If the earth suddenly shrinks to $$\frac{1}{64}$$th of its original volume with its mass remaining the same, the period of rotation of earth becomes $$\frac{24}{x}$$ h. The value of x is _______.

If the volume shrinks to $$\frac{1}{64}$$ of the original, and volume $$\propto R^3$$:

$$\frac{R'^3}{R^3} = \frac{1}{64} \Rightarrow R' = \frac{R}{4}$$

By conservation of angular momentum (no external torque):

$$I\omega = I'\omega'$$

$$\frac{2}{5}MR^2 \cdot \omega = \frac{2}{5}MR'^2 \cdot \omega'$$

$$R^2 \omega = \frac{R^2}{16} \omega'$$

$$\omega' = 16\omega$$

Since $$T = \frac{2\pi}{\omega}$$:

$$T' = \frac{T}{16} = \frac{24}{16} = \frac{24}{16} = \frac{3}{2} \text{ h}$$

So $$\frac{24}{x} = \frac{24}{16}$$, giving $$x = 16$$.

The value of x is 16.