 # The height ' h ' at which the weight of a body 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. (1) $\frac{\sqrt{5}}{2} R-R$

2. (2) $\frac{R}{2}$

3. (3) $\frac{\sqrt{5} R-R}{2}$

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

Correct Option: , 3

Solution:

(3) The acceleration due to gravity at a height $h$ is given by

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

Here, $G=$ gravitation constant

$M=$ mass of earth Given, $g=g^{\prime}$

The acceleration due to gravity at depth $h$ is

$g^{\prime}=\frac{G M}{R^{2}}\left(1-\frac{h}{R}\right)$

$\therefore \frac{G M}{(R+h)^{2}}=\frac{G M}{R^{2}}\left(1-\frac{h}{R}\right)$

$\therefore R^{3}=(R+h)^{2}(R-h)=\left(R^{2}+h^{2}+2 h R\right)(R-h)$

$\Rightarrow R^{3}=R^{3}+h^{2} R+2 h R^{2}-R^{2} h-h^{3}-2 h^{2} R$

$\Rightarrow h^{3}+h^{2}(2 R-R)-R^{2} h=0$

$\Rightarrow h^{3}+h^{2} R-R^{2} h=0$

$\Rightarrow h^{2}+h R-R^{2}=0$

$\Rightarrow h=\frac{-R \pm \sqrt{R^{2}+4(1) R^{2}}}{2}=\frac{-R+\sqrt{5} R}{2}=\frac{(\sqrt{5}-1)}{2} R$