On a solid sphere lying on a horizontal surface,a force $F$ is applied at a height of $R/2$ from the centre of mass. The initial acceleration of a point at the top of the sphere is (there is no slipping at any point).

Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring $= m$,radius $= r$)

$A$ disc of mass $M$ and radius $R$ is free to rotate about its vertical axis as shown in the figure. $A$ battery-operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass $M$ and radius $R/2$ is fixed to the motor's thin shaft. Initially,both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed $\omega$. If the angular speed at which the large disc rotates is $\omega/n$,then the value of $n$ is. . . . .

$A$ uniform cube of mass $m$ and side $a$ is placed on a frictionless horizontal surface. $A$ vertical force $F$ is applied to the edge as shown in the figure. Match the following (most appropriate choice):

$(a)$ $\frac{mg}{4} < F < \frac{mg}{2}$	$(i)$ Cube will move up
$(b)$ $F > \frac{mg}{2}$	$(ii)$ Cube will not exhibit motion
$(c)$ $F > mg$	$(iii)$ Cube will begin to rotate about $A$
$(d)$ $F = \frac{mg}{4}$	$(iv)$ Normal reaction effectively at $a/3$ from $A$,no motion

$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to a point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about $P$ when the rod becomes vertical is:

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