If the earth suddenly shrinks to $\frac{1}{64}$ of its original volume, while keeping the same mass, then…
If the earth suddenly shrinks to $\frac{1}{64}$ of its original volume, while keeping the same mass, then the duration of the day will be (Assume earth is a perfect sphere)
24 hours
1.5 hours
16 hours
48 hours
Solution
If earth shrinks to $\frac{1}{64}$ of its original volume;
$\frac{4}{3} \pi r^3=\frac{1}{64}\left(\frac{4}{3} \pi R^3\right)$
Here, $r=$ new radius and $R=$ original radius
$\begin{aligned} \Rightarrow & r^3 & =\frac{R^3}{64} \\ \text { or } & r & =\frac{R}{4}\end{aligned}$
As there is no external torque, angular momentum of earth remains constant
i.e., $\quad I_1 \omega_1=I_2 \omega_2 \quad\left[\because I_{\text {sphere }}=\frac{2}{5} M R^2\right]$
$\begin{aligned} \Rightarrow \quad & \quad \frac{2}{5} M R^2 \cdot \frac{2 \pi}{T_1}=\frac{2}{5} M r^2 \frac{2 \pi}{T_2} \\ & \quad\left(\therefore \omega=\frac{2 \pi}{T}, T=\text { time of revolution about axis }\right)\end{aligned}$
$\Rightarrow \quad T_2=\frac{r^2 T_1}{R^2}$
$\Rightarrow \quad T_2=\frac{\frac{R^2}{16} \times T_1}{R^2} \quad\left[\therefore r=\frac{R}{4}\right]$
$\begin{array}{ll}\Rightarrow & T_2=\frac{T_1}{16}=\frac{24}{16} \mathrm{~h} \\ \text { or } & T_2=1.5 \mathrm{~h}\end{array}$