$A$ ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is . . . . . . .

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ($g=$ acceleration due to gravity,$\sin 30^{\circ}=0.5$).

State the necessary condition for a solid cylinder to roll without slipping down an inclined plane with friction.

$A$ ring of radius $3a$ is fixed rigidly on a table. $A$ small ring of mass $m$ and radius $a$ rolls inside it without slipping,as shown in the figure. The small ring is released from position $A$ (the horizontal position). What will be the speed of the center of the small ring when it reaches the lowest point?

$A$ solid cylinder of diameter $30 \ cm$ is rolled down an inclined plane from a height of $2 \ m$. If there is no energy loss due to friction,the angular velocity at the base of the plane is ....... $rad/s$. (Take $g = 10 \ m/s^2$)

$A$ rigid body of mass $M$ and radius $R$ rolls without slipping on an inclined plane of inclination $\theta$, under gravity. Match the type of body in Column-$I$ with the magnitude of the force of friction in Column-$II$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$