IIT - JEE 2015 (Advanced)

CODE : 8

2

PART I : PHYSICS

Section 1 (Maximum Marks : 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened 0 In all other cases

1. The energy of a system as a function of time t is given as E(t) = A2 exp(t), where = 0.2 s1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is

1. [4] E(t) = A2 exp (t) (E) = 2A (A) exp (t) + A2 exp ( t) (E) = 2A exp ( t) A + A2 exp (t) (t) (E) = 2A exp (t) A + A2 exp (t) t t

(E) A

2 ( t) 2 0.0125 0.2 0.015 5E A

(E)

(E)

= 0.0250 + 0.015 = 0.040 % error = 4%

2. The densities of two solid spheres A and B of the same radii R vary with radial distance r

as A(r) = r

kR

and B(r) = 5

rk

R

, respectively, where k is a constant. The moments

of inertia of the individual spheres about axes passing through their centres are IA and IB,

respectively. If B

A

I

I =

n

10, the value of n is

2. [6] V = 4 x2 x (m) = 4(x) x2 . x

I = 2 42 2( m)x 4 (x) x x

3 3 I

R4

0

x (x) dx

R9

55

0BR

5A 5

0

1 1x dx RRI 6101I 101 Rx dx 6R

x x

R

3

3. Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, /3, 2/3 and . When they are superposed, the intensity of the resulting wave is nI0. The value of n is

3. [3] Some Intensity some amplitude

Resultant amplitude = 2a sin3

= 3 a

I = 3I0

4. For a radioactive material, its activity A and rate of change of its activity R are defined as

A = dN

dt and R =

dA

dt , where N(t) is the number of nuclei at time t. Two radioactive

sources P (mean life ) and Q (mean life 2) have the same activity at t = 0. Their rates of

change of activities at t = 2 are RP and RQ, respectively. If P

Q

R

R =

n

e, then the value of n

is 4. [2] P and Q have same activity at t = 0. If NPO and NQO are the number of nuclei at t= 0 Then A =P NPO = Q NQO

PO

QO

N

N= Q

P

P = 1

and Q =

1

2

Q

P

=

1

2 PO

QO

N

N=

1

2

A =dN

dt

= N

R =dA

dt

=

2

2

d N

dt=

dN

dt= 2N

P

Q

R

R=

2P P2Q Q

N

N

At t = 2

NP = PO2

N

e and NQ = QON

e

P

Q

N

N= PO

2QO

N e

e N =

1

2e

P

Q

R

R=

14

2e =

2

e n = 2

3

Advanced : Question Paper & Solution

4

5. A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle (n) with the

normal (see the figure). For n = 3 the value of is 60° and d

dn

= m. The value of m is

5. [1]

sin 60° = 13 sin r

1r = 30° 2r = 30°

2n sin r = sin

sin = n

2

dcos

dn

=

1

2

d

dn

=

1

2cos60 = 1

6. In the following circuit, the current through the resistor R (=2) is I Amperes. The value

of I is

6. [1] The given Ckt reduces to simple Ckt like this

1r 2r 60

6.5 12

6

2

10

1

2

4

2 4

6.5 12

6

2

10

2

4

5

I = 1A I = 1

7. An electron in an excited state of Li2+ ion has angular momentum 3h/2. The de Broglie wavelength of the electron in this state is pa0 (where a0 is the Bohr radius). The value of p is

7. [2]

= 3h

2 electron in quantum state n = 3

In the light of de-Broglie's hypothesis n = 2 rn

n = 2 2

0n

az

= 2 0n

az

Pa0 = 2 03

a3

P = 2

8. A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l from M, the tension in

the rod is zero for m = M

k288

. The value of k is

8. [7] For zero tension Tensile force = Compressive force F1 = 2F3 + F2

2

2 2 2

Gmm m GMm2G

9 16

M M

2m9 16

7M M

2 K144 288

K = 7

2

4

2

6

12 6.5

2

4.5

6.5 V

I

M F1 F3 F3 F2 m

3

6

Section 2 (Maximum Marks : 32) This section contains EIGHT questions. Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE

of these four option(s) is(are) correct For each question, darken the bubble(s) corresponding to all the correct option(s) in the

ORS Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 If none of the bubbles is darkened 2 In all other cases

9. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative permittivities ( 1 2 and 2 = 4) are

introduced between the two plates as shown in the figure, the capacitance becomes C2. the

ratio 2

1

C

C is

(A) 6/5 (B) 5/3 (C) 7/5 (D) 7/3

9. (D)

0 0s s

C 2 , C 4d d

These are in series, their equivalent 0 0

0

0 0

s s4 2 4 sd d

s s 3 d4 2d d

0 03

2 s sC

2 d d

2 0 0 04 s s 7 s

C3 d d 3 d

and 1 0s

Cd

2

1

C 7

C 3 …(D)

C C

d/2

C3 S/2

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (7)

7

10. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)

(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is 1 11

P V4

(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1

(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is 1 17

P V3

(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is 1 117

P V6

10. (B)

1 1

1

P V

T = 2 2

2

P V

T

(A) 1 1

1

P V

T = 1

21

2VP

3T P2 = 1

3P

2 S = Cross section Area

P2S = Kx P2Sx = kx2

21kx

2 = 2 x

1P S

2 = 2 2 2

1P (V V )

2

Energy is spring = 1 1 21 3

P (2V V )2 2 = 1 1

3P V

4 [A is incorrect]

(B) U = 2 13

nR(T T )2 = 2 2 1 1

3(P V P V )

2

1 1

1

P V

T = 2 1

1

P 2V

3T P2 = 1

3P

2

U = 1 1 1 13 3

P . 2V P V2 2

= 3P1V1 [B is correct]

(C) 1 1

1

P V

T = 1

21

3VP

4T P2 = 1

4P

3

P2S = Kx 2Sx

P2

= 21kx

2 = w

w = 1 2 11 4

P (V V )2 3

1 1 12

P (3V V )3

= 1 14

P V3

[C is incorrect]

(D) 1 1 2 2 1 14 3

P V (P V P V )3 2

= 1 1 1 1 1 14 3 4

P V P 3V P V3 2 3

= 1 1 1 14 9

P V P V3 2

= 1 1 1 18P V 27P V

6

= 1 1

35P V

6 [D is incorrect]

8

11. A fission reaction is given by 236 140 9492 54 38U Xe Sr x y, where x and y are two particles.

Considering 23692 U to be at rest, the kinetic energies of the products are denoted by KXe,

KSr, Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per nucleon of 23692 U, 140

54 Xe and 9438Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering

different conservation laws, the correct option(s) is (are) (A) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV (B) x = p, y = e, KSr = 129 MeV, KXe = 86 MeV (C) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV (D) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV

11. (A) Q Value of the reaction Q = 94 8.5 + 140 8.5 236 7.5 Q = 219 MeV X and Y share 4 MeV together remaining 215 MeV will be shared between Xe and Sr nucleus. Heavier particle will

have less kinetic energy. x = n and y = n and KSr = 129 MeV and KXe = 86 MeV …(A) 12. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts

strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are)

(A) P has more tensile strength than Q (B) P is more ductile than Q (C) P is more brittle than Q (D) The Young’s modulus of P is more than that of Q

12. (A), (B)

Strain

P Q

Stress

Q

P

Strain

Stress

9

For same strain stress in Q is more than P. Young's module of Q > Young's modulus of P. (D) is not correct. For same stress strain in P is more. P is more ductile (A), (B) 13. A spherical body of radius R consists of a fluid of constant density and is in equilibrium

under its own gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is(are)

(A) P(r = 0) = 0 (B) P(r 3R / 4) 63

P(r 2R / 3) 80

(C) P(r 3R / 5) 16

P(r 2R / 5) 21

(D)

P(r R / 2) 20

P(r R / 3) 27

13. (B), (C)

dp

g(r)dr

32

G m(r) 4g(r) ; m(r) r

3r

4

g(r) Gr3

2dp 4Gr

dr 3

p(r) p(R) = 2 2

24 R rG

3 2

p(r) p(R) = 2 2 22G R r

3

put p(R) = 0

p(r) = 2 2 22G R r

3

91p(3R / 4) 9 7 6316

4p(2R / 3) 16 5 8019

91p(3R / 5) 1625

4p(2R / 5) 21125

11p(R / 2) 9 3 274

1p(R / 3) 4 8 3219

r

IIT JEE QUESTION PAPERS PAPER 2WITH SOLUTIONS 2015

Publisher : Faculty Notes Author : Panel of Experts

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