Keeping the mass of the Earth constant, if it | vedclass

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(A) According to the law of conservation of angular momentum, $L = I \omega = 	ext{constant}$.Since the Earth is treated as a solid sphere, its momen

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$6$

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(A) According to the law of conservation of angular momentum, $L = I \omega = 	ext{constant}$.Since the Earth is treated as a solid sphere, its moment of inertia is $I = \frac{2}{5}MR^2$.Initially, $I_1 = \frac{2}{5}MR^2$ and the time period is $T_1 = 24 	ext{ hours}$.When the radius is halved, $R_2 = \frac{R}{2}$, the new moment of inertia is $I_2 = \frac{2}{5}M(\frac{R}{2})^2 = \frac{1}{4}I_1$.Using $I_1 \omega_1 = I_2 \omega_2$, we have $I_1 (\frac{2\pi}{T_1}) = I_2 (\frac{2\pi}{T_2})$.Substituting the values: $I_1 (\frac{2\pi}{24}) = (\frac{1}{4}I_1) (\frac{2\pi}{T_2})$.This simplifies to $\frac{1}{24} = \frac{1}{4T_2}$, which gives $T_2 = \frac{24}{4} = 6 	ext{ hours}$.

Keeping the mass of the Earth constant, if its radius is halved, then what will be the duration of the day (in $\text{ hours}$)?

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