$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid-point elastically. If the particle comes to rest after the collision,find the value of $M/m$.

$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?

In the given figure,a ring of mass $m$ is kept on a horizontal surface,and a body of equal mass $m$ is attached through a string wound on the ring. When the system is released,the ring rolls without slipping. Consider the following statements and choose the correct option.
$(i)$ Acceleration of the centre of mass of the ring is $\frac{g}{3}$.
$(ii)$ Acceleration of the hanging particle is $\frac{2g}{3}$.
$(iii)$ Frictional force (on the ring) acts in the forward direction.
$(iv)$ Frictional force (on the ring) acts in the backward direction.

The temperature of a thin uniform circular disc of $1 \ m$ diameter is increased by $10^{\circ} C$. The percentage increase in the moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face is: (linear coefficient of expansion $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$)

$A$ body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface,its shape can be:

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: