The earth is assumed to be a sphere of radius ' $R$ ', and mass ' $M$ ' having period of rotation ' $T$ '.…
- $\frac{2 \pi \mathrm{MR}^2}{5 \mathrm{~T}}$
- $\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}$
- $\frac{\mathrm{MR}^2 \mathrm{~T}}{2 \pi}$
- $\frac{\mathrm{MR}^2 \mathrm{~T}}{4 \pi}$
Solution
Moment of inertia of solid sphere is $\frac{2}{5} \mathrm{MR}^2$ $\begin{aligned} & \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\ & \mathrm{~L}=\left(\frac{2}{5} \mathrm{MR}^2\right)\left(\frac{2 \pi}{\mathrm{~T}}\right)=\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}} \end{aligned}$
Asked in: MHT CET 2024 (03 May Shift 2)