TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, initial volume of earth, $V_i = V$ Final volume of earth, $V_f = \frac{V}{8}$ and $R_i, R_f$ are initial and final radii of earth. From Kepler's third law of planetary motion, $T^2 \propto R^3$. . . (i) where, $T$ is time-period. Since, volume, $V = \frac{4}{3} \pi R^3$ $\therefore V \propto R^3$ $\Rightarrow \; \frac{V_i}{V_f} = \frac{V}{\frac{V}{8}} = \frac{ \frac{4}{3}\pi }{ \frac{4}{3}\pi } \frac{R_f^3}{R_f^3}$ $\Rightarrow \; 8 = \left( \frac{R_i}{R_f} \right)^3 \Rightarrow \; \frac{R_i}{R_f} = 2$ By using law of conservation of angular momentum, $L_i = L_f$ where, $L_i, L_f$ are initial and final angular momentum, respectively. $I_i \omega_i = I_f \omega_f$ where, $I_i, I_f$ and $\omega_i, \omega_f$ are initial and final moment of inertia and angular speed, respectively. $\Rightarrow \; I_i \frac{2\pi}{T_i} = I_f \frac{2\pi}{T_f}$ $\Rightarrow \; \frac{T_i}{T_f} = \frac{I_i}{I_f}$ $T_f = T_i \left( \frac{I_f}{I_i} \right)$ Moment of inertia of earth $I_{\text{(earth)}} = \frac{2}{5} M R^2$ $\therefore \; T_f = T_i \left( \frac{ \frac{2}{5} M R_f^2 }{ \frac{2}{5} M R_i^2 } \right) = T_i \left( \frac{R_f}{R_f} \right)^2$ $= 24 \left( \frac{1}{2} \right)^2$ $= 24 \times \frac{1}{4} = 6 \text{h}$