$A$ solid sphere of mass $1 \ kg$ is rolling on a rough table with a linear speed of $1 \ m/s$. Its total kinetic energy is ....... $J$.

The speed of rolling of a ring of mass $M$ changes from $V$ to $3\ V$. What is the change in its kinetic energy (in $,MV^2$)?

$A$ solid sphere is moving on a horizontal plane. The ratio of its translational kinetic energy to its rotational kinetic energy is:

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300 \, J$. The translational kinetic energy of the disc is:

$A$ boy is pushing a ring of mass $2 \ kg$ and radius $0.5 \ m$ with a stick as shown in the figure. The stick applies a force of $2 \ N$ on the ring and rolls it without slipping with an acceleration of $0.3 \ m/s^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs,and the coefficient of friction between the stick and the ring is $(P/10)$. The value of $P$ is

Three solid spheres are made to move on a rough horizontal surface. Sphere $P$ is given a spin and released. Sphere $Q$ is given a forward linear velocity. Sphere $R$ is given linear and rotational motions as shown in the figure. Directions of the friction force on spheres $P, Q, R$ are respectively