$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $30 \ m$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $[$use $g=10 \ m \ s^{-2}]$ (in $m \ s^{-1}$)

$A$ hollow cylinder and a solid cylinder start rolling down an inclined plane. Which one will take more time to reach the bottom of the plane?

$A$ 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 $\,ms^{-2}$)

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)

$A$ solid cylinder and a solid sphere of same mass and radius roll without slipping on a rough inclined plane. The force of friction is ..........

$A$ sphere rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?