$A$ rotating table completes one revolution in $10 \ s$. Its moment of inertia is $100 \ kg \cdot m^2$. $A$ man of mass $50 \ kg$ stands at the center of the rotating table. If the man moves $2 \ m$ away from the center,what will be the angular velocity of the table in $rad/s$?

Due to the melting of ice at the poles of the Earth,the moment of inertia of the Earth ....,the angular velocity ....,and the duration of the day becomes ....

If the mass of the Earth is kept constant and its radius is reduced to half, then the duration of the day will become ........ $hr$.

In the above problem,the angular velocity of the system after the particle sticks to it will be ....... $rad/s$. (in $.3$)

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

$A$ particle of mass $m$ moves with a velocity $v_0$ in a circle of radius $R_0$ on a smooth horizontal plane. The mass is tied to a string passing through a hole in the smooth plane as shown in the figure. The tension in the string is gradually increased,and finally,the mass $m$ moves in a circle of radius $\frac{R_0}{2}$. What is the ratio of the final kinetic energy to the initial kinetic energy?