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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

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