$A$ 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

$A$ flywheel is used with an engine because it

$A$ uniform rod of mass $m$ and length $\ell$ hinged at end $A$ is released from the horizontal position shown in the figure. Just after the rod is released:

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$

$A$ 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 $M/m$.

One of two identical cylinders,cylinder $A$,rotates at an angular speed of $50 \text{ revolutions per second}$. This rotating cylinder is brought into contact with a second stationary cylinder,$B$. Due to kinetic friction between the two cylinders,the stationary cylinder starts rotating with an angular acceleration,while cylinder $A$ undergoes angular deceleration. If the magnitude of the angular acceleration for both cylinders is $1 \text{ revolution per second}^2$,after how many seconds $(t)$ will the angular speeds of both cylinders become equal?

$A$ circular hoop of mass $m$ and radius $R$ rests flat on a horizontal frictionless surface. $A$ bullet,also of mass $m$ and moving with a velocity $v$,strikes the hoop and gets embedded in it. The thickness of the hoop is much smaller than $R$. The angular velocity with which the system rotates after the bullet strikes the hoop is