If the earth suddenly shrinks to frac{1}{64}t | vedclass

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Question

(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. If the volume becomes $\frac{1}{64}$ of the original volume,then $\frac{V'}{V} = \fr

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$. If the volume becomes $\frac{1}{64}$ of the original volume,then $\frac{V'}{V} = \frac{1}{64} = \left(\frac{R'}{R}ight)^3$. Thus,$\frac{R'}{R} = \sqrt[3]{\frac{1}{64}} = \frac{1}{4}$,which means $R' = \frac{R}{4}$.Since no external torque acts on the earth,its angular momentum $L = I\omega$ is conserved. The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.Applying conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$$\frac{2}{5} M R^2 \omega_1 = \frac{2}{5} M (R')^2 \omega_2$$R^2 \omega_1 = \left(\frac{R}{4}ight)^2 \omega_2$$R^2 \omega_1 = \frac{R^2}{16} \omega_2$$\omega_2 = 16 \omega_1$Since $\omega = \frac{2\pi}{T}$,we have $\frac{2\pi}{T_2} = 16 \left(\frac{2\pi}{T_1}ight)$,which implies $T_2 = \frac{T_1}{16}$.Given $T_1 = 24 	ext{ h}$,we get $T_2 = \frac{24}{16} 	ext{ h}$.Comparing this with $\frac{24}{x} 	ext{ h}$,we find $x = 16$.

If the earth suddenly shrinks to $\frac{1}{64}$th of its original volume with its mass remaining the same,the period of rotation of earth becomes $\frac{24}{x} \text{ h}$. The value of $x$ is $.......$

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