$A$ sphere is rolling up an inclined plane. The frictional force acting on it is

$A$ solid sphere of radius $R$ undergoes pure rolling down a plane inclined to the horizontal at an angle $\theta$. If the radius of gyration is $k$,then its acceleration is

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

$A$ ring takes time $t_1$ in slipping down an inclined plane of length $L$ and takes time $t_2$ in rolling down the same plane. The ratio $\frac{t_1}{t_2}$ is

$A$ spherical uniform body of radius $R$,mass $M$,and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. Identify the correct statement$(s)$.