MHT-CET 2016 : Physics - Actual Test Paper

(Solution at the end)

This is a Simulator Test. Do not attempt this Test Paper until you are ready to take this Test under strict

examination conditions.

IMPORTANT INSTRUCTIONS

1. The test is of 45 Min duration 2. The Test consists of 50 questions. The maximum marks are 50. 3. Each question is allotted 1 (one) mark for each correct response

4. There is No Negative Marking. No mark shall be awarded for marking two or more answers of same question, scratching or overwriting.

5. There is only one correct answer for each question. More than one answer indicated a question will be deemed as incorrect response and will be negatively marked.

6. All calculations / writing work are to be done on the, space provided for this purpose in the Test Booklet itself. 7. Use of Electronic / Manual Calculator and Whitener is prohibited. 8. Use Blue / Black Ball Point Pen only for writing particulars / marking responses on the OMR Sheet. (OMR Sheet attached at the end)

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(2) CET Score Booster : Physics

1. For a gas v

R

C= 0.4, where „R‟ is the universal gas constant and „Cv‟ is molar specific heat at constant

volume. The gas is made up of molecules which are

(A) rigid diatomic (B) monoatomic (C) non-rigid diatomic (D) polyatomic

2. In vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point

is

(A) 5 (B) 2 (C) 0.5 (D) 0.2

3. Two wires having same length and material are stretched by same force. Their diameters are in the ratio

1 : 3. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter) when

stretched is

(A) 3 : 1 (B) 9 : 1 (C) 27 : 1 (D) 81 : 1

4. A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have

same mass and total kinetic energy of the ring is 4 J then total kinetic energy of the disc is

(A) 3 J (B) 4 J (C) 5 J (D) 6 J

5. When the observer moves towards the stationary source with velocity, „V1‟, the apparent frequency of

emitted note is „F1‟. When the observer moves away from the source with velocity „V1‟, the apparent

frequency is „F2‟. If „V‟ is the velocity of sound in air and 1

2

F2

F then

1

V?

V

(A) 2 (B) 3 (C) 4 (D) 5

6. Wire having tension 225 N produces six beats per second when it is tuned with a fork. When tension

changes to 256 N, it is tuned with the same fork, the number of beats remain unchanged. The frequency

of the fork will be

(A) 186 Hz (B) 225 Hz (C) 256 Hz (D) 280 Hz

7. Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be

shown that pressure is

(A)

rd1

3

kinetic energy per unit volume of a gas(B)

rd2

3

kinetic energy per unit volume of a gas

(C)

th3

4

kinetic energy per unit volume of a gas(D) 3

2 kinetic energy per unit volume of a gas

8. A mass „m1‟ connected to a horizontal spring performs S.H.M. with amplitude „A‟. While mass „m1‟ is

passing through mean position another mass „m2‟ is placed on it so that both the masses move together

with amplitude „A1‟. The ratio of 1A

A is (m2 < m1)

(A)

1

21

1 2

m

m m

(B)

1

21 2

1

m m

m

(C)

1

22

1 2

m

m m

(D)

1

21 2

2

m m

m

9. A particle moves along a circle of radius „r‟ with constant tangential acceleration. If the velocity of the

particle is „v‟ at the end of second revolution, after the revolution has started then the tangential

acceleration is

(A) 2v

8 r (B)

2v

6 r (C)

2v

4 r (D)

2v

2 r

MHT-CET 2016 : Physics Paper (3)

10. Two strings A and B of same material are stretched by same tension. The radius of the string A is double

the radius of string B. Transverse wave travels on string A with speed „VA‟ and on string B with speed

„VB‟. The ratio A

B

V

V is

(A) 1

4 (B)

1

2 (C) 2 (D) 4

11. The bob of a simple pendulum performs S.H.M. with period „T‟ in air and with period „T1‟ in water.

Relation between „T‟ and „T1‟ is (neglect friction due to water, density of the material of the bob is =

39

8 kg/m3 , density of water = 1 g/cc)

(A) T1 = 3 T (B) T1 = 2 T (C) T1 = T (D) T1 = T

2

12. In a capillary tube of radius „R‟, a straight thin metal wire of radius „r‟ (R > r) is inserted symmetrically

and one end of the combination is dipped vertically in water such that the lower end of the combination

is at same level. The rise of water in the capillary tube is

[T = surface tension of water,

(A) T

(R r) g (B)

R g

2T

(C)

2T

(R r) g (D)

(R r) g

T

13. When open pipe is closed from one end then third overtone of closed pipe is higher in frequency by

150 Hz than second overtone of open pipe. The fundamental frequency of open end pipe will be

(A) 75 Hz (B) 150 Hz (C) 225 Hz (D) 300 Hz

14. A disc of radius „R‟ and thickness R

6 has moment of inertia „I‟ about an axis passing through its centre

and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a

sphere about its diameter is

(A) I

5 (B)

I

6 (C)

I

32 (D)

I

64

15. Let a steel bar of length „l‟, breadth „b‟ and depth „d‟ be loaded at the centre by a load „W‟. Then the sag

of bending of beam is (Y = Young‟s modulus of material of steel)

(A) 3

3

W

2bd Y (B)

3

3

W

4bd Y (C)

2

3

W

2bd Y (D)

3

2

W

4bd Y

16. Which of the following quantity does NOT change due to damping of oscillations ?

(A) Angular frequency (B) Time period (C) Initial phase (D) Amplitude

17. If the end correction of an open pipe is 0.8 cm then the inner radius of that pipe will be

(A) 1

3 cm (B)

2

3 cm (C)

2

cm (D) 0.2cm

18. A progressive wave is represented by y = 12 sin (5t – 4x) cm. On this wave, how far away are the two

points having phase difference of 90° ?

(A) 2

cm (B)

4

cm (C)

8

cm (D) cm

16

19. Two particles of masses „m‟ and „9m‟ are separated by a distance „r‟. At a point on the line joining them

the gravitational field is zero. The gravitational potential at that point is (G = Universal constant of

gravitation)

(A) 4Gm

r (B)

8Gm

r (C)

16Gm

r (D)

32Gm

r

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20. A black rectangular surface of area „A‟ emits energy „E‟ per second at 27°C. If length and breadth are

reduced to rd

1

3of initial value and temperature is raised to 327°C then energy emitted per second

becomes

(A) 4E

9 (B)

7E

9 (C)

10E

9 (D)

16E

9

21. A liquid drop having surface energy „E‟ is spread into 512 droplets of same size. The final surface energy

of the droplets is

A) 2E (B) 4E (C) 8E (D) 12E

22. Let „M‟ be the mass and „L‟ be the length of a thin uniform rod. In first case, axis of rotation is passing

through centre and perpendicular to the length of the rod. In second case axis of rotation is passing

through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to

second case is

(A) 1 (B) 1

2 (C)

1

4 (D)

1

8

23. A simple pendulum of length „ ‟ has maximum angular displacement „‟. The maximum kinetic energy

of the bob of mass „m‟ is (g = acceleration due to gravity)

(A) mg (1 + cos) (B) mg (1 + cos2) (C) mg (1 – cos) (D) mg (cos – 1)

24. Angular speed of hour hand of a clock in degree per second is

(A) 1

30 (B)

1

60 (C)

1

120 (D)

1

720

25. The value of gravitational acceleration „g‟ at a height „h‟ above the earth‟s surface is g/4 then

(R = radius of earth)

(A) h = R (B) h = R

2 (C)

Rh

3 (D)

Rh

4

26. In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire x

cm long. If the length of the potentiometer wire is increased without changing the cell, the balancing

length will (Driving source is not changed)

(A) increase (B) decrease (C) not change (D) becomes zero

27. An iron rod is placed parallel to magnetic field of intensity 2000 A/m. The magnetic flux through the rod

is 6 × 104 Wb and its cross-sectional area is 3 cm2. The magnetic permeability of the rod in Wb/Am

(A) 101 (B) 102 (C) 103 (D) 104

28. Alternating current of peak value 2

ampere flows through the primary coil of the transformer. The

coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced

in secondary coil is (Frequency of a.c. = 50 Hz)

(A) 100 V (B) 200 V (C) 300 V (D) 400 V

29. An electron of mass „m‟ has de-Broglie wavelength „‟ when accelerated through potential difference

„V‟. When proton of mass „M‟, is accelerated through potential difference „9V‟, the de-Broglie

wavelength associated with it will be (Assume that wavelength is determined at low voltage)

(A) M

3 m

(B)

M

3 m

(C)

m

3 M

(D)

m

3 M

30. Interference fringes are produced on a screen by using two light sources of intensities „I‟ and „9I‟. The

phase difference between the beams is 2

at point P and at point Q on the screen. The difference

between the resultant intensities at point P and Q is

(A) 2 I (B) 4 I (C) 6 I (D) 8 I

MHT-CET 2016 : Physics Paper (5)

31. Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area 'A

3' and the

separation between the plates is „d‟, „2d‟ and „3d‟ respectively. The equivalent capacity of combination

is (0 = absolute permittivity of free space)

(A) 07 A

18d

(B) 011 A

18d

(C) 013 A

18d

(D) 017 A

18d

32. In an oscillator, for sustained oscillations, Barkhausen criterion is A equal to (A = voltage gain without

feedback, = feedback factor)

(A) zero (B) 1

2 (C) 1 (D) 2

33. Light of wavelength „‟ which is less than threshold wavelength is incident on a photosensitive material.

If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then

stopping potential will

(A) increase (B) decrease (C) be zero (D) become exactly half

34. A ray of light travelling through rarer medium is incident at very small angle „i‟ on a glass slab and after

refraction its velocity is reduced by 20%. The angle of deviation is

(A) i

8 (B)

i

5 (C)

i

2 (D)

4i

5

35. The maximum frequency of transmitted radio waves above which the radio waves are no longer reflected

back by ionosphere is _______ (N = maximum electron density of ionosphere, g = acceleration due to

gravity)

A) gN (B) gN2 (C) g N (D) g2N2

36. In Bohr‟s theory of Hydrogen atom, the electron jumps from higher orbit „n‟ to lower orbit „p‟. The

wavelength will be minimum for the transition

(A) n = 5 to p = 4 (B) n = 4 to p = 3 (C) n = 3 to p = 2 (D) n = 2 to p = 1

37. Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. „V‟. If one of the

capacitor is completely filled with dielectric material of constant „K‟, then potential difference of the

other capacitor will become

(A) K

V(K 1) (B)

KV

K 1 (C)

K 1

KV

(D)

V

K(K 1)

38. The LC parallel resonant circuit

(A) has a very high impedance (B) has a very high current

(C) acts as resistance of very low value (D) has zero impedance

39. A galvanometer of resistance 30 is connected to a battery of emf 2V with 1970 resistance in series. A

full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10

divisions, the resistance in series required is

(A) 4030 (B) 4000 (C) 3970 (D) 2000

40. Two coherent sources „P‟ and „Q‟ produce interference at point „A‟ on the screen where there is a dark

band which is formed between 4th bright band and 5th bright band. Wavelength of light used is 6000A.

The path difference between PA and QA is

(A) 1.4 × 104 cm (B) 2.7 × 104 cm (C) 4.5 × 104 cm (D) 6.2 × 104 cm

41. The schematic symbol of light emitting diode is (LED)

(A) (B) (C) (D)

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42. The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is „W‟.

The work done in increasing it from 10V to 15V will be

(A) W (B) 0.6 W (C) 1.25 W (D) 1.67 W

43. Magnetic flux passing through a coil is initially 4 × 104 Wb. It reduces to 10% of its original value in „t‟

second. If the e.m.f. induced is 0.72 mV then „t‟ in second is

(A) 0.3 (B) 0.4 (C) 0.5 (D) 0.6

44. Resolving power of telescope increases when

(A) wavelength of light decreases (B) wavelength of light increases

(C) focal length of eye-piece increases (D) focal length of eye-piece decreases

45. When light of wavelength „‟ is incident on photosensitive surface, the stopping potential is „V‟. When

light of wavelength „3‟ is incident on same surface, the stopping potential is „V

6‟. Threshold

wavelength for the surface is

(A) 2 (B) 3 (C) 4 (D) 5

46. From Brewster‟s law, except for polished metallic surfaces, the polarising angle

(A) depends on wavelength and is different for different colours

(B) independent of wavelength and is different for different colours

(C) independent of wavelength and is same for different colours

(D) depends on wavelength and is same for different colours

47. Two particles X and Y having equal charges after being accelerated through same potential difference

enter a region of uniform magnetic field and describe a circular paths of radii „r1‟ and „r2‟ respectively.

The ratio of the mass of X to that of Y is

(A) 1

2

r

r (B) 1

2

r

r (C)

2

2

1

r

r

(D)

2

1

2

r

r

48. When an electron in Hydrogen atom revolves in stationary orbit, it

(A) does not radiate light though its velocity changes

(B) does not radiate light and velocity remains unchanged

(C) radiates light but its velocity is unchanged

(D) radiates light with the change of energy

49. The magnetic field (B) inside a long solenoid having „n‟, turns per unit length and carrying current „I‟

when iron core is kept in it is (0 = permeability of vacuum, = magnetic susceptibility)

(A) 0nI (1 ) (B) 0nI (C) 0nI2 (1 + ) (D) 0nI (1 + )

50. In balanced metre bridge, the resistance of bridge wire is 0.1/cm. Unknown resistance „X‟ is connected

in left gap and 6 in right gap, null point divides the wire in the ratio 2 : 3. Find the current drawn from

the battery of 5 V having negligible resistance.

(A) 1 A (B) 1.5 A (C) 2 A (D) 5 A

MHT-CET 2016 : Physics Paper (7)

Solution to MHT-CET 2016 : Physics - Actual Test Paper

ANSWER KEY

Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans. Q. No. Ans.

1. (A) 2. (D) 3. (D) 4. (A) 5. (B)

6. (A) 7. (B) 8. (A) 9. (A) 10. (B)

11. (A) 12. (C) 13. (D) 14. (A) 15. (B)

16. (C) 17. (B) 18. (C) 19. (C) 20 (D)

21. (C) 22. (B) 23. (C) 24. (C) 25. (A)

26. (A) 27. (C) 28. (B) 29. (C) 30. (C)

31. (B) 32. (C) 33. (A) 34. (B) 35. (C)

36. (D) 37. (B) 38. (A) 39. (C) 40. (B)

41. (B) 42. (D) 43. (C) 44. (A) 45. (D)

46. (A) 47. (D) 48. (A) 49. (D) 50. (A)

DETAILED SOLUTIONS

1. (A)

V

R

C = 0.4 P V

V

C C

C

= 0.4

P

V

C1

C = 0.4 P

V

C

C= 1.4 r = 1.4

As r = 1.4 the gas is diatomic in nature. For example air molecules.

2. (D)

K.E = 21mv

2

At lowest point in vertical circular motion. VL = 5rg and at highest point Vh = rg

h

L

K.E 1

K.E 5 = 0.2

3. (D)

Strain energy per unit volume u = U 1

v 2 Stress Strain

U 1 F

V 2 A

1

2 Y strain2

2 2

s s L

L L s

Strain Su Y A

u Y Strain L A

=

2 22 2L

2 2s

r 3

r 1

= 81 : 1

4. (A)

Total K.E of the rolling disc or ring is given by.

K.E = 2 21 1mv l

2 2

For ring and disk, translational kinetic energy 21mv

2 is constant.

Rolling K.E. of disc is 2 21mR

4

Rolling K.E of ring is 2 21mR

2

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As for ring, 4J = 2 2 21 1mv mR

2 2

2 2mR 4J

For disc 2 2 2 21 1mR mR

2 4 =

4 4

2 2

J = (2 + 1) J = 3J

5. (B)

1

2

f2

f , Speed of approach = Speed of leaving

The apparent frequency of sound by observer when it is approaching source is given by

f1 = 10

v vf

v

…(1)

When observer is moving away from source

f2 =1

0

v vf

v

…(2)

Here f0 is the frequency of sound.

Taking ratio of (1) with (2) we have

1 1

2 1

f v v

f v v

2 = 1

1

v v

v v

2v 2v1 = v + v1 2v v = 3v1 1

v

v = 3

6. (A)

For wire vibrating under tension, the fundamental frequency is given by

f1 = 1 T

2 , Where T is tension & is mass per unit length of the string.

It is given that, when it is sounded with a tuning fork of frequency x, 6 beats per seconds were heard

(x f1 ) = 6

f1 = 1 225

2L ; f2 =

1 256

2L

1

2

f 15

f 16

f2 = 1

16f

15 =

16(x 6)

15

f2 = (x + 6)

(x + 6) = 16

(x 6)15

15x + 90 = 16x 96

X = 186 Hz

7. (B)

The pressure exerted by the gas on the walls of container is

i.e. P = 1

3 C2 = 21 M

C3 V

= 22 1 MC

3 2 V

= 2

3[K.E. per unit volume]

8. (A)

If the velocity of mass m1 while passing through the mean position is and 1 is the velocity of

(m1 + m2) just after m2 is placed on m1, then by law of conservation of momentum we have,

m1 = (m1 + m2) 1

MHT-CET 2016 : Physics Paper (9)

1 1

1 2

m

m m

…(1)

Total energy initially is

21kA

2 = 2

1

1m

2 …(2)

Total energy after m2 is placed is

21

1kA

2 = 2

1 2 1

1(m m )

2 …(3)

By Eqs. (2) and (3):

1

21 1 1 2

1

A m m

A m

Putting value of 1

from Eq. (1),

1

21 1

1 2

A m

A m m

9. (A)

Using v2 u2 = 2as and u = 0 and s = 4r

2as = v2 2 2 2v v v

a2s 2 4 r 8 r

10. (B)

The velocity of wave travelling on string is

V = 2

T T 1 T

m rr

where is the density

V 1

r A B

B A

V r 1

V r 2

11. (A)

T = 2 effg

In air geff = g, where as in water

geff = g

3 3

3

910 10

8g9

108

= g

91

1 8 g8 g9 8 9 9

8

T1 = 2eff

9

g g T1 = 9 T 3T

12. (C)

h = 2T cos

g(R r)

For cos = 0, h = 2T

g(R r)

13. (D)

Let f0 = 0v

2L be the fundamental frequency of the open pipe

Its second overtone is 00

3v3f

2L

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Let f = 0v

4L be the fundamental frequency of closed pipe.

And third overtone of the organ pipe closed at one end only is, (f3)closed = 0v7

4 L

As given 0 07v 3V

4L 2L = 150

0v7 6

4 4 L

= 150

0v

4L = 150 0v

300 Hz2L

14. (A)

Volume of disc is 3

2 R RA d R

6 6

Moment of inertia of disc is I = 1

2MR2

When the disk is remolded in solid sphere of volume V having radius r, then

3

3R 4r

6 3

3R 3

6 4 = r3

r3 = 3R

8 r =

R

2

Moment of inertia of sphere is given by

2 2 2

22 2 R MR MR 1 Im r m

5 5 4 10 2 5 5

15. (B) 16. (C)

17. (B)

For open organ pipe, L = 1.2 r

r = L 1 0.8 2

1.2 1.2 3 cm.

18. (C)

y = 12 m(5t 4x).

Comparing in y = A sin (wt kx)

We have A = 12, w = 15 and k = 4 = 2

=

2

phase difference = 2 x

2 x

2

x =

4 8

cm

19. (C)

For the system described above

Let the point be at distance x from m.

For the gravitational field to be zero we have

2 2

Gm 9Gm

x (r x)

Solving we get x = r

4

The point is at a distance r

4 from m and

3r

4from 9m.

m

x

P

(rx)

9 m

MHT-CET 2016 : Physics Paper (11)

The potential at the point = Gm G 9m

r 3r

4 4

= 16 Gm

r

20. (D)

E = 4 4

0e T ) and A = b

When and b change to 3

and b

3

A = A

9

4

4

E A (327 273)

E A (27 273)

;

4E 1 600

E 9 300

E = 41

2 E9 E =

16 E

9

21. (C)

The surface area of the liquid drop is A1 = 4R2

Its surface energy is E

When the drop splits in 512 droplets, the surface area of droplets is A2 = 512 4r2

The volume of bigger drop is 34R

3 and Volume of small droplets is 34

512 r3

34R

3 = 512 34

r3 r =

R

8

A2 = 512 4r2 = 512 4

2R

8

= 8A1

Surface energy E = A.T (T is surface tension and A is area)

2 2 1

1 1 1

E A T 8 A

E A T A

= 8

En = 8E

22. (B)

M.I of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is

I = 2ML

12 and I =

21MK (where K1 is radius of gyration)

21MK K1 =

L

2 3 …(1)

When axis of rotation of rod is passing through one and of rod, then

I = 2

22

MLMK

3 K2 =

L

3 …(2)

Taking ratio of (1) and (2) we get

1

2

K L 3 1

K L 22 3 1

2

K 1

K 2

23. (C)

When bob is at rest at extreme position, the pendulum has only

potential energy which is given by

P.E. = mgh = mg( cos ) = mg (1 cos .

When bob comes to the mean position, the pendulum loses P.E. of bob

and it gets converted to K.E.

h

cos

(12) CET Score Booster : Physics

24. (C)

c2 360 1 1

t 12 3600 12 3600 12 10 120

deg/sec

25. (A)

2R

g gR h

When g

g4

then,

2g R

g4 R h

1 R

2 R h

2R = R + h R = h

26. (A)

For potentiometer of length x, let null point be obtained for a particular cell of emf E at L cm, (say)

v

E Lx

…(i)

When the length is increased from x to x then

v

E Lx

…(ii)

From (i) and (ii)

x L

x L

The balancing length increases when if the length of the potentiometer wire increases.

27. (C)

4

3

4 3

B 6 10 110

H AH 3 10 2 10

28. (B)

Peak value of current 0 rms

2I I 2 Amp

Coefficient of mutual inductance is M = 1 H

Induced emf in secondary is given by

2

die M

dt where 0i i sin t

Here, = 2n = 100 n = 50

2 o

de 1 i sin t

dt 0

2i cos( t) 100 cos(100 t) 200cos100 t

e0 = 200 V

29. (C)

When electron or any charged particle is accelerated through potential difference v, then kinetic energy

gained is given by E = eV …(1)

2 2

2

0 2

1 p hE mv

2 2m 2m

…(2)

2

2

h heV

2m 2meV

…(3)

When proton of mass M is accelerated through p.d. of 9V, then the deBroglie wavelength obtained is

h h

2Me 9V 3 2MeV

m

3 M

30. (C)

At point P, resultant intensity of interfering wave is

p res 1 2 1 2(I ) I I 2 I I cos

MHT-CET 2016 : Physics Paper (13)

For = p res, (I )2

= 10I

At point Q, resultant intensity of interfering wave is

2

Q resI 10I 2 9I 10I 6I 4I ( = )

I = P Qres resI I 10I 4I 6I

31. (B)

0AC

d

As given 0 0 01 2 3

A A AC ; C ; C

3d 6d 9d

eq 1 2 3C C C C C for parallel combination of capacitors

0 0eq

A 11 A1 1 1C

d 3 6 9 18d

32. (C)

For sustained oscillations according to Barkhausen criteria, A . = 1

33. (A)

According to photoelectric equation,

hc

= E = eV where E = K.E. and is work function.

If we decrease the wavelength , then stopping potential V will increase.

(Note: In the question the words “emitted photo electrons are moving with same velocity” should not be

there.)

34. (B)

As at interface, velocity of light reduces by 20% of C

Refractive index = 1

2

v sin i i

v sin r r

V2 = 0.8 V1

1

2

V 1 5

V 0.8 4

i 5

r 4

r = 4

i5

= i – r = i

5

35. (C)

For ionosphere, the maximum frequency of radio waves that can be reflected back is given by

g N (As per textbook)

where N is maximum electron density of ionosphere and g is acceleration due to gravity.

[Note : The correct formula for frequency is 9 N which is wrongly printed in the textbook.]

36. (D)

According to Bohr‟s theory the wavelength of radiation emitted is given by 2 2f i

1 1 1R

n n

For ni = 2 and nf = 1, the wavelength emitted will be minimum as energy difference E = E2 E1 is

maximum

37. (B)

If we fill one of the capacitors with dielectric material of dielectric constant K

then its capacitance C2 = KC1

charge Q = C1V1 = C2V2

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1 2

2 1

V C

V C 1 2

1 2 1 2

V C

V V C C

1 1

1 1

V KC

V C KC

V1 =

KV

1 K

38. (A)

LC parallel resonant circuit has very high impedance.

39. (C)

The total resistance in the circuit

Rtot = R5 + Rg = 1970 + 30 = 2000

To reduce the current to half, the total resistance should be doubled, i.e. Rtot = 4000

RS = Rtot Rg = 4000 30 = 3970

40. (B)

Between 4th bright band and 5th bright band there is 5th dark band. The phase difference for nth dark band

is given by = (2n 1)

For 5th dark band = 9

Path difference x = 2

= 9

2

=

9

2 = 59

6 102

cm = 2.7 104 cm

41. (B)

42. (D)

W = 1

2CV2

The work done in increasing the potential is given by

W = dU 2 22 1V V

2 22 11

2 22 3 2

V VW

W V V

here V1 = 5V, V2 = 10V and V3 = 15V

2

W 100 25 75

W 225 100 125

W2 = 1.67 W

43. (C)

1 = 4 104 wb and 2 = 0.1 1 = 4 105 wb

e = 72 105 V

t = ?

Using e = d

dt

dt

d

e

= 54 5

5

4 10 94 10 4 10

7272 10

=

1

2 = 0.5 sec

44. (A)

Resolving power of telescope is given by

P = d

1.22

When decreases, resolving power increases and vice versa.

45. (D)

Using hc

eV

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When, wavelength is incident on metallic surface, the stopping potential required to stop most

energetic electron is V and when wavelength increases to 3, stopping potential required as V

6

hc

eV

…(1)

and hc eV

3 6

…(2)

Thus equation (2) can be rewritten as 2hc

= eV + 6 …(3)

Subtracting equation (1) from (3) we get

hc

5

and 0

hc

Thus 0 = 5

46. (A)

According to Brewster‟s law, polarizing angle depends on wavelength and is different for different

colors.

47. (D)

The force acting on the particle inside magnetic field is FB = qvB sin

This provides the necessary centripetal force FC = 2mv

r

2mv

qvBr

r = mv

qB

x x x

y y y

r m v

r m v x x x

y y y

r m v

r m v

…(1)

Since they are accelerated through the same potential difference they have same kinetic energy.

2x x

1m v

2 = 2

y y

1m v

2

yx

y x

mv

v m

Putting this in Eq (1) : x x

y y

r m

r m

2 2

x x 1

y y 2

m r r

m r r

48. (A)

For an electron revolving in stationary orbits it doesn‟t radiate light through is velocity change.

49. (D)

50. (A)

x 6

1 2

5v

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x 2

6 3 x = 4

1

2

2

3 , 1 + 2 = 100 cm

1 = 40 cm, 2 = 60 cm

R1 = Resistance of 1 = 40 0.1 = 4

R2 = Resistance of 2 = 60 0.1 = 6

No current hours through the galvanometer so it can be ignored.

x = 4 and 6 in series give 10

R1 and R2 in series give 10 .

10 and 10 in parallel give 5

current I = V 5

R 5 = 1A