If the radius of the earth is reduced to three-fourth of its present value without any change in its mass,then the duration of the day on earth will be . . . . . . hours $30$ minutes.

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity:

$A$ force $\vec F = \alpha \hat i + 3\hat j + 6\hat k$ is acting at a point $\vec r = 2\hat i - 6\hat j - 12\hat k$. The value of $\alpha$ for which angular momentum about the origin is conserved is:

$A$ person is rotating on a rotating table with an angular speed of $\omega$. He holds two identical masses in his outstretched hands. Without moving his hands,he simply drops the masses. How will his angular velocity change?

$A$ torque is applied to a body,changing its angular velocity from $\omega_1$ to $\omega_2$. What is the ratio of the initial radius of gyration to the final radius of gyration?

Keeping the mass of the Earth constant, if its radius is halved, then what will be the duration of the day (in $\text{ hours}$)?