A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle of inclination ${30}^{°}$. The centre of mass of the cylinder has a speed of 4 m/s. The distance travelled by the cylinder on the inclined surface will be $[take$ $g=$ $10$ $m/s^2]:$

1.	2.2 m	2.	1.6 m
3.	1.2 m	4.	2.4 m

A disc of radius \(2~\text{m}\) and mass \(100~\text{kg}\) rolls on a horizontal floor. Its centre of mass has a speed of \(20~\text{cm/s}\). How much work is needed to stop it?

A solid cylinder of mass \(2~\text{kg}\) and radius \(4~\text{cm}\) is rotating about its axis at the rate of \(3~\text{rpm}.\) The torque required to stop after \(2\pi\) revolutions is:
1. \(2\times 10^6~\text{N-m}\)
2. \(2\times 10^{-6}~\text{N-m}\) 
3. \(2\times 10^{-3}~\text{N-m}\) 
4. \(12\times 10^{-4}~\text{N-m}\)

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:
1. $M{R}^2$

2. $\frac{2}{5}M{R}^2$

3. $\frac{3}{2}M{R}^2$

4. $\frac{1}{2}M{R}^2$

A particle of mass \(m\) moves in the XY plane with a velocity \(v\) along the straight line AB. If the angular momentum of the particle with respect to the origin \(O\) is \(L_A\) when it is at \(A\) and \(L_B\) when it is at \(B,\) then: 
         

A wheel has an angular acceleration of \(3.0\) rad/s² and an initial angular speed of \(2.0\) rad/s. In a time of \(2\) s, it has rotated through an angle (in radians) of:

The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, around their respective axes is:

If \(\vec F\) is the force acting on a particle having position vector \(\vec r\) and \(\vec \tau\) be the torque of this force about the origin, then:

Two particles that are initially at rest, move towards each other under the action of their mutual attraction. If their speeds are \(v\) and \(2v\) at any instant, then the speed of the centre of mass of the system will be:
1. \(2v\)
2. \(0\)
3. \(1.5v\)
4. \(v\)