$A$ solid metallic sphere of radius '$R$' having moment of inertia '$I$' about its diameter is melted and recast into a solid disc of radius '$r$' of uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is

According to the parallel axis theorem,$I = I_{cm} + Mx^2$. What will be the graph between $I$ and $x$?

$A$ wheel of radius $r$ rolls without slipping with a speed $v$ on a horizontal road. When it is at a point $A$ on the road,a small lump of mud separates from the wheel at its highest point $B$ and drops at point $C$ on the road. The distance $AC$ will be

State the equivalent physical quantities in linear motion corresponding to the moment of inertia and torque in rotational motion.

$A$ 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
$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$

$A$ particle is moving in a uniform circular motion with angular momentum $L$. If its angular frequency is doubled and its kinetic energy is halved,then the new angular momentum of the particle will be .......