In free space, a rifle of mass \(M\) shoots a bullet of mass \(m\) at a stationary block of mass \(M\) at a distance \(D\) away from it. When the bullet has moved through a distance \(d\) towards the block, the centre of mass of the bullet-block system is at a distance of:
1. \(\frac{D-d}{M+m}~\text{from the bullet}\)
2. \(\frac{md+ MD}{M+m}~\text{from the block}\)
3. \(\frac{2md+ MD}{M+m}~\text{from the block}\)
4. \(\frac{(D-d)M}{M+m}~\text{from the bullet}\)

Blocks A and B are resting on a smooth horizontal surface given equal speeds of 2 m/s in the opposite sense as shown in the figure.

At t = 0, the position of blocks are shown, then the coordinates of centre of mass at t = 3s will be 

1.  (1, 0)

2.  (3, 0)

3.  (5, 0)

4.  (2.25, 0)

A wheel is at rest. Its angular velocity increases uniformly and becomes 80 rad/s after 5 s. The total angular displacement is 

1. 800 rad

2. 400 rad

3. 200 rad

4. 100 rad

A force\(- F \hat k\) acts on O, the origin of the coordinate system. The torque at the point (1, -1) will be:

1. $-\mathrm{F}\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)$

2. $\mathrm{F}\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)$

3. $-\mathrm{F}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)$

4. $\mathrm{F}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right)$

Moment of inertia of an object does not depend upon

A thin uniform circular disc of mass \(M\) and radius \(R\) is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with an angular velocity $\omega$. Another disc of the same dimensions but of mass \(\frac{1}{4}M\) is placed gently on the first disc co-axially. The angular velocity of the system will be:

When a torque acting upon a system is zero, then which of the following will be constant 

A flywheel is in the form of a uniform circular disc of radius 1 m and mass 2 kg. The work which must be done on it to increase its frequency of rotation from 5 rev ${\mathrm{s}}^{-1}$ to 10 rev ${\mathrm{s}}^{-1}$ is approximately

1.  1.5 x ${10}^2$ J

2.  3.0 x ${10}^2$ J

3.  1.5 x ${10}^3$ J

4.  3.0 x ${10}^3$ J

A constant torque acting on a uniform circular wheel changes its angular momentum from ${\mathrm{A}}_0$ to $4 {\mathrm{A}}_0$ in 4s. The magnitude of this torque is 

1. $\frac{3{\mathrm{A}}_0}{4}$

2. $4 {\mathrm{A}}_0$

3. ${\mathrm{A}}_0$

4. $12 {\mathrm{A}}_0$