$A$ body slides down a smooth inclined plane having angle $\theta$ and reaches the bottom with velocity $v$. If the body is a solid sphere rolling down the same plane,then its linear velocity at the bottom of the plane is

$A$ solid cylinder $P$ rolls without slipping from rest down an inclined plane,attaining a speed $v_p$ at the bottom. Another smooth solid cylinder $Q$ of the same mass and dimensions slides without friction from rest down the inclined plane,attaining a speed $v_q$ at the bottom. The ratio of the speeds $\frac{v_q}{v_p}$ is

$A$ cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}} \ m/s^2$,where $x=$ . . . . . . . (use $g=10 \ m/s^2$)

$A$ solid sphere and a hollow cylinder roll up without slipping on the same inclined plane with the same initial speed $v$. The sphere and the cylinder reach maximum heights $h_1$ and $h_2$,respectively,above the initial level. The ratio $h_1: h_2$ is $\frac{n}{10}$. The value of $n$ is . . . . . . .

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

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. The angular velocity of the cylinder when it reaches the bottom of the plane will be