$A$ solid sphere rolls down two different inclined planes of the same height,but of different inclinations. In both cases:

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls down an inclined plane of angle $\theta$ without slipping. What is its acceleration?

$A$ cylinder of mass $M_c$ and a sphere of mass $M_s$ are placed at points $A$ and $B$ of two inclines,respectively (See Figure). If they roll on the incline without slipping such that their accelerations are the same,then the ratio $\frac{\sin \theta_c}{\sin \theta_s}$ 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

$A$ thick-walled hollow sphere has outside radius $R_0$. It rolls down an incline without slipping and its speed at the bottom is $v_0$. Now the incline is waxed,so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be $5v_0/4$. The radius of gyration of the hollow sphere about an axis through its centre is

$A$ 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 . . . . . . .