fMCCkksa esa l s , d esaBhd 3 xsans gksA

(1) 11

55 23 3

(2) 10

2553

(3) 12

12203

(4) 11

1223

Sol. 1

Success (p) = 13

Failure (q) = 23

Acc. to binomial distribution, we have to find,

P(x = 3) = 3C12 .

313

· 9

23

= 553

1123

.

87. 16 çs{k.kks ds vk¡dM+ksa ds l eqP; dk ekè; 6 gS ; fn buesa l s, d çs{k.k eku 16 dks gVk fn; k t kr k gS r Fkk u; s r hu 3,4 r Fkk 5 eku ds çs{k.kksa dks vk¡dM+ksa esa t ksM+k t kr k gS r ci fj .kkeh vk¡dM+ks dk ekè; gksxkA(1) 16.8 (2) 16.0(3) 15.8 (4) 14.0

Sol. 4

1 2 3 15a a a ...... a 1616

16

......(1)

1 2 3 15a a a ...... a (3 4 5)??

18

......(2)(1) a1 + a2 + a3 +.....+ a15 = (16)2 – 16

now

(2) 2(16) 16 12

18

= 256 4

18

= 25218

= 14

mean = 14

88. ; fn fdl h Vkoj ds f' k"kZ l s r hu l j s[ kk, sa fcUnq A, B r Fkk CVkoj ds i kn dh vksj t kus okyh j s[ kk l s mUu; u dks.k Øe' k%30º, 45º r Fkk 60º cukr s gS r c AB : BC dk vuqi krgksxk &

(1) 3 :1 (2) 3 : 2(3) 1: 3 (4) 2 : 3



Sol. 1

AB h( 3 –1) 3 :11BC h(1– )3

89. ekuk tan–1y = tan–1x + tan–12

2x1 – x

,

t gk¡ |x| < 13 r c y dk , d eku gksxkA

(1) 3

23x – x1– 3x

(2) 3

23x x1 – 3x

(3) 3

23x – x1 3x

(4) 3

23x x1 3x

Sol. 1

tan–1 y = tan–1 x + tan–1 22x

1 x

|x| < 13

tan–1 22x

1 x = 2 tan–1 x

tan–1 y = tan–1 x + 2 tan–1 x= 3 tan–1 x

= tan–1 3

23x x1 3x

y = 3

23x x1 3x

90. ~ s v(~ r ^ s) dk udkj kRed cj kcj gSA(1) s ^ ~ r (2) s ^ (r ^ ~ s)(3) s v (r v ~ s) (4) s ^ r

Sol. 4~ S V (~ r ^ S)

S r ~ r ~ r ^ S ~ S ~ S V (~ r ^ S) ~ (~ SV(~ r ^ S))T T F F F F TT F T T F T FF T F F T T FF F T F T T F

Documents

JEE MAIN EXAMINATION - 2015 QUESTION WITH SOLUTION PAPER CODE - A€¦ · (Page # 2) JEE MAIN Examination(2015) (Code - A) Rank Booster Test Series [JEE Advanced] 12th & 13th Students