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If the Earth were to suddenly contract to \(\frac{1}{n}\) th of its present radius without any change in its mass, the duration of the new day will be nearly
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Verified Answer
The correct answer is:
\(24 / \mathrm{n}^2 \mathrm{hr}\).
Hints: \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2\)
\(\begin{aligned}
& \frac{2}{5} \mathrm{MR}^2\left(\frac{2 \pi}{\mathrm{T}_1}\right)=\frac{2}{5} \mathrm{M} \cdot \frac{\mathrm{R}^2}{\mathrm{n}^2}\left(\frac{2 \pi}{\mathrm{T}_2}\right) \\
& \mathrm{T}_2=\frac{\mathrm{T}_1}{\mathrm{n}^2}=\frac{24}{\mathrm{n}^2}
\end{aligned}\)
\(\begin{aligned}
& \frac{2}{5} \mathrm{MR}^2\left(\frac{2 \pi}{\mathrm{T}_1}\right)=\frac{2}{5} \mathrm{M} \cdot \frac{\mathrm{R}^2}{\mathrm{n}^2}\left(\frac{2 \pi}{\mathrm{T}_2}\right) \\
& \mathrm{T}_2=\frac{\mathrm{T}_1}{\mathrm{n}^2}=\frac{24}{\mathrm{n}^2}
\end{aligned}\)
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