jee-advance physics 2015 paper 2

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SECTION - IThis section contains 8 questions. Each question, when workedout will result in one integer from 0 to 9 (both inclusive).1. For a radioactive material, its activity A and rate of change of

its activity R are defined as dNAdt

= - and – ,dARdt

= where

N(t) is the number of nuclei at time t. Two radioactive sourcesP (mean life t) and Q (mean life 2t) have the same activity att = 0. Their rates of change of activities at t = 2t are RP and

RQ, respectively. If ,P

Q

R nR e

= then the value of n is

2. The monochromatic beam of light is incident at 60° on oneface of an equilateral prism of refractive index n and emergesfrom the opposite face making an angle q(n) with the normal

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

q = m.

The value of m is

60° q

3. In the following circuit, the current through the resistorR (= 2 W) is I amperes. The value of I is

(=2 )W

4. An electron is an excited state of Li2+ ion has angularmomentum 3h/2p. The de Broglie wavelength of the electronin this state is pp a0 (where a0 is the Bohr radius). The valueof p is

5. A large spherical mass M is fixed at one position and twoidentical point masses m are kept on a line passing throughthe centre of M (see figure). The point masses are connectedby a rigid massless rod of length l and this assembly is freeto move along the line connecting them. All three massesinteract 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 .288Mm k æ ö= ç ÷

è ø The value

of k is

6. The energy of a system as a function of time t is given as E(t)= A2 exp(–at,) where a = 0.2 s–1. The measurement of A hasan error of 1.25%. If the error in the measurement of time is1.50%, the percentage error in the value of E(t) at t = 5 s is

7. The densities of two solid spheres A and B of the same radii

R vary with radial distance r as rA(r) = rkR

æ öç ÷è ø

and rB(r) =

JEE ADVANCED 2015 PHYSICS (PAPER 2)

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5,rk

Ræ öç ÷è ø

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 , ,10

B

A

I nI

= the

value of n is8. Four harmonic waves of equal frequencies and equal

intensities I0 have phase angles 0, 2,

3 3p p

and p. When they

are superposed, the intensity of the resulting wave is nI0.The value of n is

SECTION - II

This section contains 8 multiple choice questions. Each questionhas 4 choices (a), (b), (c) and (d) out of which ONE or MORETHAN ONE are correct.

9. In terms of potential difference V, electric current I, permittivitye0, permeability m0 and speed of light c, the dimensionallycorrect equation(s) is(are)(a) m0I2 = e0V2 (b) m0I = m0V(c) I = e0cV (d) m0cI = e0V

10. Consider a uniform spherical charge distribution of radiusR1 centred at the origin O. In this distribution, a sphericalcavity of radius R2, centred at P with distance OP = a= R1 – R2 (see figure) is made. If the electric field inside thecavity at position r

r is (r),Euruur

then the correct statement(s)is(are)

(a) Eur

is uniform, its magnitude is independent of R2 butits direction depends on r

r

(b) Eur

is uniform, its magnitude depends on R2 and itsdirection depends on r

r

(c) Eur

is uniform, its magnitude is independent of a but itsdirection depends on a

r

(d) Eur

is uniform and both its magnitude and directiondepend on a

r

11. In plotting stress versus strain curves for two materials Pand Q,a student by mistake puts strain on the y-axis andstress on the x-axis as shown in the figure. Then the correctstatement(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 spherical body of radius R consists of a fluid of constantdensity and is in equilibrium under its own gravity. If P(r) isthe pressure at r(r < R), then the correct option(s) is (are)

(a) P(r = 0) = 0 (b)( = 3 /4) 63=( = 2 /3) 80

P r RP r R

(c)( = 3 /5) 16=( = 2 /5) 21

P r RP r R (d)

( = /2) 20=( = /3) 27

P r RP r R

13. A parallel plate capacitor having plates of area S and plateseparation d, has capacitance C1 in air. When two dielectricsof different relative primitivities (e1 = 2 and e2 = 4) areintroduced between the two plates as shown in the figure,

the capacitance becomes C2. The ratio 2

1

CC is

(a) 6/5 (b) 5/3(c) 7/5 (d) 7/3

14. An ideal monoatomic gas is confined in a horizontal cylinderby a spring loaded piston (as shown in the figure). Initiallythe gas is at temperature T1, pressure P1 and volume V1 andthe spring is in its relaxed state. The gas is then heated veryslowly to temperature T2, pressure P2 and volume V2. Duringthis process the piston moves out by a distance x. Ignoringthe friction between the piston and the cylinder, the correctstatement(s) is (are)

3


(a) If V2 = 2V1 and T2 = 3T1, then the energy stored in the

spring is 14 P1V1

(b) If V2 = 2V1 and T2 = 3T1, then the change in internalenergy is 3P1V1

(c) If V2 = 3V1 and T2 = 4T1, then the work done by the gas

is 73 P1V1

(d) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the

gas is 176 P1V1

15. A fission reaction is given by 236 140 9492 54 38U Xe Sr ,x y® + + +

where x and y are two particles. Considering 23692U 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 236 14092 54U, Xe and 94

38Sr be 7.5 MeV,8.5 MeV and 8.5 MeV, respectively. Considering differentconservation 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

16. Two spheres P and Q of equal radii have densities r1 and r2,respectively. The spheres are connected by a massless stringand placed in liquids L1 and L2 of densities s1 and s2 andviscosities h1 and h2, respectively. They float in equilibriumwith the sphere P in L1 and sphere Q in L2 and the stringbeing taut (see figure). If sphere P alone in L2 has terminal

velocity PVur

and Q alone in L1 has terminal velocity QV ,ur

then

L2

L1

(a)1

2

P

Q

V

Vh

=h

ur

ur (b)2

1

P

Q

V

Vh

=h

ur

ur

(c) . 0P QV V >ur ur

(d) . 0P QV V <ur ur

SECTION - IIIThis section contains 2 paragraphs, each describing theory,experiments, data etc. four questions related to the two paragraphswith two questions on each paragraph. Each question has one ormore than one correct answer(s) among the four given options(a), (b), (c) and (d).

PARAGRAPH 1Light guidance in an optical fibre can be understood by consideringa structure comprising of thin solid glass cylinder of refractiveindex n1 surrounded by a medium of lower refractive index n2. Thelight guidance in the structure takes place due to successive totalinternal reflections at the interface of the media n1 and n2 as shownin the figure. All rays with the angle of incidence i less than aparticular value im are confined in the medium of refractive indexn1. The numerical aperture (NA) of the structure is defined assin im.

17. For two structure namely S1 with 1 45 / 4n = and n2 = 3/2,

and S2 with n1 = 8/5 and n2 = 7/5 and taking the refractiveindex of water to be 4/3 and that of air to be 1, the correctoption(s) is(are)(a) NA of S1 immersed in water is the same as that of S2

immersed in a liquid of refractive index 16

3 15

(b) NA of S1 immersed in liquid of refractive index 615

is

the same as that of S2 immersed in water(c) NA of S1 placed in air is the same as that of S2 immersed

in liquid of refractive index 415

(d) NA of S1 placed in air is the same as that of S2 placed inwater

18. If two structure of same cross-sectional area, but differentnumerical apertures NA1 and NA2(NA2 < NA1) are joined

4


longitudinally, the numerical aperture of the combinedstructure is

(a)1 2

1 2

NA NANA NA+ (b) NA1 + NA2

(c) NA1 (d) NA2

PARAGRAPH 2

In a thin rectangular metallic strip a constant current I flows alongthe positive x-direction, as shown in the figure. The length, widthand thickness of the strip are l, w and d, respectively.A uniform magnetic field B

su

is applied on the strip along thepositive y-direction. Due to this, the charge carriers experience anet deflection along the z-direction. This results in accumulationof charge carriers on the surface PQRS and appearance of equaland opposite charges on the face opposite to PQRS. A potentialdifference along the z-direction is thus developed. Chargeaccumulation continues until the magnetic force is balanced bythe electric force. The current is assumed to be uniformlydistributed on the cross-section of the strip and carried byelectrons.

19. Consider two different metallic strips (1 and 2) of the samematerial. Their lengths are the same, widths are w1 and w2and thicknesses are d1 and d2 respectively. Two points Kand M are symmetrically located on the opposite faces parallelto the x-y plane (see figure). V1 and V2 are the potentialdifferences between K and M in strips 1 and 2, respectively.Then, for a given current I flowing through them in a givenmagnetic field strength B, the correct statement(s) is(are)(a) If w1 = w2 and d1 = 2d2, then V2 = 2V1(b) If w1 = w2 and d1 = 2d2, then V2 = V1(c) If w1 = 2w2 and d1 = d2, then V2 = 2V1(d) If w1 = 2w2 and d1 = d2, then V2 = V1

20. Consider two different metallic strips (1 and 2) of samedimensions (length l, width w and thickness d) with carrierdensities n1 and n2, respectively. Strip 1 is placed in magneticfield B1 and strip 2 is placed in magnetic field B2, both alongpositive y-directions. Then V1 and V2 are the potentialdifferences developed between K and M in strips 1 and 2,respectively. Assuming that the current I is the same forboth the strips, the correct option(s) is(are)(a) If B1 = B2 and n1 = 2n2, then V2 = 2V1(b) If B1 = B2 and n1 = 2n2, then V2 = V1(c) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1(d) If B1 = 2B2 and n1 = n2, then V2 = V1

5


1. (2)( )2 – t2

2 2– – –od N edA d dN d NR

dt dt dt dt dt

lé ù= = = =ê úë û

\ R = No l2 e–lt = (Nol) l e–lt = Aole–lt

[Q Ao = Nol]

\

2– 2

– 22 2QP

Q P

ttP P

t tQ Q

R e P e eR Q eee

e

tll t

l l tt

l l t= = ´ = =

l tl

\ n = 22. (2) Here ÐMPQ + ÐMQP = 60°. If ÐMPQ = r then ÐMQP

= 60 – rApplying Snell’s law at Psin60° = n sin r ...(i)Differentiating w.r.t ‘n’ we get

O = sin r + n cos r × drdn ...(ii)

60°

60°–rrr

M

QP60° q

Applying Snell’s law at Qsin q = n sin (60° – r) ...(iii)

Differentiating the above equation w.r.t ‘n’ we get

cos q ddnq

= sin (60° – r) + n cos (60° – r) dr–dn

é ùê úë û

\cos q ddnq

= sin (60° – r) – n cos (60° – r)tan– r

né ùê úë û[from (ii)]

\d 1dn cosq

=q

[sin (60° – r) + cos (60° – r) tan r] ...(iv)

From eq. (i), substituting 3n = we get r = 30°

From eq (iii), substituting 3n = , r = 30° we get q = 60°On substituting the values of r and q in eq (iv) we getd 1dn cos 60

q=

° [sin 30° + cos 30° tan 30°] = 2

3. (1) The equivalent resistance of balanced wheatstonebridge is

Req = 3 6 23 6

´= W

+

Balancedwheatstone

bridge2W

4W10W

12W

2W

I

6.5V 4W

2W8W6W

1W

The equivalent resistance of balanced wheat stonebridge is

Req= 6 18 9

24 2´

= W

\ I = 6.5

2 4.5+ = 1A

2W

4W

10W

12W

2W

I

6.5V6W

Balancedwheat stone

bridge

4. (2) Given mvr = 3h2p

Þ n = 3

\32

h r h=

l ph

mvé ùl =ê úë ûQ

\ l = 2

02 2

3 3r na

zé ùp

= p ê úê úë û

2

0nr az

é ù=ê ú

ê úë ûQ

\ l = 0 02 3 3 23 3

a a´é ùp = pê úë û\ p = 2

5. (7) For the tension in the rod to be zero, the force on boththe masses m and m should be equal in magnitude anddirection. Therefore

3l l

M

m m

SOLUTIONS

6


2 2 2 2–(4 ) (3 )GMm Gmm GMm Gmm

+ =l l l l

\ 2m = M 1 1–9 16

é ùê úë û

\ m = 7288

M

K = 76. (4) E = A2 e–0.2t

\ loge E = 2 loge A –0.2tOn differentiating we get

2 – 0.2dE dA dt tE A t

= ´

As errors always add up therefore

100 2 100 0.2 100dE dA dttE A t

æ ö æ ö´ = ´ + ´ç ÷ ç ÷è ø è ø

\dEE

× 100 = 2 × 1.25% + 0.2 × 5 × 1.5%

\dEE

× 100 = 4%

7. (6) 2

0( )

RI dm r= ò

dr

R

r

\2 2

04

RI r dr r= r ´ p ´ò

\4

04

RI r dr= p rò

\4 5

0 0

44R R

Ar KI k r dr r drR R

p= p ´ =ò ò

6 54 46 6

K R RKR

æ öp= = pç ÷ç ÷

è ø

54

04

R

BrI K r drR

æ ö= p ç ÷è øò

= 10 5

54 4

10 10K R RK

Rp

´ = p

\6

10B

A

II

= Þ n = 6

8. (3) 02sin sin sin sin

3 3y I O p pé ù= + + + pê úë û

0 03 3 3

2 2y I I

é ù= + =ê ú

ë û\ Ir = y2 = 3I0 Þ n = 3

9. (a, c)We know that

0

00 0

1 andC Rm

= =em e

Now, m0 I2 = e0 V2

\2

2020

V RI

m= =

e Þ Option A is correct

Now, e0 I = m0 V

\0

0

1IV R

m= =

e Þ Option B is incorrect

Now, I = e0C V

\0

1 V RC I

= =e

\

00 0

11 R=

em e

\ 0

0R

m=

eÞ Option C is correct

Now, mo C I = e0 V

\0 0

0 0

0 0

11C

V RI C

m m= = = ´

e em e

= µ0

Þ Option (d) is incorrect10. (d) Assume the cavity to contain similar charge distribution

of positive and negative charge as the rest of sphere.Electric field at M due to uniformly distributed chargeof the whole sphere of radius R1

PM

ar

O

3E rr

=e

ur r

7


Electric field at M due the negative charge distributionin the cavity

23

E MPr=

e

ur uuur

\ The total electric field at M is

1 23 3

E E E r MPr r= + = +

e e

ur ur ur uuurr

\ ( )–3 3

a r r aE r MPr r é ù= + + =ë ûe e

ur uuurr r r rrQ

\ 3E ar

=e

ur r

(d) is the correct option11. (a, b) The maximum stress that P can withstand before

breaking is greater than Q. Therefore (A) is a correctoption.

Max(strain P)

Strain

Max(strain Q)

Maximumstress of Q

Stress

Maxstressof P

The strain of P is more than Q therefore P is more ductile.Therefore (B) is a correct option.

stressYstrain

=

For a given strain, stress is more for Q. ThereforeYQ > YP.

12. (b, c) Let us consider an elemental mass dm shown in theshaded portion.

drr

pp + dp

Here

P 4pr2 – (P + dP) 4pr2 = 3GMrR

r (4pr2) dr

\ 30

–P r

R

GMdp rdrR

r=ò ò

\ P = 2 2

32GM R r

Rr é ù-ë û

\

2 22

222

9 7–16(r 3R/ 4) 6316

(r 2R/ 3) 8054–99

R RRPP RRR

é ùê úê ú= ë û= = =

= é ùê úê úë û

and

22

22

9–25(r 3R/ 5) 16

(r 2R/ 5) 214–25

RRPP RR

é ùê úê ú= ë û= =

= é ùê úê úë û

B and C are correct options.13. (d)

C1

d/2

+

–

e1 = 2

s/2

C2

e2 = 4

C3

e1 = 2

s/2

01

2 s/2C =d/2e 0

24 s/2C =

d/2e

0 03

2 s/2 sC =d d

e e=

Ceq =

0 001 2

301 2

2 4

6

s ssC C d dC sC C d

d

e e´ e´

+ = +e+

= 0 0s s43 d d

e e+

8


\ Ceq = 01

7 73 3

sC

de

= 01

sC

deé ù=ê úë û

Q

14. (b) Applying combined gas law

1 1 2 2

1 2

PV P VT T

=

If V2 = 2 V1 and T2 = 3T1 then

1 1 2 11 2

1 1

2 23 3

PV P VP P

T T´

= Þ =

Now change in internal energy

DU = 2f

[nR (T2 – T1)] = 2f

[P2V2 – P1V1]

For monoatomic gas f = 3

DU = 1 1 1 1 13 3 2 – 32 2

P V PV Pé ù´ =ê úë ûV1

\ (b) is the correct option.Now assuming that the pressure on the piston on theright hand side (not considering the affect of spring)remains the same throughout the motion of the pistonthen,

Pressure of gas = 1 2 1kx kxP P PA A

+ Þ = +

where k is spring constant and A = area of piston

Energy stored = 212

kx

2 1kxP PA

= +

1 132

kxP PA

= +

12P kx

A=

\ 12

P Akx =

Also,V2 = V1 + AxV1 = Ax

\ 1Vx

A=

\ Energy = 1 112 2

P A VA

´ = 1 114

PV

\ A is correctNow

1kxW PdV P dVA

æ ö= = +ç ÷è øò ò = 1

kxP dV dVA

+ò ò

\ 1 (dx) AkxW P dVA

= + ´ò ò

\ W = 2

1 2 1(V – V )2

kxP +

1 1 2 2 12

1 2

12 1 2 1

4Here on applying we get

32and [ 3 ]

PV P V PP

T TVV V Ax x V VA

é ù= =ê úê úê ú

= + Þ = =ê úë ûQ

\ W = 2P1V1 + 1 1212 3

P A VA

´ ´ = 1 173

PV

C is correct optionHeat suppliedQ = W + DU

= 1 12 2 1 1

7 3 ( – )3 2PV P V PV+

= 1 11 1 1 1

7 3 4 3 –3 2 3

PVP V PVé ù+ ê úë û

= 1 1416

PV

15. (a) 236 140 9492 54 38U Xe Sr x y® + + +

The number of proton in reactants is equal to theproducts (leaving x and y) and mass number of product(leaving x and y) is two less than reactants\ x = p, y = e– is ruled out [B] is incorrectand x = p, y = n is ruled out [C] is incorrectTotal energy loss = (236 × 7.5) – [140 × 8.5 + 94 × 8.5]= 219 MeVThe energies of kx and ky together is 4MeVThe energy remain is distributed by Sr and Xe which isequal to 219 – 4 = 215 MeV\ A is the correct optionAlso momentum is conserved

1. .K Em

\ µ . Therefore K.Esr > K.Exe

16. (a, d) From the figure it is clear that(a) s2 > s1(b) r2 > s2 [As the string is taut](c) r1 < s1 [As the string is taut]\ r1 < s1 < s2 < r2When P alone is in L2

21 2

2

2 ( – )g9P

rV

p r s=

h is negative as r1 < s2

Where r is radius of sphere.When Q alone is in L1

22 1

1

2 ( – )g9Q

rV

p r s=

h is positive as r2 > s1

Therefore .P QV V O<ur ur

option (d) is correct

TT

s1

s2

9


Also 1 2 1

2 1 2

––

P

Q

VV

r s h= ´

r s h ...(i)

For equilibrium of Q

3 32 2

4 43 3

T r g r g+ p s = p r ...(ii)

For equilibrium of P

3 31 1

4 43 3

T r g r g+ p r = p s ...(iii)

(iii) – (ii) givesr1 – s2 = s1 – r2 ...(iv)From (i) and (iv)

P 1

Q 2

V –V

h=

h \P 1

2Q

VV

h=

h

\ A is also a correct option17. (a, c)

Applying Snell’s law at P; ns sin im = n1 sin (90° – C)ns = Refractive index of surrounding

Q n2

P 90° – Cim

n1Cns

Also sin C = 2

1

nn

n1

n2

C

Now

NA = sin im = 2

1 1 221

cos 1 –s s

n n nC

n n n=

\2 21 2–

s

n nNA

n=

For S1 (in air)

NA = 45 9 3–16 4 4

=

For S1 (in water)

NA = 34

45 9 9–16 4 16

=

For s1 s6in15

æ ö=ç ÷

è øn

NA = 15 45 9 3 156 16 4 24

- =

For S2 (in water)

3 64 49 3 15–4 25 25 4 5

NA = =

For S2 (in air)

64 49 15–25 25 5

NA = =

For S2 (in ns = 415

)

15 64 49 3–4 25 25 4

NA = =

For S2 16in

3 15æ ö

=ç ÷è ø

sn

3 15 64 49 9–16 25 25 16

NA = =

(a), (c) are correct options

18. (d) 2 21 2

1 –s

NA n nn

=

HereNA2 < NA1\ the NA of combined structure is equal to the

smaller value of the two numerical apertures.(d) is the correct option.

19. (a, d) When megnetic force balances electric forceFB = FEq vd B = q E

\ vd B = Vw [Q V = E × w]

\ V = wvdB = Iw B

newdé ù ´ê úë û

dI Iv

neA newdé ù= =ê úë û

\IV B

ned= ´

\1Vd

µ Þ V1d1 = V2d2

when d1 = 2d2, V2 = 2V1and when d1 = d2, V2 = V1(a), (d) are correct options

20. (a, d) Here

1 1 2 2

1 2

V n V nBVn B B

µ Þ =

If B1 = B2 and n1 = 2n2 Þ V2 = 2V1and of B1 = 2B2 and n1 = n2 Þ V2 = 0.5V1A and C are the correct options.

jee-advance physics 2015 paper 2

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