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

Earth's angular momentum is conserved: $$I\omega = const$$. $$I = \frac{2}{5}MR^2$$.

$$I_1\omega_1 = I_2\omega_2 \Rightarrow R^2 \omega_1 = (3R/4)^2 \omega_2 \Rightarrow \omega_2 = \frac{16}{9}\omega_1$$.

$$T_2 = \frac{2\pi}{\omega_2} = \frac{9}{16} T_1 = \frac{9}{16} \times 24 = 13.5$$ hours = 13 hours 30 minutes.

The answer is 13.