If the earth were to suddenly contract to fra | vedclass

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(C) According to the principle of Conservation of Angular Momentum $(COAM)$,since no external torque acts on the earth,the angular momentum remains co

Vedclass · Accepted Answer

$\frac{24}{n^2} \ hr$

Vedclass · Answer

(C) According to the principle of Conservation of Angular Momentum $(COAM)$,since no external torque acts on the earth,the angular momentum remains constant: $L = I_1 \omega_1 = I_2 \omega_2$.For a solid sphere,the moment of inertia is $I = \frac{2}{5} MR^2$. Let the initial radius be $R$ and the final radius be $R' = \frac{R}{n}$.Substituting the values:$\frac{2}{5} MR^2 \left( \frac{2\pi}{T} \right) = \frac{2}{5} M \left( \frac{R}{n} \right)^2 \left( \frac{2\pi}{T'} \right)$Simplifying the equation:$R^2 \cdot \frac{1}{T} = \frac{R^2}{n^2} \cdot \frac{1}{T'}$$\frac{1}{T} = \frac{1}{n^2 T'}$$T' = \frac{T}{n^2}$Given the initial duration of the day $T = 24 \ hr$,the new duration is $T' = \frac{24}{n^2} \ hr$.

If the earth were to suddenly contract to $\frac{1}{n}^{th}$ of its present radius without any change in its mass,then the duration of the new day will be:

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