$A$ solid sphere at rest rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of the plane is ($g=$ acceleration due to gravity).

$A$ solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{b} g \sin \theta$ where $b$ is $........$. (Round off to the Nearest Integer)
($g =$ acceleration due to gravity)
($\theta =$ angle as shown in figure)

$A$ solid sphere rolls down from the top of an inclined plane,$7 \ m$ high,without slipping. Its linear speed at the foot of the plane is $\left(g = 10 \ m/s^2\right)$

$A$ solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$,respectively,then

$A$ solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '$V_1$'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '$V_2$'. The ratio $V_1 : V_2$ is (neglect friction).

$A$ disc rolls without slipping down an inclined plane of length $L$ and inclination $\theta$. What will be its velocity at the bottom?