A solid sphere rolls down without slipping on an inclined plane, then percentage of rotational kinetic energy of total energy will be ........ $\%.$

A disc of mass $1\,kg$ and radius $R$ is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $4 \sqrt{\frac{x}{3 R}}$ rad s$^{-1}$ where $x=$

$\left( g =10\,ms ^{-2}\right)$

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively as shown is the fig. The velocities of two discs at the points $A$ and $B$ of the inclined planes are related as

The rotational kinetic energy of a solid sphere of mass $3 \;kg$ and radius $0.2\; m$ rolling down an inclined plane of height $7\; m$ is 

A body of moment of inertia of $3\ kg-m^2$ rotating with an angular velocity of $2\ rad/sec$ has the same kinetic energy as a mass of $12\ kg$ moving with a velocity of .......... $m/s$

The moment of inertia of a body about a given axis is $1.2 \;kg m^{2}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500\; joule$, an angular acceleration of $25 \;rad s^{-2}$ must be applied about that axis for a duration of