Modul Physics 2015 Paper 2

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JEE ADVANCED 2015 PHYSICS (PAPER 2) SECTION - I This section contains 8 questions. Each question, when worked out 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 A = -

(=2W)

dN dA and R = – , where dt dt

N(t) is the number of nuclei at time t. Two radioactive sources P (mean life t) and Q (mean life 2t) have the same activity at t = 0. Their rates of change of activities at t = 2t are RP and RQ, respectively. If 2.

RP n = , then the value of n is RQ e

The 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 q(n) with the normal

4.

5.

dq (see the figure). For n = 3 the value of q is 60° and = m. dn The value of m is

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

æ M ö M, the tension in the rod is zero for m = k ç ÷ . The value è 288 ø 60°

3.

q

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

of k is

6.

7.

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 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 The densities of two solid spheres A and B of the same radii r R vary with radial distance r as rA(r) = k æç ö÷ and rB(r) = èRø

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2 5

ærö k ç ÷ , respectively, where k is a constant. The moments è Rø

of inertia of the individual spheres about axes passing through their centres are IA and IB, respectively. If , 8.

IB n = , the I A 10

value of n is Four harmonic waves of equal frequencies and equal p 2p , and p. When they 3 3 are superposed, the intensity of the resulting wave is nI0. The value of n is

intensities I0 have phase angles 0,

SECTION - II This section contains 8 multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE or MORE THAN ONE are correct. In terms of potential difference V, electric current I, permittivity e0, permeability m0 and speed of light c, the dimensionally correct equation(s) is(are) (a) m0I2 = e0V2 (b) m0I = m0V (c) I = e0cV (d) m0cI = e0V 10. Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R 2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the electric field inside the uruur r cavity at position r is E (r), 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 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

9.

ur E is uniform, its magnitude is independent of R2 but r its ur direction depends on r (b) E is uniform, its magnitude depends on R2 and its r direction depends on r ur (c) E is uniform, its magnitude is independent of a but its r direction depends on a ur (d) E is uniform and both its magnitude and direction r depend on a 11. 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)

13. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative primitivities (e1 = 2 and e2 = 4) are introduced between the two plates as shown in the figure, the capacitance becomes C2. The ratio

C2 is C1

(a)

(a) 6/5 (b) 5/3 (c) 7/5 (d) 7/3 14. 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) (c)

(a)

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

(b)

1 PV 4 1 1 If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1 If V2 = 3V1 and T2 = 4T1, then the work done by the gas

spring is

(c)

7 PV 3 1 1 If V2 = 3V1 and T2 = 4T1, then the heat supplied to the

is (d)

gas is

17 PV 6 1 1

15. A fission reaction is given by

236 92 U

® 140 54 Xe +

94 38 Sr

where x and y are two particles. Considering

ur VP h ur = 1 h VQ 2 ur ur V P .V Q > 0

3

ur VP h2 (b) ur = h VQ 1 ur ur (d) V P .V Q < 0

SECTION - III This section contains 2 paragraphs, each describing theory, experiments, data etc. four questions related to the two paragraphs with two questions on each paragraph. Each question has one or more than one correct answer(s) among the four given options (a), (b), (c) and (d). PARAGRAPH 1 Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

+ x + y,

236 92 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

236 140 92 U, 54 Xe

and

94 38 Sr

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 16. Two spheres P and Q of equal radii have densities r1 and r2, respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities s1 and s2 and viscosities h1 and h2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal ur ur velocity VP and Q alone in L1 has terminal velocity V Q , then

17. For two structure namely S1 with n1 = 45 / 4 and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are) (a) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index

L2

6

(b)

NA of S1 immersed in liquid of refractive index

(c)

the same as that of S2 immersed in water NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index

L1

16 3 15

(d)

15

is

4 15

NA of S1 placed in air is the same as that of S2 placed in water

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


4 longitudinally, the numerical aperture of the combined structure is (a) (c)

NA1 NA2 NA1 + NA2

NA 1

(b) NA1 + NA2 (d) NA 2 PARAGRAPH 2

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively. su A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

19. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2 respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic 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 same dimensions (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both 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

SOLUTIONS 1.

(

\

2.

)

2 –l t dA d é dN ù d 2 N d N o e =– = = – (2) R = – dt dt êë dt úû dt 2 dt 2 \ R = No l2 e–lt = (Nol) l e–lt = Aole–lt [Q Ao = Nol]

RP l P = RQ l Q

e – l Pt e

– lQ t

l t

=

2W

dr dn

10W 12W

P

Req=

...(ii)

\

60°–rr

6 ´ 18 9 = W 24 2

I=

6.5 = 1A 2 + 4.5

2W

Q r

4W

The equivalent resistance of balanced wheat stone bridge is

60° 60°

2W

4W

\ n=2 (2) Here ÐMPQ + ÐMQP = 60°. If ÐMPQ = r then ÐMQP = 60 – r Applying Snell’s law at P sin60° = n sin r ...(i) Differentiating w.r.t ‘n’ we get O = sin r + n cos r ×

8W

6W

6.5V

2t 2 lP e Q ´ = = 2 t l t lQ e P t e et

I

q

6W

6.5V

M

é dr ù êë – dn úû

dq \cos q = sin (60° – r) – n cos (60° – r) dn \

4.

é tan r ù ê– n ú ë û [from (ii)]

dq 1 = [sin (60° – r) + cos (60° – r) tan r] ...(iv) dn cos q

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

3.

dq 1 = dn cos 60° [sin 30° + cos 30° tan 30°] = 2 (1) The equivalent resistance of balanced wheatstone bridge is Req =

3´ 6 = 2W 3+ 6

4W

...(iii)

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

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

2W

Balanced wheat stone bridge

10W 12W

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

Balanced wheatstone bridge

2W

I

2t e 2t

1W

5.

(2) Given mvr =

3h Þn=3 2p

h ù é êQ l = mv ú ë û

\

hr 3h = l 2p

\

2pr 2 é n2 ù l = 3 = 3 p ê a0 z ú êë ûú

\

l =

é n2 ù êQ r = a0 ú z úû ëê

2 é3 ´ 3ù p a0 ê ú = 2pa0 3 ë 3 û

\ p=2 (7) For the tension in the rod to be zero, the force on both the masses m and m should be equal in magnitude and direction. Therefore M m 3l


m l

6 GMm (4l)

\

+

Gmm l

=

2

GMm (3l)

2

–

Gmm l

é1 1 ù 2m = M ê – ú ë 9 16 û

(3)

9.

é 3 3ù + y = I0 ê ú = 3 I0 2 û ë 2 \ Ir = y2 = 3I0 Þ n=3 (a, c)We know that 1

C=

\

dE × 100 = 2 × 1.25% + 0.2 × 5 × 1.5% E

\

dE × 100 = 4% E

m0 V 2 = = R2 e0 I 2 Now, e0 I = m0 V m0 I 1 = = e0 V R Now, I = e0C V

0

\

Þ

1 1 m0 e0

m0 =R e0

Now,

mo C I = e0 V

Þ

\

R

I = 4 p ò r r 4 dr 0

R

\

I A = 4p ò k 0

=

4pK R

r 4 pK ´ r 4 dr = R R

R

òr

5

M

dr

P

0

r a

æ R6 ö R5 ç ÷ = 4pK ç 6 ÷ 6 è ø R

1 = µ0 1 m 0 e0

Þ Option (d) is incorrect 10. (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 charge of the whole sphere of radius R1

0

\

Option C is correct

m0 V m0 R = = = ´ e0 I C C e0

R

I = ò r ´ 4 pr 2 dr ´ r 2

Option B is incorrect

=R

\

\

Option A is correct

1 V = =R e0 C I

e0

R

m0 e0

Þ

\

\

r dr

m0e 0

\

R

I = ò (dm)r 2

and R =

m0 I2 = e0 V2

Now,

dE æ dA ö æ dt ö ´ 100 = 2 ç ´ 100 ÷ + 0.2t ç ´ 100 ÷ E è A ø è t ø

(6)

p 2p é ù + sin pú y = I 0 êsin O + sin + sin 3 3 ë û

8.

m=

dE dA dt =2 – 0.2 ´ t E A t As errors always add up therefore

7.

IB 6 = Þn=6 I A 10

\

2

7M 288 K=7 (4) E = A2 e–0.2t \ loge E = 2 loge A –0.2t On differentiating we get

\

6.

2

O

5

ærö I B = 4p ò K ç ÷ r 4 dr èRø 0

=

4pK R5

´

R10 R5 = 4pK 10 10

ur r r E= r 3e


7 Here

Electric field at M due the negative charge distribution in the cavity ur r uuur E2 = MP 3e \ The total electric field at M is ur ur ur r r r uuur E = E1 + E 2 = r+ MP 3e 3e \

11.

P 4pr2 – (P + dP) 4pr2 = P

\

0

ur r r r r r r uuur r a – r éQ r + MP = a ù E= r+ ë û 3e 3e

(

– ò dp =

)

ur r r \ E= a 3e (d) is the correct option (a, b) The maximum stress that P can withstand before breaking is greater than Q. Therefore (A) is a correct option.

GM r

GMr R3

r (4pr2) dr

r

ò rdr

R3 R

GM r é 2 2 ù R -r û 2 R3 ë

\

P=

\

é 2 9R 2 ù 2 êR – ú 7R 16 úû P (r = 3R/ 4) ëê 63 = = 16 = P (r = 2 R/ 3) é 2 4 R 2 ù 5R 2 80 êR – ú 9 9 úû ëê é 2 9 R2 ù êR – ú 25 úû 16 P (r = 3R/ 5) êë = = and P (r = 2 R/ 5) é 2 4R 2 ù 21 êR – ú 25 úû ëê

Strain Max (strain P)

B and C are correct options. 13.

Max (strain Q)

(d) d/2

Stress Maximum stress of Q

Max stress of P

C1

C2

e1 = 2

e2 = 4 s/2 –

+

C3 e1 = 2

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

s/2

stress strain For a given strain, stress is more for Q. Therefore YQ > YP. (b, c) Let us consider an elemental mass dm shown in the shaded portion. Y=

12.

r p + dp

p

C1 =

2 e0 s/2 d/2

C3 =

dr

4 e0 s/2 d/2

2 e 0 s/2 e0s = d d

C1 ´ C2 + C3 = Ceq = C1 + C2

=

C2 =

2e 0 s 4 e 0 s ´ d d + e0 s 6e 0 s d d

4 e 0 s e0 s + 3 d d


8 7 e0 s 7 = C1 3 d 3 (b) Applying combined gas law

\

14.

Ceq =

e0 s ù é êQ C1 = d ú ë û

4 P1 ù PV P2V2 é 1 1 ê Here on applying T = T we get P2 = 3 ú 1 2 ê ú ê ú 2V1 [QV2 = 3V1 ] ê and V2 = V1 + Ax Þ x = ú A ë û

PV PV 1 1 = 2 2 T1 T2 If V2 = 2 V1 and T2 = 3T1 then

1 P1 A 2V1 7 ´ ´ = PV 1 1 2 3 3 A C is correct option Heat supplied Q = W + DU

\ W = 2P1V1 +

PV P ´ 2V1 2 1 1 = 2 Þ P1 = P2 T1 3T1 3 Now change in internal energy f f [nR (T2 – T1)] = [P2V2 – P1V1] 2 2 For monoatomic gas f = 3

DU =

3 é3 ù P1 ´ 2V1 – PV 1 1 ú = 3P1 V1 ê 2 ë2 û \ (b) is the correct option. Now assuming that the pressure on the piston on the right hand side (not considering the affect of spring) remains the same throughout the motion of the piston then, DU =

15. (a)

kx kx Þ P2 = P1 + A A where k is spring constant and A = area of piston

Pressure of gas = P1 +

Energy stored = P2 = P1 +

1 2 kx 2

kx A

16.

P1 kx = 2 A kx =

P1 A 2

Also, V2 = V1 + Ax V1 = Ax \ \

\ A is correct Now

=

7 PV 3 1 1 + 3 2

é4 ù 41 PV 1 1ú = 1 1 ê 3 P1 3V1 – PV 6 ë û

236 92 U

94 ® 140 54 Xe + 38 Sr + x + y The number of proton in reactants is equal to the products (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 incorrect and x = p, y = n is ruled out [C] is incorrect Total energy loss = (236 × 7.5) – [140 × 8.5 + 94 × 8.5] = 219 MeV The energies of kx and ky together is 4MeV The energy remain is distributed by Sr and Xe which is equal to 219 – 4 = 215 MeV \ A is the correct option Also momentum is conserved

1 . Therefore K.Esr > K.Exe m (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 < r2 When P alone is in L2

VP =

2pr 2 (r1 – s 2 ) g is negative as r1 < s2 9h2

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

V x= 1 A

Energy =

7 PV 3 1 1 + ( P2V2 – PV 1 1) 3 2

\ K .E. µ

kx 3 P1 = P1 + 2 A

\

=

1 P1 A V1 1 ´ = PV 1 1 2 2 A 4

2pr 2 (r2 – s1 ) g is positive as r2 > s1 9h1 ur ur Therefore V P . V Q < O option (d) is correct VQ =

kx ö æ kx W = ò PdV = ò ç P1 + ÷ dV = ò P1dV + ò dV Aø è A kx ´ (dx) A A

\

W = ò P1dV + ò

\

kx 2 W = P1 (V2 – V1 ) + 2

s1 T T s2


9 VP r1 – s 2 h1 = ´ VQ r2 – s1 h2

Also

15 45 9 3 15 - = 6 16 4 24 For S2 (in water)

...(i)

NA =

For equilibrium of Q 4 3 4 pr s 2 g = pr 3 r 2 g 3 3 For equilibrium of P T+

4 3 4 pr r1 g = pr 3 s1 g 3 3 (iii) – (ii) gives r1 – s2 = s1 – r2 From (i) and (iv) T+

VP h =– 1 VQ h2

\ 17. (a, c)

VP

\

VQ

3 64 49 3 15 – = 4 25 25 4 5 For S2 (in air) NA =

...(ii)

NA =

...(iii) ...(iv) =

For S2 (in ns =

h1 h2

NA =

A is also a correct option

ns

P

NA =

n2 n1

C 90° – C

Also sin C =

n2 n1

19. n1 n2

(d)

NA =

Now

\

NA =

1 ns

n12 – n22

V w

\

vd B =

\

é I ù V = wvdB = w ê ú´B ë newd û

\

[Q V = E × w]

20.

V =

I ´B ned

1 Þ V1d1 = V2d2 d when d1 = 2d2, V2 = 2V1 and when d1 = d2, V2 = V1 (a), (d) are correct options (a, d) Here

\

NA =

æ 6 ö For s1 ç in ns = ÷ 15 ø è

64 49 9 – = 25 25 16

I I ù é êvd = neA = newd ú ë û

ns

45 9 9 – = 16 4 16

3 15 16

n12

For S1 (in air)

3 NA = 4

64 49 3 – = 25 25 4

n22

n12 – n22

45 9 3 – = 16 4 4 For S1 (in water)

)

Here NA2 < NA1 \ the NA of combined structure is equal to the smaller value of the two numerical apertures. (d) is the correct option. (a, d) When megnetic force balances electric force FB = FE q vd B = q E

C

n1 n1 NA = sin im = n cos C = n 1 – s s

15

(a), (c) are correct options 18.

im

15 4

4

æ 16 ö ÷ For S2 ç in ns = 3 15 ø è

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

64 49 15 – = 25 25 5

Vµ

Vn V n B Þ 11= 2 2 n B1 B2 If B1 = B2 and n1 = 2n2 Þ V2 = 2V1 and of B1 = 2B2 and n1 = n2 Þ V2 = 0.5V1 A and C are the correct options. Vµ


Modul Physics 2015 Paper 2

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