NSEP_2011

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PHYSICS CAREER POINTNSEP 2011-2012 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN PHYSICS 2011-2012

Total time : 120 minutes (A-1, A-2 & B) PART-A (Total Marks : 180)

SUB-PART A-1 : ONLY ONE OUT OF FOUR OPTIONS IS CORRECT

N.B. Physical constants are given at the end SUB-PART A-1

1. A piece of n-type semiconductor is subjected to an electric field Ex. The left end of the semiconductor is exposed to a radiation so that electron-hole pairs are generated continuously. Let n be the number density of

electrons. The electron current density Je, is given by Je = enµeEx + eDe dxdn . The dimensions of electron drift

mobility (µe) and electron diffusion coefficient (De) are respectively. (a) [M–1 T–2 I1] and [L2T–1] (b) [M1 T–2 I–1] and [M1 L2 T–1] (c) [M–1 T2 I1] and [L2T–1] (d) [M–1 T2 I2] and [L1T–2 I1] Ans. [c]

Sol. dxdneDEenJ exee +µ=

4

232 1)(

)(L

DITIT

MLTITLIL e

e +µ

=−−

−

µe = [M–1 I T2] De = [L2T–1] 2. A metal sample carrying a current along X axis with density Jx is subjected to a magnetic field Bz (along Z

axis). The electric field Ey (Hall field) developed along Y axis is directly proportional to Jx as well as Bz. The constant of proportionality (Hall coefficient) has SI unit

(a) C/m2. (b) m2s/C (c) m2/C (d) m3/C Ans. [d] Sol.

xe–

z

y

B

Jx = AIJ x

I = JxA evdBz = eEy

Ey = vdBz

neAvd = JxA

neBJ

E zxy =

Constant of proportionality = ne1 =

Cm3

2 / 31



3. A vibratory motion is represented by x = 2A cos ωt + A cos

π

+ω+π+ω+

π

+ω2

3cos2

)cos(2

tAtAt . The

resultant amplitude of the motion is

(a) 2

9A (b) 25A (c)

25A (d) 2A

Ans. [b] Sol.

A/2

2AA

A

A/2

R

A

25

4

22 AAAR =+=

4. A force (F) acting on a body is dependent on its displacement s as F ∝ s–1/3. Therefore, the power delivered

by the force varies with its displacement as (a) s2/3 (b) s1/2 (c) s–5/3 (d) s0 Ans. [d] Sol. F = k s–1/3

a = 3/1−smk

∫∫ −= dssmkvdv 3/1

=

23

2

3/22 smkv …. (i)

v ∝ s1/3

P = Fv ∝ s–1/3 × s1/3 P ∝ s0

5. Bamboo strips are hinged to form three rhombi as shown. Point A0 is fixed to a rigid support. The lengths of

the side of the rhombi are in the ratio 3 : 2 : 1. Point A3 is pulled with a speed v. Let vAl and vA2 be the speeds

with which points A1 and A2 move. Then, the ratio vAl : vA2 is -

(a) 2 : 3 (b) 3 : 5 (c) 3 : 2 (d) 5 : 2

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Ans. [b] Sol.

3x 3x2x 2x x x

6x cos θθθ θ θ

θ

4x cos θ2x cos θ

x

As per constant relation

53

2

1 =A

Avv

6. A particle of mass m is made to move with uniform speed v along the perimeter of a regular hexagon.

Magnitude of impulse applied at each corner of the hexagon is

(a) mv (b) 3mv (c) 2

mv (d) zero

Ans. [a] Sol.

60ºv

v

Change = vv =

2º60sin2 , Imp. = ∆p = mv

7. Two chambers of different volumes, one containing m1 g of a gas at pressure p1 and other containing m2 g of the same gas at pressure p2 are joined to each other. If the temperature of the gas remains constant the common pressure reached is

(a) 21

2211mm

pmpm++ (b)

21

1221mm

pmpm++ (c) 2

221

2121 )(mm

pppm+

+ (d) 1221

2121 )(pmpmppmm

++

Ans. [d] Sol.

m1g m2g

Mmn 1

1 = ; Mmn 2

2 =

No. of mole remain constant

RTpV

RTpV

RTVp

RTVp 212211 +=+

p = 21

2211VV

VpVp++

p1 × V1 = RTMm1 ; p2V2 = RT

Mm2

2

2

1

1

21

pRTm

pRTm

RTmRTmp+

+= ⇒

1221

2121 )(pmpm

mmpp+

+

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8. Two liquid drops of equal radii are falling through air with the terminal velocity v. If these two drops coalesce to form a single drop, its terminal velocity will be

(a) v2 (b) 2v (c) v3 4 (d) v3 2 Ans. [c] Sol. Terminal velocity, 2rvT ∝ Let radius of single drop = r and radius of bigger drop = R

3

43

4233 Rr π

=π

×

⇒ R = (2)1/3 r As vT ∝ r2 23/12' )2( rRvT ∝∝

3/13/1' )4(1

)2(1

==T

T

vv

TT vv 3' 4=

9. An elevator of mass M is accelerated upwards by applying a force F. A mass m initially situated at a height of 1m above the floor of the elevator is falling freely. It will hit the floor of the elevator after a time equal to

(a) mgF

M+2 (b)

mgFM

−2 (c)

FM2 (d)

MgFM

+2

Ans. [d]

Sol. mFae = ↑

gas −= ↓

as/e = as – ae = M

MgFgmF +

=+

221 atuts +=

2211 t

MMgF

+

=

MgF

Mt+

= 2

10. The formation of solid argon is due to vander Waals bonding. In this case the potential energy as a function of interatomic separation can be written as (Lennard Jones 6-12 potential energy) E(r) = – Ar–6 + Br–12 where A and B are constant > Given that A = 8.0 × 10–77 Jm6 and B = 1.12 × 10–133 Jm12 , the bond length for solid argon is

(a) 3.75 nm (b) 0.0375 nm (c) 0.750 nm (d) 0.375 nm Ans. [d]

Sol. 0126)( 137 =−−= −− BrArdr

rdE

0126137 =−

rB

rA

AB

AAr 2

6126 == = 77

133

1081012.12−

−

×

×× = 0.28 × 10–56

r6 = 0.28 × 10–2 × 10–54 = 0.0028 × 10–54 r = 0.375 nm

5 / 31



11. Let A and B be the points respectively above and below the earth's surface each at a distance equal to half the radius of the earth. If the acceleration due to gravity at these points be gA and gB respectively, then gB : gA

(a) 1 : 1 (b) 9 : 8 (c) 8 : 9 (d) zero Ans. [b]

Sol. 22 94

2R

GM

RR

GMg A =

+

=

233R

RGM

RGMrgB ×== = 22R

GM

89

49

21

=×=A

Bgg

12. Let vrms, v* and vavg represent the root mean square, the most probable and the average velocities respectively,

in case of a gaseous system in equilibrium at certain temperature. Then, vrms : v* : vavg is (a) 8 : 3π : 2π (b) 8 : 2π : 3π (c) 3π : 2π : 8 (d) 3 : 2 : 8 Ans. [c]

Sol. MRTvrms

3=

MRTv pm

2.. =

MRTvav π

=8

.

vrms : vm.p. : vav. = 3π : 2π : 8 13. In the arrangement of resistance shown below, the effective resistance between points A and B is

(a) 23.5 ohm (b) 38.0 ohm (c) 19.0 ohm (d) 25.0 ohm Ans. [c] Sol.

A

12Ω

B

3

1

2

4

6

5

24Ω7

25Ω

24Ω

12Ω15Ω

30Ω

30Ω25Ω

15Ω

10Ω

10Ω10Ω

10ΩBA

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10Ω

10Ω56

12Ω

12Ω

24Ω

25Ω

34

25Ω

15/2

730Ω1 2

B30Ω10Ω

10ΩA

24Ω

5Ω 15Ω

15/2 Ω

25/2 Ω6 Ω

12 Ω

5 ΩBA

5Ω 15Ω

15/2 Ω

12 Ω

5 ΩBA

6 Ω

Req. = 19Ω

14. A block of material of specific gravity 0.4 is held submerged at a depth of 1 m in a vessel filled with water.

The vessel is accelerated upwards with acceleration of a0 = g/5. If the block is released at t = 0 s, neglecting

viscous effects, it will reach the water surface at t equal to (g = 10 ms–2)

(a) 0.60 s (b) 0.33 s (c) 3.3 s (d) 1.2 s

Ans. [b]

Sol. In frame of vessel

aVggVggV ss ρ=

+ρ+

+ρ−

55 l

( )

aggs

s =

+

ρρ−ρ

5l

a=)12(4.06.0

a = 18 m/s2

221 atuts +=

218211 t×=

912 =t ⇒ st

31

=

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15. The maximum tension in the string of a simple pendulum is 1.2 times the minimum tension. If θ0 is the angular amplitude, then θ0 is

(a)

−

54cos 1 (b)

−

43cos 1 (c)

−

1615cos 1 (d)

−

87cos 1

Ans. [c] Sol.

T2

θ0

T1

v

l

2

1mvmgT =−

[ ])cos1(21 θ−+= ll

gmmgT

T1 = 3mg – 2 mg cos θ T2 = mg cos θ

56

2

1 =TT

56

coscos23

=θ

θ−mg

mgmg

15 mg – 10 mg cos θ = 6 mg cos θ 15 mg = 16 mg cos θ cos θ = 15/16 16. A uniform line charge with density λ = 50 µC/m lies along X axis. The electric flux per unit length crossing

the portion of the plane z = – 3 m bounded by y = ± 3 m is (a) 4.68 µC/m (b) 9.36 µC/m (c) 50 µC/m (d) 18.7 µC/m Ans. [ ] Sol.

dx

xθ

yλ

θ3

x = 0

dA2

04

2x+

πελ

z

223 x+

θ=φ cosEdAd

= 22

0

3

92 xyx

dx

+×

+πε

×λ

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∫ ∫+

− +πε

×λ=φ

2

22

0 )9(3

xdxd =

2

0

1

0 3tan

332

πελ − x

×

πελ

× −

32tan2 1

0

IAPT has given (b) as the correct answer to this question. But The dimensions of the answer & value of answer is not matching to any of the given options Hence the

question is wrong. 17. A plane mirror perpendicular to XY plane makes an angle of 30º with the X axis. An object placed at (–20, 0)

forms an image in the mirror. The point of incidence is (0, 0) and the plane of incidence is the XY plane. The coordinates of the image are :

(a) )10,310( (b) )10,310(− (c) )310,10( −− (d) )10,310( −− Ans. [b,c] Sol.

y

x

A

30º30º30º60º

xI

(–20, 0)20

2030º

sin 30º =

20x

102021x =×=

cos 30º = 20y

3102023y =×=

10x −= , y = – 10 3 18. Magnetic flux through a stationary loop with a resistance R varies during the time interval τ as φ = at(τ – t)

where a is a constant. The amount of heat generated in the loop during the time interval τ is

(a) R

a6

32τ (b) R

a4

32τ (c) R

a3

32τ (d) R

a2

32τ

Ans. [c]

Sol. atadtde 2−τ=φ

=

Re

dtdH 2

= = R

ata 2)2( −τ

dtR

ataH ∫τ

−τ=

0

2)2( =τ

−−τ

0

3

)2)(3()2(aR

ata = Ra

aaa)6(

)()2( 33

−τ−τ−τ =

Raaa)6)(1(

3333

−τ−τ− =

Ra3

32τ

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19. Four functions given below describe motion of a particle. (I) y = sin ωt – cos ωt, (II) y = sin3 ωt,

(III)

ω−

π= ty 3

43cos5 , (IV) y = 1 + ωt + ω2t2. Therefore, simple harmonic motion is represented by

(a) only (I) (b) (I), (II) and (III) (c) (I) and (III) (d) (I) and (II) Ans. [c] Sol. I & III are showing S.H.M. on the basis of superposition principle. 20. A magnetic field is established with the help of a pair of north and south poles as shown. A small bar magnet

placed freely in the field will undergo

(a) pure translational motion (b) pure rotational motion (c) rotational motion superimposed on translational motion (d) oscillatory motion Ans. [c] Sol.

B in Non uniform ∴ Rotational motion super imposed on translation motion Q.21 In a hydrogen atom, the magnetic field at the centre of the atom produced by an electron in the nth orbit is

proportional to -

(a) 21n

(b) 31n

(c) 41n

(d) 51n

Ans. [d]

Sol. B = RI

πµ2

0 = RRev

π

πµ

22

0

B ∝ 2Rv ∝ 22 )(

)/1(n

n

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B ∝ 51n

Q.22 A particle of mass m carries a charge +q. It enters into a region of uniform magnetic field →B existing below

the line l l' as shown. The time spent by the particle in the magnetic field is -

(a) (π –2θ)

qBm (b) infinite as the particle gets trapped

(c) 2θqBm (d) (π + 2θ)

qBm

Ans. [d] Sol.

90–θ 90–θ

π + 2θ

90–θ

90

C

θ

B

× ×

× × ×

+q

l l'

Time spend =

wθ2 =

qBm)2( θ+π .

Q.23 A 2µF capacitor is charged as shown in the figure. The change in its stored energy after the switch S is turned to position 2 is -

(a) 96% (b) 20% (c) 4% (d) 80% Ans. [a]

Sol. U2µF = 21 (2)(v)2

Vcommon = 82082

+×+×V =

5V

∴ '2 FU µ =

21 (2)

2

5

V

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∴ % Change = U

UU '– × 100 = 96%

Q.24 An infinite number of charges each equal to 0.2 µC are arranged in a line at distances 1, 2, 4, 8 … meter from a fixed point. The potential at the fixed point is -

(a) 1800 volt. (b) 2000 volt. (c) 3600 volt. (d) 2250 volt. Ans. [c] Sol.

0.2 0.2 0.2

V = k 0.2 6–10.......81

41

21

11

×

+++

V =

2/1–11 × 9 × 109 × 0.2 × 10–16

V = 3600 volt Q.25 A ball of mass m moving with a speed u along a direction making an angle θ with the vertical strikes a

horizontal steel plate. The collision lasts for a time interval t. If e is the coefficient of restitution between the ball and the plate, the average force exerted by the plate on the ball is -

(a) t

emu (b) t

emu θcos (c) tmue θ+ cos)1(2 (d)

tmue θ+ cos)1(

Ans. [d] Sol.

φ θ v u

e =

θφ

coscos

uv

⇒ v cos φ = eu cos θ …… (i)

u sin θ = v sin φ ….. (ii)

Av. force = tp∆

⇒ t

uvm ]coscos[ θ+φ ⇒ t

emu )1(cos +θ

Q.26 In a Young's double slit experiment sources of equal intensities are used. Distance between slits is d and wavelength of light used is λ(λ << d). Angular separation of the nearest points on either side of central maximum where intensities become half of the maximum value is -

(a) dλ (b)

d2λ (c)

d4λ (d)

d6λ

Ans. [b] Sol.

x

xθ

I + I + 2 I I cos φ

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2I = 2I + 2I cosφ cos φ = 0

φ = 2π

λ∆π x2 =

2π

∆x = 4λ

Ddx =

4λ

x = d

D4

λ

θ = Dx2 =

D2 ×

dD4

λ = d2

λ

Q.27 The variation of magnetic field along the axis of a solenoid is graphically represented by (O is the centre with

l, l' as the extremities of the solenoid along the axis)

(a) (b)

(c) (d)

Ans. [d]

Q.28 A wooden cube is placed on a rough horizontal table. A force is applied to the cube. Gradually the force is

increased. Whether the cube slides before toppling or topples before sliding is independent of -

(a) the position of point of application of the force ..

(b) the length of the edge of the cube.

(c) mass of the cube.

(d) coefficient of friction between the cube and the table.

13 / 31



Ans. [c]

Sol. IAPT has given (b) as the correct answer to this question. But the most appropriate answer to this question should be (c). For more details refer the solution given below

a

mg

N

F

h

f

The block will slide if F > µ mg

The block will topple if Fh > mg

2a or F >

hmga2

The sliding will occur earlier if µmg < h

mga2

⇒ µ < h

a2

& topping will occur earlier if µ > h

a2

So independent of mass. [This answer is correct but the answer provided by IAPT is wrong] Q.29 There are two organ pipes of the same length and the same material but of different radii. When they are

emitting fundamental notes - (a) broader pipe gives note of smaller frequency (b) both the pipes give notes of the same frequency

(c) narrower pipe gives note of smaller frequency (d) either of them gives note of smaller or larger frequency depending on the wavelength of the wave. Ans. [a]

Sol. f = )2(4 xl

v+

where x = 0.6 r

∴ higher the radius lower the frequency Q.30 The wavelength of sodium line observed in the spectrum of a star is found to be 598 nm, whereas that from

the sodium lamp in the laboratory is found to be 589 nm. Therefore, the star is moving with a speed of about - (a) 2.7 × 106 m/s away from the earth (b) 5.4 × 106 m/s towards the earth

(c) 1.6 × 106 m/s away from the earth (d) 4.6 × 106 m/s away from the earth Ans. [d] Sol. λ = 589 λ' = appearant = 598 at is higher than original λ ∴ some is moving away.

'

'–λ

λλ =Cv

598

589–598 = 8103×v

14 / 31



V = 598

9103 8 ××

= 4.6 ×106 m/s away from earth. Q.31 In a series LCR circuit, impedance Z is the same at two frequencies f1 and f2. Therefore, the resonant

frequency of this circuit is -

(a) 2

21 ff + (b) 21

212ff

ff+

(c) 2

22

21 ff +

(d) 21 ff

Ans. [d]

Sol. 1fZ =

2fZ

Q R is same in both series

so, 1fX =

2fX

⇒ 1

)–( fCL XX = 2

)–( fLC XX

⇒ 1

)( fLX + 2

)( fCX = 2

)( fCX + 1

)( fCX

⇒ 2πL(f1 + f2) =

+

π 21

112

1ffC

⇒ 2πL(f1 + f2) = Cπ2

1

+

21

21

ffff

⇒ 4π2LC = 21

1ff

⇒ 2π LC = 21

1ff

⇒ Resonant frequency = LCπ2

1 = 21 ff

Q.32 Two particles are moving along X and Y axes towards the origin with constant speeds u and v respectively. At time t = 0, their respective distances from the origin are x and y. The time instant at which the particles will be closest to each other is -

(a) 22

22

vu

yx

+

+ (b) 22 vu

uyvx++ (c) 22 vu

vyux++ (d)

vuyx

++ 222

Ans. [c] Sol.

ut x – ut

vt

y – vt

Z2 = (x – ut)2 + (y – vt)2

2Z dtdZ = – 2u(x – ut) + 2v(y – vt) = 0

15 / 31



ux – u2t + vy – v2t = 0

t = 22 vuvyux

++

Q.33 A container of volume 0.1 m3 is filled with nitrogen at a temperature of 47°C and a gauge pressure of

4.0 × 105 Pa. After some time, due to leakage, the gauge pressure drops to 3.0 ×105 Pa and the temperature to 27°C. The mass of nitrogen that has leaked out is about -

(a) 128 g (b) 84 g (c) 154 g (d) 226 g Ans. [b] Sol. P1 = 5 × 105 V1 = 0.1 m3 T1 = 320 k

n1 = 1

11

RTVP =

3203.8105 4

××

n2 = 3003.8

104 4

××

∆n = n1 – n2 = 3.8

104

3004–

3205

⇒ 3.8

104

× 3203001280–1500

⇒ 2

4

103.8332010220

××××

⇒ 2.76 mole Now of N2 leaked out = 2.76 × 28 ⇒ 77.28 gm Most probable answer = 84 gm Q.34 Ninety percent of a radioactive sample is left over after a time interval t. The percentage of initial sample

that will disintegrate in an interval 2t is - (a) 38% (b) 19% (c) 9% (d) 62% Ans. [b] Sol. In time interval of 't', 10% decays so in next interval of 't' again 10% of remaining sample will decayed hence

total 81% sample in left so 19% will decay. Q.35 A circuit is arranged as shown. At time t = 0 s, switch S is placed in position l. At t = 5 s, contact is changed

from 1 to 2. The voltage across the capacitor is measured at t = 5 s and at t = 6 s. Let these voltages be V1 and V2 respectively. Then, V1 and V2 respectively are -

16 / 31



(a) 10 volt and 0 volt (b) 9.18 volt and 3.67 volt

(c) 9.18 volt and 3.37 volt (d) 10 volt and 3.67 volt Ans. [c] Sol. At t = 0 Charging is started At t = 5 sec Voltage across capacitor Vc = V0 ( )RC/te1 −−

= 10

− −×××

− 63 1010010205

e1

= 10 ( )5.2e1 −− = 9.18 volt At t = 5 sec discharging is started At t = 6 sec capacitor has been discharged for 1 sec so Vc = V0

RC/te−

= 9.18 63 1010010101

e −×××−

= 9.18 × e = 9.18 × 0.37 = 3.37 volt Q.36 There are two thermocouples A and B made of the same pair of metals. In A each wire is 50 cm long and in B

each wire is 150 cm long. Both the thermocouples are maintained between the same lower temperature θ1 and higher temperature θ2. Each of the two thermocouples is connected to the same microammeter successively. Let ε be the thermoemf and I be the thermoelectric current. Then which of the following statements is true ?

(a) both ε and I are equal for A and B (b) Both ε and I are greater for B than those for A (c) ε is the same for both but I is greater for A (d) ε is the same for both but I is greater for B.

Ans. [c] Sol. (i) Temperature difference is same thermo emf will be same. (ii) Length of loop B is more so resistance of loop B is more. ⇒ Thermoelectric current in loop B is less. Q.37 Two identical particles move at right angles to each other, possessing de Broglie wavelengths λ1 and λ2. The

de Broglie wavelength of each of the particles in their centre of mass frame will be -

(a) 2

22

21 λ+λ (b)

221 λ+λ (c)

21

212λ+λ

λλ (d) 22

21

212

λ+λ

λλ

Ans. [D] Sol. Let m be the mass of each particle and thus verticals are v1 and v2 respectively

17 / 31



v2

v1 y

x

2222

)(v

mmv

v xcm ==

2211

)(v

mmv

v ycm ==

2222

2)(2vvvv xcm =−=

21

)(2vv ycm −=

So magnitude of velocity of 2nd particle is C.O.M frame becomes 44

21

22 vv

+ = 2

22

21 vv +

λ = 1mv

h ⇒ v1 = 1λm

h

λ2 = 2mv

h ⇒ v2 = 2λm

h

So De broglie wavelength of 2nd partical in COM frame is

λ =

2

22

21 vvm

h

+ =

22

22

2

21

2

2

2

λ+

λ nh

mhm

h =

22

21

112

λ+

λ

⇒ 22

21

212

λ+λ

λλ

Q.38 The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 400 nm

is 500 mV. When the incident wavelength is changed to a new value, the stopping potential is found to be 800 mV. New wavelength is about -

(a) 365 nm (b) 250 nm (c) 640 nm (d) 340 nm Ans. [a]

Sol. Energy of photon (1) E1 = )(

12400Åλ

eV

I case

E1 = 4000

12400 = 3.1 eV

Q Vstopping = 500 m ⇒ KE of emitted e– = 0.5 eV Q Ephoton = W + KEe–

⇒ Work function w = 2.6 eV II case Vstopping = 800 mv ⇒ –eKE = 0.8 eV

18 / 31



∴ Ephoton = W + –eKE

= 2.6 + 0.8 = 3.4 eV

Q Ephoton = )(

12400Åλ

eV

3.4 = λ

12400

λ = 3650 Å = 365 nm Q.39 for the logic circuit given below, the outputs Y for A = 0, B = 0 and A = 1, B = 1 are -

(a) 0 and 1 (b) 0 and 0 (c) 1 and 0 (d) 1 and 1

Ans. [b]

Sol.

P

Z

XY

y

A

B A B X Y Z P y

0 0 1 1 0 0 0

1 1 0 0 0 0 0

Q.40 In a hydrogen like atom electron makes transition from an energy level with quantum number n to another

with quantum number (n –1). If n >> 1, the frequency of radiation emitted is proportional to -

(a) 21n

(b) 31n

(c) n2 (d) 41n

Ans. [b]

Sol. En = – 2

26.13n

Z

E(n –1) = – 2

2

)1–(6.13

nZ

∆E = 13.6 Z2

22

1–)1–(

1nn

19 / 31



⇒ 13.6 Z2

22

22

)1–()1–(–

nnnn

⇒ 13.6 Z2

22 )1–(

1–2nnn

⇒ ∝ 31n

as (n >> 1).

20 / 31



Sub-Part A-2 In question 41 to 50 any number of options (1 or 2 or 3 or all 4) may be correct. You are to identify all of them correctly to get 6 marks. Even if one answer identified is incorrect or one correct answer is missed, you get zero score. 41. Consider an electron orbiting the nucleus with speed v in an orbit of radius r. The ratio of the magnetic

moment to the orbital angular momentum of the electron is independent of (a) radius r (b) speed v

(c) charge of electron e (d) mass of electron me

Ans. [a, b]

Sol. momentangularmomentmagentic =

mq

2

is valid for all charge moving on a circular path = em

e2

42. A current I0 enters into a parallel combination of resistors R1 and R2. Current I1 flows through R1 and I2

through R2. The current I0 distributes in such a way that

(a) power consumed in R1 and in R2 is the same

(b) total power consumed in R1 and R2 is minimum

(c) I1 is proportional of R2 and I2 is proportional to R1

(d) the power consumed in each of R1 and R2 is minimum

Ans. [c, b]

Sol.

R1

R2

I1

I2

I0

(i) In parallel I ∝ 1/R

⇒ I1 : I2 = R2 : R1

(ii) Total power = I12 + I2

2R2

P = I12R1 + (I0 – I1)2 R2

1dI

dP = 0 for minima

2I1R1 + 2(I0 – I1) × (–1)R2 = 0

I1(R1 + R2) – I0R2 = 0

I1 = 21

20

RRRI+

It mean total power is minima.

21 / 31



43. Weight of a body on the surface of the earth depends on

(a) distance of the body from the centre of the earth

(b) the latitude of the place on the earth surface where the body is placed

(c) the longitude of the place on the earth surface where the body is placed

(d) the angular speed of rotation of the earth about its own axis

Ans. [a,b.d]

Sol. Theory based

44. Which of the following is /are involved in the formation of rain drops in a cloud ?

(a) saturation of vapour pressure (b) temperature

(c) viscosity (d) surface tension

Ans. [a,b,d] Sol. Theory based 45. A cyclic process on p-V diagram is as shown below. The same process can be shown on p-T or V-T diagrams.

Choose the correct alternative /s. 1 2

3

4V

p

(a)

p

T

3

2 1

4

(b)

V

T

2

34

1

(c)

p

T

2

34

1 (d)

V

T

3

21

4

Ans. [a, b]

Sol.

1 2

3

4 V

P Isothermal

Isothermal

22 / 31



P

T

3

2 1

4

V

T

2

3 4

1

46. When a bright light source is placed 30 cm in front of a thin lens, an erect image is formed at 7.5 cm from the

lens. A faint inverted image is also formed at 6 cm in front of the lens due to reflection from the front surface

of the lens. When the lens is turned around, this weaker inverted image is now formed at 10 cm in front of

the lens. Therefore,

(a) the lens is diverging biconcave

(b) the refractive index of the glass of the lens is 1.6

(c) radii of curvature of surfaces of the lens are 10 cm and 15 cm respectively.

(d) the lens behaves as a converging lens of focal length 30 cm when immersed in a liquid of refractive index 2

Ans. [a,b,c,d]

Sol.

Incident zone – + Refraction zone

f1 =

v1 –

u1

f1 =

5.7–1 –

30–1

f1 =

5.7–1 +

301

f1 =

101–

f = – 10

lens is diverting as f is in refraction zone

23 / 31



+ –

Image due to reflect

O

30

R1

f1 =

u1 +

v1

f1 =

301 +

61

f1 =

51

f = – 5

21R = – 5

R1 = 10 cm

30

O

R2

f1 =

301 +

101

f1 =

304

f = 4

30

2

2R = 4

30

R2 = 15 cm

– +

f1 =

1

12 –n

nn

21

1–1RR

24 / 31



10–1 =

11–2n

151–

10–1

–101 = n2 – 1

+

30)23(–

–101 =

30)1–( 2n × 5

53 = n2 –1

n2 = 1.6

f1 =

22–6.1

151–

101–

f1 =

305–

24.0–

f = 2

60 = 30 cm

47. Let n1 and n2 moles of two different ideal gases be mixed. If ratio of specific heats of the two gases are γ1 and γ2 respectively, then the ratio of specific heats γ of the mixture is given through the relation

(a) (n1 + n2)γ = n1γ1 + n2γ2 (b) 1

)( 21

−γ+ nn =

11

1

−γn +

12

2

−γn

(c) (n1 + n2) 1−γγ = n1 11

1

−γγ + n2 12

2

−γγ (d) (n1 + n2)(γ – 1) = n1(γ1 – 1) + n2(γ2 – 1)

Ans. [b,c]

Sol. γ = 221

221

1

1

VV

pp

CnCn

CnCn

+

+ =

11

11

2

2

1

1

2

22

1

11

−+

−

−+

−

γγ

γγ

γγ

nn

nn

48. A resistance of 4 ohm is connected across a cell. Then it is replaced by another resistance of 1 ohm. It is

found that power dissipated in resistance in both the cases is 16 watt. Then, (a) internal resistance of the cell is 2 ohm (b) emf of the cell is 12 volt (c) maximum power that can be dissipated in the external resistance is 18 watt (d) short circuit current from the cell is infinite Ans. [a,b,c]

Sol.

E r

4Ω

2

4

+ rE × 4 = 16

E r

1Ω

2

1

+ rE × 1 = 16

25 / 31



2)4(4r+

= 2)1(1r+

2 = rr

++

14

2 + 2r = 4 + r r = 2Ω (a)

2

24

+E × 4 = 16

6E = 2

E = 12 (b) Max power ⇒ when R = r

∴ Pmax = 2

2

RE × R =

rE4

2

= 241212

×× = 18 watt. (c)

Short circuit current = rE =

212 = 6 Amp

∴ Ans (a, b, c) 49. Two solid spheres A and B of equal volumes but of different densities dA and dB respectively, are connected

by a string. They are fully immersed in a fluid of density dF. They get arranged in an equilibrium state as shown in the figure with the tension in the string. The arrangement is possible only if

A

B

(a) dA < dF (b) dB > dF (c) dA > dF (d) (dA + dB) = 2dF Ans. [a,b,d]

Sol.

VdFg

T VdAg

VdBg

VdFg

For equilibrium of entire system 2VdFg = VdAg + VdBg

2dF = dA + dB

To keep the string having tension, dF > dA & dB > dF

26 / 31



50. A particle Q is moving along + Y axis. Another particle P is moving in XY plane along a straight line x = –d(d > 0) with a uniform speed v parallel to that of Q. At time t = 0, particles P and Q happen to be along X axis whereas a third particle R situated at x = + d starts moving opposite to P with a constant acceleration a. At all further instants the three particles happen to be collinear. Then Q

(a) has an initial speed 2v

(b) will come to rest after a time interval av

(c) has an acceleration –2a

(d) will return to its initial position after a time interval av2

Ans. [a,b,c,d]

Sol. y – y1 = )(–

112

12 xxxxyy

−−

y – vt = )(2

21 2

dxd

vtat+

−−

on differentiating for R (x = 0)

v′ – v = – )(2

dd

vat −− ⇒ v′ – v = 2

vat −− ....(1)

(i) at t = 0 ; v′ – v = 2

0 v− ⇒ v′ = 2v

(ii) for v′ = 0; 0 – v = 2

vat −− ⇒ t = av

(iii) diff (1) a′ – 0 = – 2a ⇒ a′ = –

2a

(iv) will return to its position after time av2

27 / 31



PART B MARKS : 60 All question are compulsory. All question carry equal marks.

1. (a) A 40 watt, 120 volt incandescent bulb has a tungsten filaments 0.381 m long. The diameter of the

filament is 33 µm. Tungsten has a resistivity 5.51 × 10–8 ohm-m at room temperature (20ºC). Given that the resistivity of the tungsten filament varies as T6/5, estimate the temperature of the filament when it is operated at its rated voltage.

(b) Assume that the electrical power dissipated in the filament is radiated from the surface of the filament. If emissivity of the filament surface is 0.35, determine the temperature of the filament and compare it with that obtained in part (a)

Sol. (a) Resistance at rating = 40

)120( 2 = 360 Ω

Initial resistance = Alρ = 26

8

)105.16(381.01051.5

−

−

×××

π = 24.55 Ω

R = ALρ

ρ = L

RA = 26 )103.3(4381.0

360 −×π

×

ρ = 8.08 × 10–7 ohm-m ρ ∝ T6/5

0ρ

ρ = 5/6

0TT

8

7

105.51008.8

−

−

×× =

5/6

293I

T = 2764 K (b) eσΑΤ4 = 40

T4 = )381.0)(105.16(21067.535.0

4068 −− ×××× π

= 51 × 1012

T = 2672 kelvin 2. A parallel plate capacitor with a separation d = 1.0 cm is subjected to a potential difference of 20 kV with air

as a dielectric. Assume that air behaves as a dielectric (insulator) upto a maximum electric field (called dielectric strength) of 30 kV / cm (after which is breaks down). Now, a thin plate of glass (dielectric constant K = 6.5 and dielectric strength = 290 kV/ cm) is inserted. Determine the maximum thickness of glass plate to avoid breakdown in the capacitor.

Sol.

V1 V2

k=1

t

k=6.5

d –t

20 kV

t

28 / 31



2

1

VV =

1

2

CC V1 + V2 = 20 kV

V1 = kV20CC

C

21

2 ×+

V2 = kV20CC

C

21

1 ×+

E1 = td

V1

− E2 =

tV2

kV30td

V1 <−

kV290t

V2 <

C1 = td

A0

−ε

C2 = t

A5.6 0ε

)td(t5.6

CC

1

2 −×=

V2 = 20

CC1

1

1

2×

+=

)td(5.6t20t

−+× =

t5.5d5.6t20

−

V1 = 1

CC

201

2

1 +

× = 1

)td(5.6t

201

+−

× = t5.5d5.620)td(5.6

−×−

290t

V2 <

290t5.5d5.6

20<

−

2 < 188.5d – 159.5t 159.5t < 186.5

t < 5.1595.186

13 < 19.5d –16.5t 16.5t < 19.5 –13

t < 5.165.6

t < 0.394 0.394 cm Ans.

3. (a) Cauchy's empirical formula for refractive index of a transparent medium is n – 1 = A

+ 21

λΒ

Hence obtain the condition for achromatic combination of two lenses made from different glasses. Refractive indices of flint glass and crown glass are given below.

Red light λ = 640 nm Blue light λ = 480 nm Flint glass 1.644 1.664

Crown glass 1.514 1.524 (b) Determine the focal length of the two lenses (one of flint glass and the other of crown glass) such that

their combination has focal length of + 40 cm for all colours.

29 / 31



Sol. f1 =

−

−

21

111

1RR

µ

loge1 – logef = log (µ – 1) + log

−

21

11RR

fdf− =

1−µdµ

fdf− = ω ⇒ dispersive power of material

eqf1 =

1

1f

+ 2

1f

As feq should not change with wavelength for achromatic combination

∴ 0. =λd

dfeq

0 = – 22

22

1

1

fdf

fdf

−

1

1

fω +

2

2

fω = 0

ω1 =

−

+−

12

Rv

Rv

µµµµ =

−

+−

12

644.1664.1644.1664.1 =

654.002.0 = 0.03058

ω2 = 1

2514.1524.1

514.1524.1

−+

− = 519.001.0 = 0.01927

401 =

1

1f

+ 2

1f

....(1)

1

03058.0f

+ 2

01927.0f

= 0 .....(2)

⇒ f1 = –1.58692f2

⇒ 401 =

258692.11

f− +

2

1f

⇒ 401 =

−

58692.1111

2f

⇒ f2 = 40 × 0.36985 = 14.794 cm f1 = –23.47689 cm 4. Obtain an expression for the magnetic moment associated with a solenoid of length L and number of turns N

carrying I. The inner and outer radii of the solenoid are r1 and r2.

Sol.

x

dx Area

dM = AidxN××

l

30 / 31



M = ll

×NiA = NiA

Magnetic moment of solenoid of area A = NiA

dr

r

No. of turns in element = 12 rr

drN−×

Magnetic moment = ∫−× 2

1

2

12

r

r

drrrrNI

π

M =

−−

××3

31

32

12

rrrr

NI π

5. One end of a string is attached to a rigid wall at point O, passes over a smooth pulley and carries a hanger S

of mass M at its other end. Another objects P of mass M is suspended from a light ring that can slide without

friction, along the string, as shown is figure. OA is horizontal. Find the additional mass to be attached to the

hanger S so as to raise the object P by 10 cm.

P

S

340 cm•

O A

Sol.

T T

M M

θ θ

340

320 • P

Q

initially

2T cos θ = Mg 2(Mg) cos θ = Mg cos θ = 1/2 θ = 60°

31 / 31



tan 60 = PQ

320

PQ = 20 cm

10

P′

MM + m

Q′

θ

320 •

1300

Now P′Q′ = 20 – 10 = 10 cm

2T cos θ = Mg

2(M + m)g × 130010 = Mg

2(M + m) × 131 = M

2M + 2m = 13 M 2m = M( 13 – 2)

m = 2

)213(M − = 0.9 M

Physical constants you may need –

1. Charge on electron e = 1.6 × 10–19C

2. Mass of an electron me = 9.1 × 10–31 kg

3. Universal gravitational constant G = 6.67 × 10–11 N m2/kg2

4. Permittivity of free space ε0 = 8.85 × 10–12 C2 / N m2

5. Gas constant R = 8.31 J/ K mol

6. Planck constant h = 6.62 × 10–34 Js

7. Stefan constant σ = 5.67 × 10–8 W/ m2 K4

8. Boltzmann constant k = 1.38 × 10–23 J/K

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