Question

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

• A. $\frac{\sqrt{5}}{2}R-R$
• B. $\frac{R}{2}$
• C. $\frac{\sqrt{5}R-R}{2}$
• D. $\frac{\sqrt{3}R-R}{2}$

Answer 1

The correct option is C $\frac{\sqrt{5}R-R}{2}$

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

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

Here,
$G=$ gravitation constant
$M=$ mass of earth
The acceleration due to fravity at depth h is

$g^{\prime }=\frac{GM}{R^2}(1-\frac{h}{R})$

Given, $g=g^{\prime }$

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

$\therefore R^3=(R+h{)}^2(R-h)=(R^2+h^2+2hR\left)\right(R-h)$
$\Rightarrow R^3=R^3+h^{2}R+2h{R}^2-R^{2}h-h^3-2h^{2}R$
$\Rightarrow h^3+h^2(2R-R)-R^{2}h=0$
$\Rightarrow h^3+h^{2}R-R^{2}h=0$
$\Rightarrow h^2+hR-R^2=0$
$\Rightarrow h=\frac{-R\pm \sqrt{R^2+4\left(1\right)R^2}}{2}=\frac{-R+\sqrt{5}R}{2}=\frac{(\sqrt{5}-1)}{2}R$