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 Multiple Choice QuestionsMultiple Choice Questions

21.

Two weights w1 and w2 are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration g, then the tension in the string is :

  • 4 w1w2w1 + w2

  • w1w2w1 + w2

  • 2 w1 w2w1 w2

  • w1 + w22


A.

4 w1w2w1 + w2

Equation of motion for first weight

This is a frictionless and inextensible pulley.

for first weight

T - m1g = m1 ( a - g)

⇒ T - 2m1 g = m1a

For second weight 

m2g -T = m2 ( a - g)

2 mg - T = m2 a

on solving equations (1) & (2)

T =4 m1 m2m1 + m2T = 4 m1 m2w1 + w2


22.

Figure shows two blocks A and B pushed against the smooth wall with the force F. The wall is smooth but the surfaces in contact of A and B arerough. Which of the following is true for the system of blocks ?

  • The system can not be in equilibrium

  • F should be equal to weight of A and B

  • F should be less than theweight of A and B

  • F should be more than the weight of A and B


A.

The system can not be in equilibrium

Because the wall is smooth so whatever maybe the applied force, the system will move down due to its owriweight. Hence the system cannot be in equilibrium.


23.

A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is fraction numerator straight m l squared over denominator 3 end fraction)

  • fraction numerator 3 straight g over denominator 2 straight l end fraction
  • fraction numerator 2 straight l over denominator 3 straight g end fraction
  • fraction numerator 3 straight g over denominator 2 straight l squared end fraction
  • fraction numerator 3 straight g over denominator 2 straight l squared end fraction

A.

fraction numerator 3 straight g over denominator 2 straight l end fraction

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is
           straight I space equals space fraction numerator straight m l squared over denominator 3 end fraction
where m is mass of rod and l its length.
Torque left parenthesis straight t space equals space Iα right parenthesis acting on centre of gravity of rod is given by
                  straight t equals mg l over 2
or               Iα space equals space mg l over 2
or space space fraction numerator straight m l squared over denominator 3 end fraction straight alpha space equals space mg l over 2
therefore space space space space straight alpha space equals fraction numerator 3 straight g over denominator 2 l end fraction


 

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24.

The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is

  • MR squared
  • 2 over 5 MR squared
  • 3 over 2 MR squared
  • 3 over 2 MR squared

C.

3 over 2 MR squared

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
                     straight I subscript CM space equals space 1 half MR squared
where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at it diameter and normal to the disc is
                   straight I equals space straight I subscript CM space plus space MR squared
                     equals 1 half MR squared plus space MR squared
equals 3 over 2 MR squared

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25.

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be

  • 30°

  • 45°

  • 60°


C.

45°

From force diagram shown in figure, 

                 T = mg    .... (i)

and  2T cos θ =  2 mg     .... (ii)

Combining Eqs. (i) and (ii), we have
   

    2 mg cos θ = 2 mgor      cos θ = 12or             θ = 45°


26.

A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time:

  • v/gμ

  • gμ/v

  • g/v

  • g/v


A.

v/gμ

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,

force applied = frictional force
i.e., μmg = ma
μmg = m (v/t)
t =v/μg

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27.

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If u (O) = 0,  the graph of u(x) versus x will be (where u is potential energy function)


D.

Given,   F = k x where k is positive constant.Force  , F = -dUxdxor  dUxdx = -kxor   dUx  = -kx dxIntegrating both the sides,Ux = -kx dx = -12k x2or                      x2 = -2kUx       ..... (i)Comparing Eq. (i) with the standard equation of parabola, which is given by                         x2 = 4ayWe conclude that the graph of U (x) versus x satisfies the graph of parabola. Since the equation (i) has minus sign, therefore the graph will be open downwards along U (x).


28.

A small block is shot into each of the four tracks as shown below. Each of the  frictionless track rises to the same height. The speed, with which the block enters the tracks, is same in all cases. At the highest point of the track; normal reaction is maximum in


A.

Since the block rises the same heights in all the four cases, from conservation of energy, speed of  the block at highest point will be same in all four cases. Say it is v0. Equation of motion will b

  

N + mg = mv02Ror       N = mv02R - mg

R (the radius of curvature) in first case is minimum: Therefore, normal reaction N will be maximum in first case.


29.

A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is

  • 48 N

  • 24 N

  • 12 N

  • 12 N


B.

24 N

The vector OA with rightwards arrow on top represents the momentum of the object before the collision, and the vector OB with rightwards arrow on top that after the collision. The vector AB with rightwards arrow on top represents the change in momentum of the object increment straight P with rightwards arrow on top.



As the magnitudes of  OA with rightwards arrow on top space and space OB with rightwards arrow on top are equal, the components of OA with rightwards arrow on top space and space OB with rightwards arrow on top along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due only to the change in direction of the perpendicular components.
Hence,  increment straight p space equals space OB space sin space 30 degree space minus space left parenthesis negative OA space sin space 30 degree right parenthesis
                    equals mv space sin space 30 degree space minus space left parenthesis negative mv space sin space 30 degree right parenthesis
equals 2 space mv space sin space 30 degree
Its time rate will appear in the form of average force acting on the wall.
 therefore space space space space space space space space space straight F space cross times space straight t space equals space 2 mv space sin space 30 degree
or space space space space space space space space space space space space space space straight F space equals space fraction numerator 2 mv space sin space 30 degree over denominator straight t end fraction
   Given comma space straight m space equals space 0.5 space kg comma space space straight v space equals space 12 space straight m divided by straight s comma space space space straight t equals space 0.25 space straight s
space space space space space space space space space space straight theta space equals space 30 degree
Hence comma space space space straight F space equals space fraction numerator 2 cross times 0.5 cross times 12 space sin space 30 degree over denominator 0.25 end fraction space equals space 24 space straight N

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30.

Which is true for rolling frictrion   μr , static friction μs  and  kinetic friction  μk :

  • μs > μk >  μr

  • μs <  μk <  μr

  • μs <  μk >  μr

  • μs > μr  > μk


A.

μs > μk >  μr

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. When external force exceeds the maximum limit of static friction body begins to move.
 
Once the body is in motion it is subjected to sliding or kinetic friction which opposes relative motion between the two surfaces in contact. At every instant, there is just one point of contact
between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. 
In practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling. Static friction is always greater than kinetic friction and rolling friction is less than kinetic friction.
                        μs >  μk > μr