Question

A solid cylinder of mass $M$ and radius $R$ is mounted on frictionless horizontal axle so that it can freely rotate about its axis. A string of negligible mass is wrapped round the cylinder and a body of mass $m$ is hung from the string,and if angular velocity with which cylinder rotates is proportional to $h^a$ where h is height with which mass descends downward then the value of $2a$ is

Answer 1

$\frac{1}{2}{I\omega }^2=mgh$ (kinetic energy is equal to loss of potential energy)

$\omega =\sqrt{\frac{2mg}{I}}\sqrt{h}\omega \propto \sqrt{h}\sqrt{h}=h^a\Rightarrow a=\frac{1}{2}$

Therefore 2a = 1

Hence the correct answer is $1$