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Question
CBSEENPH11020716
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An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are 1 kg first part moving with a velocity of 12 ms-1 and 2 kg second part moving with a velocity of 8 ms-1.If the third part flies off with a velocity of 4 ms-,its mass would be

  • 5 kg 

  • 7 kg

  • 17 kg

  • 3 kg

Solution

A.

5 kg 

apply the law of conservation of linear momentum. 
momentum of first part = 1 x 12 = 12 kg ms-1
Momentum of the second part  = 2 x 8 = 16 kg ms-1 '
Resultant monmentum
= [(12)2 +(16)2]1/2 = 20 kg ms-1
The third part should also have the same momentum.


Let the mass of third part be M, then 
4 x M = 20
M = 5 kg

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Question
CBSEENPH11020717
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If the dimensions of a physical quantity are given by Ma Lb Tc, the physical quantity will be 

  • pressure if a = 1, b = -1, c= 2

  • velocity if a = 1, b = 1, c = -2

  • acceleration if a = 1, b = 1, c =-2

  • force if a = 0, b = -1, c =- 2

Solution

A.

pressure if a = 1, b = -1, c= 2

Dimensions of velocity = [ M0L1T-1]
Here, a = 0, b =1, c = -1
Dimension of acceleration = [ M0L1T-2]
Dimension of force = [ M1L1T-2]
Dimension of pressure = [M1L-1T-2
The physical quantity is pressure.

Question
CBSEENPH11020718
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Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle

  • depends on v and not on R

  • depends on R and not on v

  • is independent of both v and R

  • depends on both v and R

Solution

C.

is independent of both v and R

The time period of circular motion of their charged particle is given by 

straight T space equals space fraction numerator 2 πr over denominator straight v end fraction space equals space fraction numerator 2 straight pi over denominator straight v end fraction space straight x mv over Bq straight T space equals space fraction numerator 2 πm over denominator Bq end fraction
Hence, time period of rotation of the charged particle in uniform magnetic field is independent of both v and R.

Question
CBSEENPH11020721
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A body mass 1 kg is thrown upwards with velocity 20 ms-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?  (g = 10 ms-2)

  • 20 J

  • 30 J

  • 40 J

  • 10 J

Solution

A.

20 J

The energy lost due to air friction is equal to the difference of initial kinetic energy and final potential energy.
Initially, body possess only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy = KE - PE
=1 half mv squared minus mgh equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10 space equals space 200 minus 80 space straight J space equals space 20 space straight J 1 half mv squared minus mgh equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10 space equals space 200 minus 80 space straight J space equals space 20 space straight J

Question
CBSEENPH11020722
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The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is 

  • 30 ms-2 downward

  • 4 ms-2  upwards

  •  4 ms-2  downwards

  • 14 ms-2  upwards

Solution

B.

4 ms-2  upwards

Apparent weight > actual weight, then the lift is accelerating upward.
Lift is accelerating upward at the rate of a 
Hence, equation of motion is written as
R - mg = ma
28000-20000 = 2000a
a = 8000/2000 = 4 ms-2 upwards.

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