$A$ thin hollow sphere of mass $m$ is completely filled with a liquid of mass $m$. When the sphere rolls with a velocity $v$, the kinetic energy of the system is (neglect friction):

$A$ particle performs uniform circular motion with an angular momentum $L$. If the angular frequency of the particle is doubled and its kinetic energy is halved,what will be its new angular momentum?

If the earth suddenly stops revolving and all its rotational $KE$ is used up in raising its temperature and if $s$ is taken to be the specific heat of the earth's material,the rise of temperature of the earth will be: ($R =$ radius of the earth and $\omega =$ its angular velocity,$J =$ Joule's constant)

$A$ wheel of radius $0.5 \ m$ and a moment of inertia of $10 \ kg \cdot m^2$ is rotating freely at an angular speed of $70 \ rev/min$. The wheel can be stopped in $5.0 \ s$ by pressing a wet cloth against the rim and exerting a radially inward force of $88 \ N$. The coefficient of kinetic friction between the wheel and wet cloth is:

The moment of inertia of a sphere of mass $M$ and radius $R$ is $I$. If the mass is kept constant,what will be the nature of the graph between $I$ and $R$?

The total kinetic energy of a body of mass $10 \ kg$ and radius $0.5 \ m$ moving with a velocity of $2 \ m/s$ without slipping is $32.8 \ J$. The radius of gyration of the body is .......... $m$.