The centre of a wheel rolling on a plane surface moves with a speed \(v_0.\) A particle on the rim of the wheel at the same level as the centre will be moving at speed:
1. zero
2. \(v_0\)
3. \(\sqrt{2}v_0\)
4. \(2v_0\)

Moment of inertia of a uniform cylinder of mass M, radius R and length l about an axis passing through its center and normal to its axis would be

1.  $\frac{{\mathrm{Ml}}^2}{12}$

2.  $\frac{{\mathrm{MR}}^2}{2}$

3.  $\mathrm{M}\left[\frac{{\mathrm{l}}^2}{12}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{{\mathrm{R}}^2}{4}\right]$

4.  $\mathrm{M}\left[\frac{{\mathrm{l}}^2}{12}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{{\mathrm{R}}^2}{2}\right]$

A particle moves along a circle of radius $\frac{20}{\mathrm{\pi }}m$ with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is

1. 640 $\mathrm{\pi }<mspace\; width="1em"></mspace>\mathrm{m}/{\mathrm{s}}^2$

2. 160 $\mathrm{\pi }<mspace\; width="1em"></mspace>\mathrm{m}/{\mathrm{s}}^2$

3. 40 $\mathrm{\pi }<mspace\; width="1em"></mspace>\mathrm{m}/{\mathrm{s}}^2$

4. 40 $\mathrm{m}/{\mathrm{s}}^2$

If the radius of the earth is suddenly contracted to half of its present value, then the duration of the day will be of:

A wheel has angular acceleration of 3.0 rad/$se{c}^2$ and an initial angular speed of 2.00 rad/sec.  In a time of 2 sec it has rotated through an angle (in radian) of 

1. 10               

2. 12

3. 4                 

4. 6

If a force acts on a body at a point away from the centre-of-mass, then:

A sphere can not roll on:

1.  a smooth horizontal surface 

2.  a rough horizontal surface 

3.  a smooth inclined surface 

4.  a rough inclined surface 

Two balls are thrown simultaneously in the air. The acceleration of the centre-of-mass of the two balls while in the air:

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizotal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

1. $\sqrt{\frac{2gh}{I+mr}}$

2. $\sqrt{\frac{2mgh}{I+m{r}^2}}$

3. ${\left[\frac{2mgh}{I+2m}\right]}^{\frac{1}{2}}$

4. $\sqrt{2gh}$