If the radius of the Earth becomes 1/n^{th} o | vedclass

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(C) According to the law of conservation of angular momentum,$L = I\omega = 	ext{constant}$.Since the mass $M$ remains constant,the moment of inertia

Vedclass · Accepted Answer

$24/n^2 \; 	ext{hr}$

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(C) According to the law of conservation of angular momentum,$L = I\omega = 	ext{constant}$.Since the mass $M$ remains constant,the moment of inertia $I = \frac{2}{5}MR^2$ changes as the radius $R$ changes.Let the initial radius be $R$ and the final radius be $R' = R/n$.Let the initial time period be $T_1 = 24 \; 	ext{hr}$ and the final time period be $T_2$.Using $I_1 \omega_1 = I_2 \omega_2$,we have:$\frac{2}{5}MR^2 \left( \frac{2\pi}{T_1} ight) = \frac{2}{5}M(R/n)^2 \left( \frac{2\pi}{T_2} ight)$.Simplifying this,we get $R^2 / T_1 = (R^2 / n^2) / T_2$.Therefore,$T_2 = T_1 / n^2$.Substituting $T_1 = 24 \; 	ext{hr}$,we get $T_2 = 24/n^2 \; 	ext{hr}$.

If the radius of the Earth becomes $1/n^{th}$ of its present value,without any change in its mass,then the duration of the day becomes:

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