Question

If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(\mathrm{T}\) is..... \(\\{\mathrm{A}\\}\left(\pi \mathrm{MR}^{3} / \mathrm{T}\right)\) \(\\{\mathrm{B}\\}\left(\operatorname{MR}^{2} \pi / \mathrm{T}\right)\) \(\\{C\\}\left(2 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\) \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\)

Short answer

The short answer is: The angular momentum of the Earth with given conditions is \(L = \dfrac{4\pi MR^2}{5T}\). Thus, the correct option is D.

Step-by-step solution

Recall the formula for angular momentum of a solid sphere

The formula for angular momentum (L) of a solid sphere of mass M and radius R, rotating about an axis through its center with an angular velocity ω is: \(L = \dfrac{2}{5}MR^2 ω\)

Find the relationship between period and angular velocity

The period (T) of rotation of a rotating object is the time taken for one complete revolution. The angular velocity (ω) is the rate of change of angular displacement. The relationship between angular velocity (ω) and period (T) can be written as: \(ω = \dfrac{2\pi}{T}\)

Substitute the relationship between ω and T in the angular momentum formula

Now, let's substitute the relationship between ω and T in the formula for the angular momentum of a solid sphere: \(L = \dfrac{2}{5}MR^2 \cdot \dfrac{2\pi}{T}\)

Simplify the expression

Simplify the expression for the angular momentum: \(L = \dfrac{4\pi MR^2}{5T}\)

Compare the simplified expression with the given options

Now, we compare our simplified expression with the given options: \(L = \dfrac{4\pi MR^2}{5T}\) matches with option D. So the correct answer is: \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\).