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.

<p>Given that the radius of the Earth is decreased to three-fourths of its current value while its mass remains unchanged, we need to determine the new duration of the Earth's day.</p> <p>First, using the principle of conservation of angular momentum, we have:</p> <p>$ \tau_{\text{ext}} = 0 \implies \text{Angular momentum is conserved} $</p> <p>Therefore,</p> <p>$ \frac{2}{5} M R^2 \cdot \omega_i = \frac{2}{5} M \left(\frac{3R}{4}\right)^2 \cdot \omega_f $</p> <p>Simplifying this equation, we find:</p> <p>$ \omega_f = \frac{16}{9} \omega $</p> <p>Since the period $ T $ of rotation is given by:</p> <p>$ T = \frac{2\pi}{\omega} $</p> <p>For the new period $ T_1 $, we have:</p> <p>$ T_1 = \frac{2\pi}{\omega_f} = \frac{2\pi}{\frac{16}{9}\omega} = \frac{9}{16} \times T $</p> <p>Given that the initial period $ T $ is 24 hours:</p> <p>$ T_1 = \frac{9}{16} \times 24 \text{ hours} $</p> <p>$ T_1 = 13 \text{ hours} ~30 \text{ minutes} $</p> <p>Thus, the new duration of the Earth's day would be 13 hours and 30 minutes.</p>