An inclined plane makes an angle of $30^o$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

When a uniform solid sphere and a disc of the same mass and of the same radius roll down a rough inclined plane from rest to the same distance,then the ratio of the time taken by them is

$A$ solid sphere rolls down without slipping on a $30^{\circ}$ inclined plane. If $g = 10\,m/s^2$,the acceleration of the rolling sphere is

$A$ solid sphere of mass $M$ and radius $R$ is rolling down an inclined plane of length $L$ and height $h$ without slipping. What will be the velocity of its center of mass upon reaching the bottom?

Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (see figure) starting at $A$ and $C$ with linear speeds $v_1$ and $v_2$,respectively,and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_1 = 3 \ m/s$,then $v_2$ in $m/s$ is $(g = 10 \ m/s^2)$.

$A$ sphere of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be