A solid sphere of mass \(M\) and radius \(R\) is in pure rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of the angular momentum of the sphere about the origin \(O\) is:
             

1.  $\frac{7}{5}M{R}^2\omega$

2.  $\frac{3}{2}M{R}^2\omega$

3.  $\frac{1}{2}M{R}^2\omega$

4.  $\frac{2}{3}M{R}^2\omega$

A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity $\omega$.  Four objects of mass m are held gently at the opposite ends of the ring's two perpendicular diameters. The angular velocity of the ring will be:

1. $\frac{\mathrm{M\omega }}{\mathrm{M}+ 4\mathrm{m}}$

2. $\frac{(\mathrm{M}+ 4\mathrm{m})\mathrm{\omega }}{\mathrm{M}}$

3. $\frac{(\mathrm{M}- 4\mathrm{m})\mathrm{\omega }}{\mathrm{M}+ 4\mathrm{m}}$

4. $\frac{\mathrm{M\omega }}{4\mathrm{m}}$

A boy is standing on a disc rotating about the vertical axis passing through its centre. He pulls his arms towards himself, reducing his moment of inertia by a factor of m. The new angular speed of the disc becomes double its initial value. If the moment of inertia of the boy is I₀ , then the moment of inertia of the disc will be:

1.  $2{\mathrm{I}}_{0}m$

2.  ${\mathrm{I}}_0\left(1-\frac{2}{m}\right)$

3.  ${\mathrm{I}}_0\left(1-\frac{1}{m}\right)$

4.  $\left(\frac{{\mathrm{I}}_0}{2m}\right)$

A rod is falling down with constant velocity \(V_0\) as shown. It makes contact with hinge A and rotates around it. The angular velocity of the rod just after the moment when it comes in contact with hinge A is:

The law of conservation of angular momentum is valid when:

The position of a particle is given by \(\vec r = \hat i+2\hat j-\hat k\) and momentum \(\vec P = (3 \hat i + 4\hat j - 2\hat k)\). The angular momentum is perpendicular to:

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