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JEE Advanced Exam 2019 (Paper & Solution)
PAPER-2
PART-I (PHYSICS) SECTION – 1 (Maximum Marks : 32)
This section contains EIGHT (08) questions Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
option(s). For each question, choose(s) corresponding to (all) the correct answer(s). Answer to each question will be evaluated according to the following marking scheme : Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the option is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases. Q.1 Three glass cylinders of equal height H = 30 cm and same refractive index n = 1.5 are placed on a
horiozntal surface as shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same (R = 3m). If H1, H2 and H3 are the apparent depth of a point X on the bottom of the three cylinders, respectively, the correct statement (s) is / are –
(1) 0.8 cm < (H2 – H1) < 0.9 cm (2) H3 > H1 (3) H2 > H3 (4) H2 > H1
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Ans. [3, 4] Sol. For fig. I
H1 = 5.1
H = 5.1
30 = 20 cm
300
5.11305.1
H1
2
For fig. II
600
1300
2\130
5.1H1
2
60029
3011
201
6001
305.1
H1
2
H2 = 20.6 cm. For fig. III
300
5.11305.1
H1
3
600
1201
H1
2
= 31
600 = 19.3
H2 – H1 = 20.6 – 20 = 0.6 H2 > H1, H3 < H1, H2 > H3
Q.2 An electric dipole with dipole moment 2
p0 ( i + j ) is held fixed at the origin O in the presence of an
uniform electric field of magnitude E0. If the potential is consatnt on a circle of radius R centered at the origin as shown in figure, then the correct statement (s) is/are : (0 is permittivity of free space R > > dipole size)
(1) Total electric field at point B is BE
= 0
(2) Total electric field at point A is AE
= )ji(E2 0
(3) R = 3/1
00
0
E4p
(4) The magnitude of total electric field on any two points of the circle wil be same.
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Ans. [1,3] Sol.
P0 45°
E0
A2kp0/R3
ETangential at all points on circle should be zero to make potential to be constant
EA = 030 E
Rkp2
= 2E0 + E0 = 3 E0
EB = 0 030 E
Rkp
E0 = 30
Rkp R =
31
00
031
0 E4p
Ekp
Q.3 A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is
released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement (s) is/are corerct, when the rod makes an angle 60º with vertical ?
[g is the acceleration due to gravity]
(1) The angular acceleration of the rod will be Lg2
(2) The radial accelration of the rod's center of mass will be 4g3
(3) The angular speed of the rod will be L2g3
(4) The normal reaction force from the floor on the rod will be 16Mg .
Ans. [2,3,4] Sol.
Mg L/2
= I (Mg) 3
MLsin2L 2
sinL2g3
At = 60° L4
g3323
L2g3
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a T = 2L
2L
L4
g33 = 8
g33
Mg 22
22
32ML
4MgL
3ML
21
4L
2L
L2/g3L2g32
Radial acceleration ac = 4g3
L2g3
2L
2L 2
averticaal = 23
8g33
8g360sin
8g3360cos
4g3
= 16
g15231
8g3
Mg – N = M16
g15 16Mg
Q.4 In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1m. A
parallel beam of light of wavelength 600 nm is incident on the slits at angle as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. which of the follownig statement(s) is/are correct ?
(1) For = 36.0 degree, there will be destructive interference at point O
(2) For = 0, there will be constructive interference at point P
(3) For = 36.0 degree, there will be destructive interference at point P.
(4) Fringe spacing depends on . Ans. [3]
Sol.
d
d sin
P
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Path difference at p = dsin Ddy small sin
= dDdy
= (0.3 × 10–3)
11011 3
Option (3) 2
1310132
1010600
1011180
36.01033
3
9
34
5.62
13
desttructive interferace.
(2) )eDestructiv(5.52
11106
10111037
34
)veConstructi(1106106102103d
7
734
Q.5 A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P0, volume V0, and temperature T0. If the gas mixture is adiabatically compressed to a volume V0/4, then the correct statement(s) is/are, (Given 21.2 = 2.3 ; 23.2 = 9.2 : ; R is gas constant)
(1) The final presure of the gas mixture after compression is in betwen 9P0 and 10P0. (2) The work |W| done during the proces is 13 RT0. (3) Adiabatic constant of the gas mixture is 1.6. (4) The average kinetic energy of the gas mixture after compression is in betwen 18RT0 and 19RT0. Ans. [1,2,3]
Sol. 2V21V1
2211mix CnCn
CpnpCn
2R5Cp
2R3Cv 1l
2R7Cp
2R5Cv 22
= 58
2/R202/R32
2R5)1(
2R35
2R7)1(
2R55
4VPVP 0
0 0 )4(PP 0 0
5/80 P2.9)4(PP
W = 1
VPVP 2211
= 1
58
4VP)4(VP 0
05/8
00
= 0000
5/3
00 VP15.26.029.1VP
5/341VP
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|W| = 2.15 (6) RT0 = 13RT0
9P0 < Pf < 10P0 |W| = 13 RT0 6.158
KE = nRT2f
11
2f
f21
= 433RT62.95
4V)P2.9(
5/31PV
2f 00
0
= 23RT0 options : (1) (2) (3) Q.6 A small particle of mass m moving inside a heavy, hollow and stright tube along the tube axis
undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is L = L0 the particle speed is v = v0. The piston is moved inward
at a very low speed V and such that V << L
dL v0, where dL is the infinitesimal displacement of the
piston. Which of the following statement (s) is/are correct ?
(1) The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to
0L21 .
(2) After each collision with the piston, the particle speed increases by 2V. (3) The rate at which the particle strikes the piston is v/L.
(4) If the piston moves inward by dL, the particle speed increases by 2v L
dL .
Ans. [1,2] Sol.
V0 V e = 1 Vap = Vsep
V0 +2V
V = small
Frequency of Collision =
L2V
V/L21 0
0
Speed of particle After 1st collision = V0 + 2 V Speed of particle After 2nd collision = V0 + 4 V Speed of particle After nth collision = V0 + 2 nV
time taken by piston to move L0/2 = V
2/L0
no of collision = LV4VL
V2L
L2V 0000
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For dL displacement tVdL
L2VN
VdL
0
0
Speed = V0 + 2 LdLVVV
VL2dLV 0
00
0
Option (1) and (2) Q.7 Ablock of mass 2M is attached to a massless spring with spring-consant k. This block is connected to
two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the block sare a1, a2 and a3 as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x0. Which of the following ooptin (s) is/are corect ? [g is the accelration due to gravity. Neglect friction]
(1) When spring achieves an extension of 2
x0 for the first time, the speed of the block connected to the
spring is 3g k5
M .
(2) x0 = kMg4
(3) At an extension of 4
x0 of the spring, the magnitude of accelration of the block connected to the
spring is 10
g3 .
(4) a2 – a1 = a1 – a3 Ans. [4]
Sol. ap = 2
aa 32 or a2 – a1 = a1 – a3
Also
2T
2T
Reduced mass =
m2mm2m
=
3m2
Tension = 21
21
mmmm2
geff = 3m4 geff
2T = 2 × 3m4 geff
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3m8
gT2
eff
2m
3m8
3m8 . x0g + =
21 k 2
0x
x0 = k3mg16
Energy conservation
21
4xk
21
2xg
3m8 2
00 (2m)v2
222
v3m4mv
k3mg16k
81
3mg4
k3mg16
37
kgm
932 22
mv2
v2 = k
mg2132 2
v = k
mg2132g
Q.8 A free hydrogen atom after absorbing a photon of wavelength a gets excited from the state n = 1 to the
state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelegnth 0. Let the change in momentum of atom due to the adsorption and the emission are pa and pe,
respectively. If a/e = 51 , which of the option (s) is/are correct ? [Use hc = 1242 eV nm; 1nm = 10–9 m,
h and c are Planck's consatnt and speed of light, resepctively] (1) e = 418 nm
(2) The ratio of kinetic energy fo the electron in the sate n = m to the state n = 1 is 41 .
(3) m = 2
(4) pa/pe = 21
Ans. [2,3]
Sol. E
1
51
411
41
m1
2
2
2
e
163
1615
51
161
m1
2
2m164
m1
2
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161
416.13hc
e
416126.131240
e
nm4876.1312
4161240e
KE 41
KEKE
n1
1m
2m2
SECTION – 2 (Maximum Marks : 18)
This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value using the mouse and the on-screen virtual numeric
keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value ot TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme. Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. Q.1 A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal
rails of zero ressistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1 as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1T perpendicular to the plan. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x × 10–3 A, where the value of x is ………. .
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given : e–1 = 0.37, where e is base of the natural logarithm]
Ans. [0.63] Sol.
E = vB = (10–2 m/s)
10010 (1) = 10–3 volt
i = i0 /te1 = 1
10RL 3
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= 33 10/103
e11
10
= 10–3
e11 = 0.63×10–3
Q.2 Suppose a Ra226
88 nucleus a rest and in ground state undergoes -decay to a Rn22286 nucleus in its excited
state. The kinetic energy of the emitted a particle is found to be 4.44 MeV. Rn22286 nucleus then goes to
its ground state by -decay. The energy of the emitted photon is ………….. keV. [Given : atomic mass of Ra226
88 = 226.005 u, atomic mass of Rn22286 = 222.000 u, atomic mass of
particle = 4.000 u, 1 u = 931 MeV/c2, c is speed of the light] Ans. [135]
Sol.
HeRnRa 42
*22286
22688
4.44 MeV
Rn22286 + decay
Mass defect m = (226.005) – [222.000 + 4.000] = 226.005 – 226.000 = 0.005 Q – value = 0.005 × 931.5 = 4.6575 MeV
HeRn PP KE m1
000.222
000.4mm
)KE()KE(
Rn
Rn
(KE)Rn = 2224 × 4.44 MeV
(KE)Rn 0.08 MeV Energy of particale = 4.65 – [4.44 + 0.08] = 0.135 MeV = 135 KeV Q.3 A monochromatic light is incident from air on a refracting surface of a prism of angle 75º and refractive
index n0 = 3 . The other refracting surface of the prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of 60º. The value of n2 is …………… .
Ans. [1.5] Sol.
75°
C 75–C
n
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sinC = 3
n
sin 3 sin (75 – c)
= 60°, 21 = sin(75 – c)
sin30° = sin(75 – c) 75 – c = 30 c = 45°
2
13
n n =
23 n2 =
23
n2 = 1.50 Q.4 An optical bench has 1.5 m long scale having four equal divisions in each cm. While measurnig the
focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ……………….. .
Ans. [1.38 or 1.39]
Sol. Least Count = 4cm1
Object distance u = ulens – upin du = du1 + du2
= 75 – 45 = 30 cm du = 21
41
41
dv = 21
v = (135 – 75)cm = 60 cm
f1
u1
v1
f1
301
601
f = 20 cm
222222 )20(df
)30(2/1
)60(2/1
fdf
udu
vdv
df = 21 × 400
22 )30(
1)60(
6
= 365 × 2
f
df × 100 = 20
36/10 × 100 = 3650 = 1.388
Q.5 A ball is thrown from ground at an angle with horizontal and with an initial speed u0. For the resulting
projectile motion, the magntiude of average velocity of the ball up to the point when it hits the ground for the first time is V1. After hitting the ground, the ball rebounds at the same angle but with a reduced speed of u0/. Its motion continues for a long time as shown in figure. If the magnhitude of average velocity of the ball for entire duriation of motion is 0.8 V1, the value of is …………… .
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Ans. [4]
Sol. Vavg = takenTime
ntdisplacemeTotal
=
.....g
sinu2
gsinu2
......g
sinucosu
2g
sinucosu2
0
0
00
00
= m2
m2420
1.....111
1.....111cosu
m
= V1
.....111
.....111
2
42
= V1
1
1v
11
1
11
1
2
2
12
Vavg = V1 1
112
2
V1
0.8V1 = V1 = 1
= 0.8 + 0.8
0.2 = 0.8 = 4 Q.6 A perfectly reflecting mirror of mass M mounted on a spring constitues a spring-mass system of angular
frequency such that hMW4 = 1024 m–2 with h as Planck's constant. N photons of wavelength = 8
× 10–6 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1m. If the value of N is x × 1012, then the value of x is …………… . [Consider the spring as massless]
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Ans. [1] Sol. A = 1 m N = No. of photons
Ma = Mvmax = 2N h
M(10–6)() = 2N 6108h
10–12 = M4
Nh
N =
hM4 10–12
N = 1024 × 10–12 = 1012
SECTION – 3 (Maximum Marks : 12)
This section contains Two (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Question based on List-I and List-II and ONLY ONE of
these four options satisfies the condition asked in the Multiple Choice Question. For each question, marks will be awarded according to the following marking scheme : Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered.) Negative Marks : –1 In all other cases. Answer the following by appropriately matching the lists based on the inforamtion given in the paragraph. Q.1 A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unti length
, 2, 3 and 4 respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 () at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while List-II lists the magnitude of some quantity. LIST-I LIST-II
(I) String-1 () (P) 1 (II) String-2 (2) (Q) 1/2 (III) String-3 (3) (R) 1/ 2 (IV) String-4 (4) (S) 1/ 3 (T) 3/16 (U) 1/16
If the tension in each string is T0, the correct match for the highest fundamental frequency in f0 units will be,
(1) I Q; II S, III R ; IV P (2) I P; II R, III S ; IV Q (3) I P; II Q, III T ; IV S (4) I Q; II P, III R ; IV T
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Ans. [2]
Sol. f0 = T
L21
0
String (1) f0 =
0
0
TL21 P
String (2) f2 = 2
f2T
L21 00
0
R
String (3) f3 = 3
f3T
L21 0
0
S
String (4) f4 = 2f
4T
L21 0
0
Q
I P, II R, III S, IV Q Q.2 Answer the following by appropriately matching the lists based on the inforamtion given in the
paragraph. A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length
, 2, 3 and 4 resopectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 () at free length L0 and tension T0 the fundamental mode frenecy is f0.
List-I gives the above four stirngs while list-II lists the magnitude of some quantity. LIST-I LIST-II
(I) String-1 () (P) 1 (II) String-2 (2) (Q) 1/2 (III) String-3 (3) (R) 1/ 2 (IV) String-4 (4) (S) 1/ 3 (T) 3/16 (U) 1/16
The length of strings 1, 2, 3 and 4 are kept fixed at L0, 2L3 0 ,
4L5 0 and
4L7 0 , respectively. String 1, 2, 3
and 4 are vibrated at their 1st, 3rd, 5th, and 14th harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the untis of T0 will be.
(1) I P; II Q, III T ; IV U (2) I P; II Q, III R ; IV T (3) I T; II Q, III R ; IV U (4) I P; II R, III T ; IV U Ans. [1]
Sol. f0 = T
L21
0
f0 =
0
0
2
0
2
0
TL21T
L1
21
2T
2L32
3
T2 = 2T0
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f0 =
0
0
3
0
3
0
TL21T
L1
32
3T
4L52
5
T3 = 0T163
f0 =
0
0
4
0
TL21
4T
4L72
14
2T
T4 04 T4 =
16T0
I P, II Q, III T, IV U Q.3 Answer the following by appropriately matching the lists based on the inforamtion given in the
paragraph. In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given
by TX, where T is temperature of the sytsem and X is the infinitesimal change in a thermodynamic
quantity X of the system. For a mole of monatomic ideal gas X =
ATTlnR
23 + R ln
AVV . Here, R is gas
constant, V is volume of gas, TA and VA ar constants. The List-I below gives some quantities invovled in a process and List-II gives some possible values of
these quantities. LIST-I LIST-II
(I) Work done by the system in process 1 2 3 (P) 31 RT0ln2
(II) Change in internal energy in process 1 2 3 (Q) 31 RT0
(III) Heat absorbed by the system in process 1 2 3 (R) RT0 (IV) Heat absorbed by the system in process 1 2 (S)
34 RT0
(T) 31 RT0 (3 + ln 2)
(U) 65 RT0
If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V0 = 31 RT0,
the correct match is.
(1) I P; II R, III T ; IV P (2) I P; II T, III Q ; IV T (3) I P; II R, III T ; IV S (4) I S; II T, III Q ; IV U
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Ans. [1]
Sol. (1) W123 = nR 3
T0 n2 + 0 = 3
RT0 n2 1 P
(2) dU123 = dU12 + dU23 = 0 + nCv dT
= 0 + 1 × 2R3
3TT 0
0 = RT0
2 R (3) dQ123 = dQ12 + dQ23
= 3
RT0 n2 + nCV dT
= 3
RT0 n2 + RT0
= 3
RT0 (n2 + 3)
3 T (4) dQ12 = dW
= 3
RT0 n2
4 P Q.4 Answer the following by appropriately matching the lists based on the inforamtion given in the
paragraph. In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given
by TX, where T is temperature of the sytsem and X is the infinitesimal change in a thermodynamic
quantity X of the system. For a mole of monatomic ideal gas X =
ATTlnR
23 + R ln
AVV . Here, R is gas
constant, V is volume of gas, TA and VA ar constants. The List-I below gives some quantities invovled in a oproces and List-II gives some possible values of
these quantities. LIST-I LIST-II
(I) Work done by the system in process 1 2 3
(P) 31 RT0ln2
(II) Change in internal energy in process 1 2 3
(Q) 31 RT0
(III) Heat absorbed by the system in process 1 2 3
(R) RT0
(IV) Heat absorbed by the system in process 1 2
(S) 34 RT0
(T) 31 RT0 (3 + ln 2)
(U) 65 RT0
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If the process carried out on one mole of monatomic ideal gas is as shown in the PV-diagram with
P0V0 = 31 RT0, the correct match is.
(1) I Q; II R, III S ; IV U (2) I Q; II S, III R ; IV U (3) I S; II R, III Q ; IV T (4) I Q; II R, III P ; IV U Ans. [1] Sol. (1) W123 = W12 + W23 = PdV + 0
= P0 V0 = 3
RT0
(2) dU123 = dU12 + du23
= nCv (T3 – T1)
= n23 R(T3 – T1)
= 23 (RT3 – RT1)
= 23 (3P0V0 – P0V0) = 3 P0V0 = RT0
2 R (3) dQ123 = dQ12 + dQ23
= nCp (T2 – T1) + nCV (T3 – T2)
= 25 (RT2 – RT1) +
23 (RT3 – RT2)
= 25 (2P0V0 – P0V0) +
23 (3P0V0 – 2P0V0)
= 25 P0V0 +
23 P0V0 = 4P0V0 =
34 P0V0
3 S
(4) dQ12 = 5P0V0 = 65 RT0
4 U