If the Earth shrinks to 1/8 of its original v | vedclass

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(C) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. Since the mass remains constant, $V \propto R^3$.If the new volume $V_2 = \frac{1}{8

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$6$

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(C) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. Since the mass remains constant, $V \propto R^3$.If the new volume $V_2 = \frac{1}{8} V_1$, then $\frac{V_2}{V_1} = \frac{R_2^3}{R_1^3} = \frac{1}{8}$.Taking the cube root, we get $\frac{R_2}{R_1} = \frac{1}{2}$, so $R_2 = \frac{1}{2} R_1$.By the law of conservation of angular momentum, $L = I \omega = 	ext{constant}$.Since $I = \frac{2}{5} M R^2$ and $\omega = \frac{2 \pi}{T}$, we have $I_1 \omega_1 = I_2 \omega_2$.Substituting the values, $\frac{2}{5} M R_1^2 \cdot \frac{2 \pi}{T_1} = \frac{2}{5} M R_2^2 \cdot \frac{2 \pi}{T_2}$.This simplifies to $\frac{R_1^2}{T_1} = \frac{R_2^2}{T_2}$, which gives $T_2 = T_1 \left( \frac{R_2}{R_1} ight)^2$.Given $T_1 = 24$ hours, $T_2 = 24 \cdot (1/2)^2 = 24 \cdot (1/4) = 6$ hours.

If the Earth shrinks to $1/8$ of its original volume, while maintaining the same mass, then the duration of the day will be (in $hrs$)

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