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 $,MV^2$)?

$A$ force of $49 \ N$ acts tangentially at the highest point of a solid sphere of mass $20 \ kg$,kept on a rough horizontal plane. If the sphere rolls without slipping,then the acceleration of the center of the sphere is (in $m/s^2$)

$A$ circular ring of mass $10 \text{ kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed of $1.5 \text{ m/s}$. The work required to stop the ring is: (in $\text{ J}$)

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

$A$ solid sphere is rolling with a linear velocity $v$. Its total kinetic energy is:

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is