A wheel is rotating about an axis through its centre at \(720\) r.p.m. It is acted upon by a constant torque opposing its motion for \(8\) seconds to bring it to rest finally. The value of torque in N-m is: (given \(I\) = $\frac{24}{\mathrm{\pi }}$ kg ${\mathrm{m}}^2$)
1. \(48\)
2. \(72\)
3. \(96\)
4. \(120\)

A string is wrapped along the rim of a wheel of moment of inertia \(0.10\) kg-m² and radius \(10\) cm. If the string is now pulled by a force of \(10\) N, then the wheel starts to rotate about its axis from rest. The angular velocity of the wheel after \(2\) s will be:

A rigid body rotates with an angular momentum of \(L.\) If its kinetic energy is halved, the angular momentum becomes:
1. \(L\)
2. \(L/2\)
3. \(2L\)
4. \(L/\)$\sqrt{2}$

Two rotating bodies \(A\) and \(B\) of masses \(m\) and \(2m\) with moments of inertia $I_A$ and $I_B (I_B>I_A)$ have equal kinetic energy of rotation. If $L_A$ $$and $L_B$ be their angular momenta respectively, then:
1. $L_A=\frac{L_B}{2}$
2. $L_A=2L_B$
3. $L_B>L_A$
4. $L_A>L_B$

A uniform rod of length l is hinged at one end and is free to rotate in the vertical plane. The rod is released from its position, making an angle $\theta$ with the vertical. The acceleration of the free end of the rod at the instant it is released is:
    

A solid cylinder of mass \(50\) kg and radius \(0.5\) m is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and the other end hanging freely. The tension in the string required to produce an angular acceleration of \(2\) revolutions/s² will be:
1. \(25\) N
2. \(50\) N
3. \(78.5\) N
4. \(157\) N

A uniform rod of length 1 m and mass 2 kg is suspended by two vertical inextensible strings as shown in following figure. Calculate the tension T (in newtons) in the left string at the instant when the right string snaps (g = 10 m/$s^2$).
       
1. 2.5 N

2. 5 N

3. 7.5 N

4. 10 N