# Solve this following

Question:

The height ' $h$ ' at which the weight of a body will be the same as that at the same depth 'h' from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

1. $\frac{\sqrt{5} R-R}{2}$

2. $\frac{\sqrt{5}}{2} R-R$

3. $\frac{\mathrm{R}}{2}$

4. $\frac{\sqrt{3} R-R}{2}$

Correct Option: 1

Solution:

- $\mathrm{M}=$ mass of earth

$\mathrm{M}_{1}=$ mass of shaded portion

$\mathrm{R}=$ Radius of earth

$\mathrm{M}_{1}=\frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^{3}} \cdot \frac{4}{3} \pi(\mathrm{R}-\mathrm{h})^{3}$

$=\frac{\mathrm{M}(\mathrm{R}-\mathrm{h})^{3}}{\mathrm{R}}$

- Weight of body is same at $\mathrm{P}$ and $\mathrm{Q}$

i.e. $m g_{P}=m g_{Q}$

$\mathrm{g}_{\mathrm{P}}=\mathrm{g}_{\mathrm{Q}}$

$\frac{\mathrm{GM}_{1}}{(\mathrm{R}-\mathrm{h})^{2}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\frac{G M(R-h)^{3}}{(R-h)^{2} R^{3}}=\frac{G M}{(R+h)^{2}}$

$(\mathrm{R}-\mathrm{h})(\mathrm{R}+\mathrm{h})^{2}=\mathrm{R}^{3}$

$\mathrm{R}^{3}-\mathrm{h} \mathrm{R}^{2}-\mathrm{h}^{2} \mathrm{R}-\mathrm{h}^{3}+2 \mathrm{R}^{2} \mathrm{~h}-2 \mathrm{Rh}^{2}=\mathrm{R}^{3}$

$\mathrm{R}^{2}-\mathrm{Rh}^{2}-\mathrm{h}^{3}=0$

$\mathrm{R}^{2}-\mathrm{Rh}-\mathrm{h}^{2}=0$

$h^{2}+R h-R^{2}=0 \Rightarrow h=\frac{-R \pm \sqrt{R^{2}+4 R^{2}}}{2}$

ie $h=\frac{-R+\sqrt{5} R}{2}=\left(\frac{\sqrt{5}-1}{2}\right) R$