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 Multiple Choice QuestionsMultiple Choice Questions

131.

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


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


132.

Two blocks of masses 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. The tension (T) in the string is

  • 48.56 N

  • 72.22 N

  • 32.56 N

  • 24.65 N


D.

24.65 N

Let a be the acceleration of the system in the direction as shown in the figure and T is the tension in the string.

∴ Equations of motion for 5 kg and 2 kg blocks can be written as 

                       5g − T = 5a      .... (i)

and     T − 2g sin θ − f = 2a      .... (ii)

f is the limiting friction and is given by

               f = µR = µmg cos θ

                 = 0.3 × 2g cos θ

Adding Eqs. (i) and (ii), we get

                        5g − 2g sin θ − f = 7a

or 5g − 2g sin 30° − 0.6g cos 30° = 7a

or  g (5 − 2 x 0.5 − 0.6 x 0.8660) = 7a

or         a = 9.8 × 3.48047    [  g = 9.8 ms-2 ]

Now from Eq. (i), we have

     T = 5g − 5a

        = 5 (9.8 − 4.87)

        = 5 × 4.93 = 24.65 N


133.

A long block A of mass M is at rest on a smooth horizontal surface.  A small block B of mass M/2 is placed on A at one end and projected along A with same velocity v. The coefficient of friction between the block is μ. Then the acceleration of blocks A and B before reaching a common velocity will be respectively

  • μg2 (towards right), μg2 (towards left)

  • μg (towards right), μg (towards left)

  • ug2 (towards right), μg (towards left)

  • μg (towards right), μg2 (towards left)


B.

μg (towards right), μg (towards left)

The force causing the motion of A is a frictional force between A and B,

So, acceleration of A

μMB g = MAaA aA = μ MBMA g = μg2 (towards right)

Block B experiences frictions force towards the left.

MBaB = μMBg ⇒aB = μg towards left


134.

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.


C.

Coefficient of sliding friction has dimensions of length.

Coefficient of friction or sliding friction has no dimension.

f = μsN μs = fN = [M0L0T0]


135.

A car is speeding on a horizontal road curving round with a radius of 60 m. The coefficient of friction between the wheels and the road is 0.5. The height of centre of gravity of the car from the road level is 0.3 m and the distance between the wheels is 0.8 m. Will the vehicle skid or topple ?

  • Yes, it will skid

  • No, it will not skid

  • Sometimes skid, sometimes not

  • Data insufficient to calculate speed for toppling


B.

No, it will not skid

 Given, r = 60 m , μ = 0.5 , h = 0.3 m , b = 0.82 = 0.4 tan θ = v2rg        v = rg tan θFor skidding tan θ = μ      vs = μrg              = 0.5 × 60 × 9.8              = 17.15 ms-1For toppling,     tan θ = bhor       vt = brgh              = 0.4 × 60 × 9.80.3              = 28 ms-1Since, vs < vt , hence vehicle will not skid.


136.

A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throuhgout these motions). The directions of the frictional force acting on the cylinder are

  • Up the incline, while ascending and down the incline while descending

  • Up the incline, while ascending as well as descending

  • down the incline, while ascending and up the incline while descending

  • Down the incline while ascending as well as descending.


B.

Up the incline, while ascending as well as descending

It is obvious that during ascending, a retarding i.e. anticlockwise moment is required. I should be remembered that torque due to friction has the same sense the angular acceleration.


137.

A body starts rolling down on an inclined plane, the top half of which is perfectly smooth and lower half is rough. The angle of the plane being 30°, if the body is brought to rest just when it reaches the bottom, then the ratio of force of friction and weight of the body is

  • 1 : 1

  • 1 : 2

  • 2 : 1

  • 3 : 2


A.

1 : 1

AB is an inclined plane of length 2l (for convenience) with centre C

  ABO = θ = 30°

    

The top half AC of the plane is smooth and the lower half CB is rough. Let F be the force of friction in this part.

For the top half of inclined plane,

     u = 0 , a = g sin 30° = g2 , s = l , v = ?As v2 - u2 = 2as v2 - 0  = 2 g2l             v   = gl For the bottom half of inclined plane,  u = gl , v = 0 , s = l , a = ?As     v2 - u2 = 2as       0 - gl = 2al                   a = -g2

In this part, acceleration, a = net forcemass         a = mg sin 30° - Fm    -g2 = g × 12 - Fmor  F m = g2 + g2 = gor      F= mg Fmg = 11


138.

Assertion:  Pseudo force is an imaginary force which is recognised only by a non-inertial observer to explain the physical situation according to Newton's laws.

Reason: Pseudo force has no physical origin, i.e., it is not caused by one of the basic interactions in nature. It does not exist in the action-reaction pair.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


B.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Pseudo force is an imaginary force which is recognised only by a non-inertial observer to explain the physical situation according to Newton's laws. The magnitude of this force F is equal to the mass m of the object and acceleration α of the frame of reference. The direction of force is opposite to the direction of acceleration.

      F = - m α


139.

The figure shows a horizontal force F acting on a block of mass M on an inclined plane (angle θ).  What is the normal reaction N on the block?

     

  • mg sinθ + F cosθ

  • mg sinθ - Fcosθ

  • mg cosθ - F sinθ

  • mg cosθ + F sinθ


D.

mg cosθ + F sinθ

  

From the figure, the normal reaction on the block is

    N = mg cosθ + F sinθ


140.

Two masses 10 kg and 20 kg respectively are connected by a massless springs as shown in the figure. A force of 200 N acts on the 20 kg mass. At the instant shown is a figure the 10 kg mass has an acceleration of 12 m/s2. The value of the acceleration of 20 kg mass is

  • 4 m/s2

  • 10 m/s2

  • 20 m/s2

  • 30 m/s2


A.

4 m/s2

The equation of motion of m1 = 10 kg mass is

F1 = m1a1 = 10 x 12  = 120 N

Force on 10 kg mass is 120 N to the right. As action and reaction are equal and opposite, the reaction force F- on 20 kg mass F = 120 N to the left.

therefore, equation of motion of mass m2 = 20 kg is

200 - F = 20 a2

200-120 = 20a2

80 = 20a2

a2 = 80 /20 = 4 m/s2