nsep solved paper 2011

8/10/2019 NSEP Solved Paper 2011

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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 CORRECTN.B. Physical constants are given at the end

SUB-PART A-11. 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 nbe the number density of

electrons. The electron current densityJe, is given byJe= eneEx+ eDedx

dn. The dimensions of electron drift

mobility (e) and electron diffusion coefficient (De) are respectively.(a) [M 1T 2 I1] and [L2T 1] (b) [M1T 2 I 1] and [M1L2 T 1](c) [M 1T2 I1] and [L2T 1] (d) [M 1T2 I2] and [L1T 2I1]

Ans. [c]

Sol. dx

dneDEenJ exee

4

232 1)(

)(

LDIT

IT

MLTITLIL e

e

e= [M1I T2]

De= [L2T 1]

2. A metal sample carrying a current alongXaxis with density Jxis subjected to a magnetic field Bz(alongZaxis). The electric fieldEy(Hall field) developed along Yaxis is directly proportional toJxas well asBz. Theconstant of proportionality (Hall coefficient) has SI unit(a) C/m2. (b) m2s/C (c) m2/C (d) m3/C

Ans. [d]

Sol.

xe

zB

Jx=A

IJx

I =JxAevdBz= eEy

Ey= vdBzneAvd= JxA

ne

BJE zxy

Constant of proportionality =ne

1=

C

m3

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3. A vibratory motion is represented byx= 2Acos t+A cos

2

3cos

2)cos(

2t

AtAt . The

resultant amplitude of the motion is

(a)2

9A (b)

2

5A (c)

2

5A (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 displacementsasFs 1/3. Therefore, the power deliveredby 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= 3/1sm

k

dssmk

vdv 3/1

2

3

2

3/22 s

m

kv . (i)

vs1/3

P=Fvs 1/3s1/3P s0

5. Bamboo strips are hinged to form three rhombi as shown. PointA0is fixed to a rigid support. The lengths of

the side of the rhombi are in the ratio 3 : 2 : 1. Point A3is pulled with a speed v. Let vAland vA2be the speeds

with which pointsA1andA2move. Then, the ratio vAl: vA2is -

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

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ns.

Sol.

3x 3x2x 2x x x

6x cos

4xcos

2xcos

x

As per constant relation

5

3

2

1 A

A

v

v

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.

60

v

v

Change = vv

2

60sin2 , Imp. = p= mv

7. Two chambers of different volumes, one containing m1gof a gas at pressurep1and other containing m2gofthe same gas at pressure p2 are joined to each other. If the temperature of the gas remains constant thecommon pressure reached is

(a)21

2211

mm

pmpm

(b)21

1221

mm

pmpm

(c)22

21

2121 )(

mm

pppm

(d)

1221

2121 )(

pmpm

ppmm

Ans. [d]

Sol.

m1g m2g

M

mn 11 ;

M

mn 22

No. of mole remain constant

RT

pV

RT

pV

RT

Vp

RT

Vp 212211

p=21

2211

VV

VpVp

p1 V1= RTM

m1 ; p2V2 = RTM

m2

2

2

1

1

21

p

RTm

p

RTmRTmRTm

p

1221

2121 )(

pmpm

mmpp

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. wo qu rops o equa ra are a ng roug a r w e erm na ve oc y v. ese wo ropscoalesce to form a single drop, its terminal velocity will be

(a) v2 (b) 2v (c) v3 4 (d) v3 2Ans. [c]

Sol. Terminal velocity, 2rvT Let radius of single drop = r

and radius of bigger drop =R

3

4

3

42

33 Rr

R= (2)1/3rAs vTr

223/12' )2( rRvT

3/13/1' )4(

1

)2(

1

T

T

v

v

TT vv3' 4

9. An elevator of mass Mis accelerated upwards by applying a force F. A mass minitially 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)mgF

M

2

(c)F

M2 (d)

MgF

M

2

Ans. [d]

Sol.m

Fae

gas

as/e= as ae=M

MgFg

m

F

2

2

1atuts

2

2

11 t

M

MgF

MgF

Mt

2

10. The formation of solid argon is due to vander Waals bonding. In this case the potential energy as a functionof interatomic separation can be written as (Lennard Jones 6-12 potential energy)E(r) = Ar6+Br12whereAandBare constant > Given that A= 8.0 10 77Jm6andB= 1.12 10 133Jm12, the bond length for solidargon is(a) 3.75 nm (b) 0.0375 nm (c) 0.750 nm (d) 0.375 nm

Ans. [d]

Sol. 0126)( 137

BrArdr

rdE

0126

137 r

B

r

A

A

B

A

Ar

2

6

126 =77

133

108

1012.12

= 0.28 10 56

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

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. et an e t e po nts respect ve y a ove an e ow t e eart s sur ace eac at a stance equa to a t eradius of the earth. If the acceleration due to gravity at these points begAandgBrespectively, thengB:gA(a) 1 : 1 (b) 9 : 8 (c) 8 : 9 (d) zero

Ans. [b]

Sol.22 9

4

2

R

GM

RR

GMgA

233R

R

GM

R

GMrgB = 22R

GM

8

9

4

9

2

1

A

B

g

g

12. Let vrms, v* and vavgrepresent 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* : vavgis(a) 8 : 3: 2 (b) 8 : 2: 3 (c) 3: 2: 8 (d) 3 : 2 : 8

Ans. [c]

Sol. M

RTvrms

3

M

RTv pm

2..

M

RTvav

8

.

vrms:vm.p.:vav.= 3: 2: 8

13. In the arrangement of resistance shown below, the effective resistance between pointsAandBis

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

Sol.

A

12

B

3

1

2

4

6

5

247

25

24

1215

30

3025

15

10

1010

10B

A

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10

1056

12

12

24

25

3

425

15/2

73012

B

3010

10

A

24

5 15

15/2

25/2

6

12

5

BA

5 15

15/2

12

5 BA6

Req.= 19

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 att= 0 s, neglecting

viscous effects, it will reach the water surface at tequal 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

aVg

gVg

gV ss

55

a

gg

s

s

5

a)12(4.0

6.0

a= 18 m/s2

2

2

1atuts

2182

11 t

9

12 t st3

1

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15. T e max mum tens on n t e str ng o a s mp e pen u um s 1.2 t mes t e m n mum tens on. I 0 s t eangular amplitude, then 0is

(a)

5

4cos 1 (b)

4

3cos 1 (c)

16

15cos 1 (d)

8

7cos 1

Ans. [c]

Sol.

T20

T1

v

2

1mv

mgT

)cos1(21

gm

mgT

T1= 3mg 2 mg cos T2= mg cos

5

6

2

1 T

T

5

6

cos

cos23

mg

mgmg

15 mg 10 mg cos = 6 mg cos 15 mg = 16 mg cos cos = 15/16

16. A uniform line charge with density = 50 C/m lies alongXaxis. The electric flux per unit length crossingthe portion of the planez = 3 m bounded byy= 3 m is(a) 4.68 C/m (b) 9.36 C/m (c) 50 C/m (d) 18.7 C/m

Ans. [ ]

Sol.

dx

x

3

x = 0

dA

2

0

42

x

z

223 x

cosEdAd

=22

0

3

92 xyx

dx

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2

22

0 )9(

3

x

dxd =

0

1

0 3tan

3

32

x

3

2tan2 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 thequestion is wrong.

17. A plane mirror perpendicular toXYplane makes an angle of 30 with theXaxis. 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 theXYplane. Thecoordinates of the image are :

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

Sol.

x

A

30

30

30

xI

(20, 0)

20

2030

sin 30 =20

x

1020

2

1x

cos 30 =20

y

310202

3y

10x ,y= 10 3

18. Magnetic flux through a stationary loop with a resistance R varies during the time interval as = at( t)where ais a constant. The amount of heat generated in the loop during the time interval is

(a)R

a

6

32 (b)

R

a

4

32 (c)

R

a

3

32 (d)

R

a

2

32

Ans. [c]

Sol. atadt

de 2

R

e

dt

dH 2 =

R

ata 2)2(

dtR

ataH

0

2)2(=

0

3

)2)(3(

)2(

aR

ata=

Ra

aaa

)6(

)()2( 33

=Ra

aa

)6)(1(

3333

=R

a

3

32

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

(III)

ty 34

3cos5 , (IV)y= 1 + t+ 2t2. 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 magnetplaced 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.

Bin 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 nthorbit is

proportional to -

(a)2

1

n (b)

3

1

n (c)

4

1

n (d)

5

1

n

Ans. [d]

Sol. B=R

I

20 =

R

R

ev

2

20

B2R

v

22 )(

)/1(

n

n

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B5

1

n

Q.22 A particle of mass mcarries a charge +q. It enters into a region of uniform magnetic field

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

(a) (2)qB

m (b) infinite as the particle gets trapped

(c) 2

qB

m (d) (+ 2)

qB

m

Ans. [d]Sol.

90 90

90

90

C

B

+q

l l'

Time spend =w

2=

qB

m)2( .

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

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

Sol. U2F=2

1(2)(v)2

Vcommon=82

082

V

=5

V

'2 FU =2

1(2)

2

5

V

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Change =U

' 100 = 96%

Q.24 An infinite number of charges each equal to 0.2 C are arranged in a line at distances 1, 2, 4, 8 meter froma 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 = k0.2 610.......8

1

4

1

2

1

1

1

V=

2/11

1 9 109 0.2 10 16

V= 3600 volt

Q.25 A ball of mass mmoving with a speed ualong a direction making an angle with the vertical strikes ahorizontal steel plate. The collision lasts for a time interval t. If eis the coefficient of restitution between the

ball and the plate, the average force exerted by the plate on the ball is -

(a) temu (b) temu cos (c) tmue cos)1(2 (d) tmue cos)1(

Ans. [d]

Sol.

vu

e=

cos

cos

u

vvcos = eu cos (i)

usin = vsin .. (ii)

Av. force =tp

t

uvm ]coscos[

t

emu )1(cos

Q.26 In a Young's double slit experiment sources of equal intensities are used. Distance between slits is d andwavelength of light used is (


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2 = 2 + 2 coscos = 0

=2

x2

=2

x =4

D

dx=

4

x=d

D

4

=D

x2=

D

2

d

D

4

=

d2

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

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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 ifF> mg

The block will topple if Fh> mg

2

a or F>

h

mga

2

The sliding will occur earlier if mg < h

mga

2 < ha

2

& topping will occur earlier if >h

a

2

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 areemitting 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

wherex= 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 fromthe sodium lamp in the laboratory is found to be 589 nm. Therefore, the star is moving with a speed of about -

(a) 2.7 106m/s away from the earth (b) 5.4 106m/s towards the earth(c) 1.6 106m/s away from the earth (d) 4.6 106m/s away from the earth

Ans. [d]Sol. = 589

' = appearant = 598 at is higherthan original some is moving away.

'

'

=C

v

598

589598=

8103

v

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V=598

9103

= 4.6 106m/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

212ffff

(c)

2

22

21 ff (d) 21ff

Ans. [d]

Sol.1f

Z =2f

Z

R is same in both series

so,1f

X =2f

X

1

)( fCL XX = 2)( fLC XX

1

)( fLX + 2)( fCX = 2)( fCX + 1)( fCX

2L(f1+f2) =

21

11

2

1

ffC

2L(f1+f2) =C2

1

21

21

ff

ff

42LC =21

1

ff

2 LC =21

1

ff

Resonant frequency =LC2

1= 21ff

Q.32 Two particles are moving alongX and Yaxes towards the origin with constant speeds uand vrespectively. Attime t= 0, their respective distances from the origin are xandy. 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)

vu

yx

222

Ans. [c]

Sol.

utx ut

vt

vt

Z2= (x ut)2+ (y vt)2

2Zdt

dZ= 2u(x ut) + 2v(y vt) = 0

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ux u t+ vy v t=

t=22 vu

vyux

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

4.0 105Pa. After some time, due to leakage, the gauge pressure drops to 3.0 105Pa and the temperature to

27C. 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

RT

VP=

3203.8

105 4

n2=3003.8

104 4

n= n1 n2=3.8

104

300

4

320

5

3.8

104

320300

12801500

2

4

103.83320

10220

2.76 mole

Now of N2leaked out = 2.76 28

77.28 gmMost 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 2tis -

(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 Sis placed in position l. At t= 5 s, contact is changedfrom 1 to 2. The voltage across the capacitor is measured at t= 5 s and at t= 6 s. Let these voltages be V1and

V2respectively. Then, V1and V2respectively are -

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(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 startedAt t= 5 sec Voltage across capacitor

Vc = V0 RC/te1

= 10

63 101001020

5

e1

= 10 5.2e1

= 9.18 voltAt t= 5 sec discharging is startedAt t= 6 sec capacitor has been discharged

for 1 sec so

Vc = V0RC/te

= 9.1863 101001010

1

e

= 9.18 e= 9.18 0.37= 3.37 volt

Q.36 There are two thermocouplesAandBmade 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 1and higher temperature 2. Each of the two thermocouples is connected to the same microammetersuccessively. Let be the thermoemf and I be the thermoelectric current. Then which of the followingstatements is true ?

(a) both andIare equal for A and B (b) Both andIare greater for B than those for A(c) is the same for both butIis greater for A (d) is the same for both butIis 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 1and 2. Thede Broglie wavelength of each of the particles in their centre of mass frame will be -

(a)2

22

21 (b)

221 (c)

21

212

(d)22

21

212

Ans. [D]

Sol. Let m be the mass of each particle and thus verticals are v1and v2respectively

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v2

v1y

x

22

22

)(

v

m

mv

v xcm

2211

)(

v

m

mvv ycm

2222

2)(2

vvvv xcm

21

)(2

vv ycm

So magnitude of velocity of 2ndparticle is C.O.M frame becomes44

21

22 vv =

2

22

21 vv

=1mv

hv1=

1mh

2=2mv

hv2=

2mh

So De broglie wavelength of 2ndpartical in COM frame is

=

2

22

21 vvm

h

=

22

22

2

21

2

2

2

nh

m

hm

h=

22

21

11

2

2

2

2

1

212

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=)(

12400

eV

I case

E1=4000

12400 = 3.1 eV

Vstopping= 500 m KEof emitted e = 0.5 eV

Ephoton= W+KEe

Work function w = 2.6 eV

II case

Vstopping= 800 mv eKE = 0.8 eV

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Ephoton= W+ eKE

= 2.6 + 0.8

= 3.4 eV

Ephoton=)(

12400

eV

3.4 =

12400

= 3650 = 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 1Ans. [b]

Sol.

P

Z

y

A

B

A B X Y Z P y

0 0 1 1 0 0 01 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)2

1

n (b)

3

1

n (c) n2 (d)

4

1

n

Ans. [b]

Sol. En= 2

26.13

n

Z

E(n 1)= 2

2

)1(

6.13

n

Z

E = 13.6Z2

22

1

)1(

1

nn

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13.6Z2

22

22

)1(

)1(

nn

nn

13.6Z2

22 )1(

12

nn

n

31n as (n >> 1).

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u - art -

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 magneticmoment 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.momentangular

momentmagentic=

m

q

2

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

e

2

42. A current I0 enters into a parallel combination of resistors R

1 and R

2. Current I

1 flows through R

1 and I

2

through R2. The currentI0distributes in such a way that

(a) power consumed inR1and inR2is the same

(b) total power consumed inR1andR2is minimum

(c)I1is proportional ofR2andI2is proportional toR1

(d) the power consumed in each ofR1andR2is 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)

2R2

1

dI

dP= 0 for minima

2I1R1+ 2(I0 I1) (1)R2= 0

I1(R1+ R2) I0R2= 0

I1=21

20

RR

RI

It mean total power is minima.

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21/31

. e g t o a o y on t e sur ace o t e eart epen s 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 onp-Vdiagram is as shown below. The same process can be shown onp-Tor V-Tdiagrams.Choose the correct alternative /s.

1 2

3

4

V

p

(a)

p

T

3

21

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

4V

PIsothermal

Isothermal

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22/31

T

3

21

4

V

T

2

34

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

f

1=

v

1

u

1

f

1=

5.7

1

30

1

f

1=

5.7

1+

30

1

f

1=

10

1

f= 10

lens is diverting asfis in refraction zone

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23/31

+

mage ue to re ect

O

30

R1

f

1=

u

1+

v

1

f

1=

30

1+

6

1

f

1=

5

1

f= 5

21R = 5

R1= 10 cm

30

O

R2

f

1=

30

1+

10

1

f

1=

30

4

f=4

30

22R =

4

30

R2= 15 cm

+

f

1=

1

12

n

nn

21

1

1

RR

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24/31

10

1=

1

12n 15

1

10

1

10

1= n2 1

30

)23(

10

1=

30

)1( 2n 5

5

3= n21

n2= 1.6

f

1=

2

26.1

15

1

10

1

f

1=

30

5

2

4.0

f=2

60= 30 cm

47. Let n1and n2moles of two different ideal gases be mixed. If ratio of specific heats of the two gases are 1and2respectively, then the ratio of specific heats of the mixture is given through the relation

(a) (n1+ n2)= n11+ n22 (b)1

)( 21 nn

=11

1

n

+12

2

n

(c) (n1+ n2)1

= n1

11

1

+ n212

2

(d) (n1+ n2)( 1) = n1(1 1) + n2(2 1)

Ans. [b,c]

Sol. =221

221

1

1

VV

pp

CnCn

CnCn

=

11

11

2

2

1

1

2

22

1

11

nn

nn

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

4

2

4

r

E 4 = 16

E r

1

2

1

r

E 1 = 16

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25/31

2)4(

4

r=

2)1(

1

r

2 =r

r

1

4

2 + 2r = 4 + rr = 2 (a)

2

24

E 4 = 16

6

E= 2

E = 12 (b)Max power when R = r

Pmax=2

2

R

E R =

r

E

4

2

=24

1212

= 18 watt. (c)

Short circuit current =r

E=

2

12= 6 Amp

Ans (a, b, c)

49. Two solid spheres A and B of equal volumes but of different densities dAand dBrespectively, are connectedby a string. They are fully immersed in a fluid of density dF. They get arranged in an equilibrium state asshown 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) = 2dFAns. [a,b,d]

Sol.

VdFg

TVdAg

VdBg

VdFg

For equilibrium of entire system 2VdFg = VdAg+ VdBg

2dF= dA+ dB

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

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26/31

. part c e s mov ng a ong + ax s. not er part c e s mov ng n p ane a ong a stra g t nex= d(d> 0) with a uniform speed vparallel to that of Q. At time t= 0, particles P and Q happen to be alongX axis whereas a third particle R situated at x= + dstarts moving opposite to P with a constant accelerationa. At all further instants the three particles happen to be collinear. Then Q

(a) has an initial speed2

v

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

v

(c) has an acceleration 2

a

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

v2

Ans. [a,b,c,d]

Sol. y y1= )(

112

12 xxxx

yy

y vt = )(2

2

1 2

dxd

vtat

on differentiating for R (x = 0)

v v = )(2

dd

vatv v =

2

vat....(1)

(i) at t = 0 ; v v =2

0 vv=

2

v

(ii) for v= 0; 0 v =2

vatt =

a

v

(iii) diff (1) a 0 = 2

aa=

2

a

(iv) will return to its position after time

a

v2

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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 thefilament is 33 m. Tungsten has a resistivity 5.51 10 8ohm-m at room temperature (20C). Given that the

resistivity of the tungsten filament varies as T6/5

, estimate the temperature of the filament when it is operatedat its rated voltage.(b) Assume that the electrical power dissipated in the filament is radiated from the surface of the filament. Ifemissivity of the filament surface is 0.35, determine the temperature of the filament and compare it with thatobtained in part (a)

Sol. (a) Resistance at rating =40

)120( 2= 360

Initial resistance =A

=

26

8

)105.16(

381.01051.5

= 24.55

R =A

L

= LRA = 26 )103.3(

4381.0360

= 8.08 10 7ohm-mT6/5

0

=

5/6

0T

T

8

7

105.5

1008.8

=

5/6

293

I

T = 2764 K(b) e

T4

= )381.0)(105.16(21067.535.0

4068

= 51 1012

T = 2672 kelvin2. 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 (calleddielectric strength) of 30 kV / cm (after which is breaks down). Now, a thin plate of glass (dielectric constantK = 6.5 and dielectric strength = 290 kV/ cm) is inserted. Determine the maximum thickness of glass plate toavoid breakdown in the capacitor.

Sol.

V1V2

k=1

t

k=6.5

d t

20 kV

t

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28/31

2

1

V

V=

1

2

C

C V1+ V2= 20 kV

V1= kV20CC

C

21

2

V2= kV20CC

C

21

1

E1=

td

V1

E

2=

t

V2

kV30td

V1

kV290t

V2

C1=td

A0

C2=t

A5.6 0

)td(t

5.6

C

C

1

2

V2= 20

C

C

1

1

1

2

=)td(5.6t

20t

=t5.5d5.6

t20

V1=1

C

C201

2

1

=

1)td(5.6

t201

=

t5.5d5.6

20)td(5.6

290t

V2

290t5.5d5.6

20

2 < 188.5d 159.5t159.5t < 186.5

t

nsep solved paper 2011

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