Q107 / 150 · MHT CET 2024 (03 May Shift 2) · Physics
The earth is assumed to be a sphere of radius ' $R$ ', and mass ' $M$ ' having period of rotation ' $T$ '. The angular momentum of earth about its axis of rotation is
- A$\frac{2 \pi \mathrm{MR}^2}{5 \mathrm{~T}}$
- Correct answerB$\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}}$
- C$\frac{\mathrm{MR}^2 \mathrm{~T}}{2 \pi}$
- D$\frac{\mathrm{MR}^2 \mathrm{~T}}{4 \pi}$
Solution
Angular momentum,
$\mathrm{L}=\mathrm{I} \omega$
Moment of inertia of solid sphere is $\frac{2}{5} \mathrm{MR}^2$ $\begin{aligned} & \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\ & \mathrm{~L}=\left(\frac{2}{5} \mathrm{MR}^2\right)\left(\frac{2 \pi}{\mathrm{~T}}\right)=\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}} \end{aligned}$
Moment of inertia of solid sphere is $\frac{2}{5} \mathrm{MR}^2$ $\begin{aligned} & \mathrm{L}=\frac{2}{5} \mathrm{MR}^2 \omega \\ & \mathrm{~L}=\left(\frac{2}{5} \mathrm{MR}^2\right)\left(\frac{2 \pi}{\mathrm{~T}}\right)=\frac{4 \pi \mathrm{MR}^2}{5 \mathrm{~T}} \end{aligned}$
Hence, the correct answer is option: 'B'.