$A$ solid sphere and a thin uniform circular disc of same radius are rolling down an inclined plane without slipping. If the acceleration of the sphere is $3 \,ms^{-2}$, then the acceleration of the disc is (in $\,ms^{-2}$)

$A$ rolling wheel of $12 \, kg$ is on an inclined plane at position $P$ and connected to a mass of $3 \, kg$ through a string of fixed length and a pulley as shown in the figure. Consider $PR$ as a friction-free surface. The velocity of the centre of mass of the wheel when it reaches the bottom $Q$ of the inclined plane $PQ$ will be $\frac{1}{2} \sqrt{xgh} \, m/s$. The value of $x$ is.............

$A$ solid cylinder is placed on a rough surface of inclination $\theta$. The minimum value of the coefficient of static friction between the cylinder and the surface so that the cylinder rolls without slipping is:

$A$ solid sphere and a disc of same mass and radius start rolling down a rough inclined plane from the same height. The ratio of the time taken in the two cases is:

$A$ body starts rolling down an inclined plane of length $L$ and height $h$. This body reaches the bottom of the plane in time $t$. The relation between $L$ and $t$ is:

$A$ 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	$(IV)$ $\frac{Mg \sin \theta}{2.5}$