$A$ 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 uniform sphere having a mass $M$,radius $R$ and moment of inertia of $\frac{2}{5} M R^2$ rolls down a plane inclined at an angle $\theta$ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is $\mu_s$. Then,

$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

$A$ ring and a disc are initially at rest,side by side,at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \ s$,then the height of the top of the inclined plane,in metres,is. . . . . . . . Take $g=10 \ m \ s^{-2}$.

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

$A$ 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