$Assertion :$ $A$ hollow shaft is found to be stronger than a solid shaft made of the same material and having the same mass per unit length.
$Reason :$ The torque required to produce a given twist in a hollow cylinder is greater than that required to twist a solid cylinder of the same length and material.

$A$ uniform sphere of mass $m$ and radius $R$ is placed on a rough horizontal surface. The sphere is struck horizontally at a height $h$ from the floor. Match the following:

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ mass $M = 40 \ kg$ is fixed at the very edge of a long plank of mass $80 \ kg$ and length $1 \ m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a mass of $100 \ kg$ be attached so that the plank starts rotating with an angular acceleration of $1 \ rad/s^2$?

$A$ rod of mass $m$ and length $L$,pivoted at one of its ends,is hanging vertically. $A$ bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x=x_M$. Then
$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$

$A$ cube of side $a$ is moving with velocity $v$ on a smooth horizontal surface. It hits a linear raised obstacle $O$ on the horizontal surface (as shown in the figure). The angular speed of the block after hitting $O$ will be

$A$ block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially,the right edge of the block is at $x=0$,in a coordinate system fixed to the table. $A$ point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block,its position is $x$ and the velocity is $v$. At that instant,which of the following options is/are correct?
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$.
$[B]$ The position of the point mass is: $x=-\sqrt{2} \frac{mR}{M+m}$.
$[C]$ The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
$[D]$ The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.