Laws of Motion

Question 549
CBSEENPH11020825

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire ‘B’ is given by

  • fraction numerator straight mu subscript straight o straight I over denominator 2 πd end fraction
  • fraction numerator 2 straight mu subscript straight o straight I over denominator πd end fraction
  • fraction numerator square root of straight mu subscript straight o straight I end root over denominator πd end fraction
  • fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end fraction

Solution

D.

fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end fraction

Force between BC and AB will be same in magnitude.

straight F subscript BC space equals straight F subscript BA space equals space fraction numerator straight mu subscript 0 straight I squared over denominator 2 πd end fraction
straight F space equals space square root of 2 straight F subscript BC
space equals space square root of 2 fraction numerator straight mu subscript 0 straight I squared over denominator 2 πd end fraction
straight F space equals space fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end root end fraction

Question 550
CBSEENPH11020849

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

  • 3D/2

  • D

  • 5D/4

  • 7D/5

Solution

C.

5D/4

As track is frictionless, so total mechanical energy will remain constant,

i.e., 0 + mgh = 12mvL2 + 0using v2 - u2 = 2gh,h = vL22g ( u = 0)For completing the vertical circle VL 5gRor, h = 5gR2g = 52R = 54D

Question 551
CBSEENPH11020851

Which one of the following statements is incorrect?

  • Rolling friction is smaller than sliding friction.

  • Limiting value of static friction is directly proportional to normal reaction.

  • Coefficient of sliding friction has dimensions of length.

  • Frictional force opposes the relative motion.

Solution

C.

Coefficient of sliding friction has dimensions of length.

Coefficient of friction or sliding friction has no dimension.

f = μsN μs = fN = [M0L0T0]

Question 552
CBSEENPH11020852

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

  • 0.5

  • 0.25

  • 0.4

  • 0.8

Solution

B.

0.25

According to law of conservation of linear momentum,

mv + 4m x 0 = 4mv' + 0

v' = v4e = relative velocity of separation Relative velocity of approach =v4ve = 14 = 0.25

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