The rotational kinetic energy of a solid sphere of mass $3 \;kg$ and radius $0.2\; m$ rolling down an inclined plane of height $7\; m$ is
$60$
$36$
$70 $
$42$
A rod of length $l$ is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in the vertical position is
Two coaxial discs, having moments of inertia $I_1$ and $\frac{I_1}{2}$ are a rotating with respectively angular velocities $\omega_1$ and $\frac{\omega_1}{2}$, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ are the final and initial total energies, then $(E_f -E_i)$ is
A uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta $ its angular velocity $\omega $ is given as
Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \;kg - m ^{2}$ and $10\; rad \,s^{-1}$ respectively while those for the second one are $0.2 \;kg - m ^{2}$ and $5\; rad \,s ^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is ...........$J$
$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then the maximum angle $‘\theta ’$ made by the hinged upper half with the vertical is ......... $^o$.