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) :

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

Answer 1

Answer: (A)

$♦ M =$ mass of earth
$M _{1}=$ mass of shaded portion
$R =$ Radius of earth
$♦ M_{1}=\frac{M}{\frac{4}{3} \pi R^{3}} \cdot \frac{4}{3} \pi(R-h)^{3}$
$=\frac{M(R-h)^{3}}{R}$
$♦$ Weight of body is same at $P$ and $Q$
i.e. $mg _{ P }= mg _{ Q }$
$g_{P}=g_{Q}$
$\frac{G M_{1}}{(R-h)^{2}}=\frac{G M}{(R+h)^{2}} $
$\frac{G M(R-h)^{3}}{(R-h)^{2} R^{3}}=\frac{G M}{(R+h)^{2}}$
$(R-h)(R+h)^{2}=R^{3}$
$R^{3}-h R^{2}-h^{2} R-h^{3}+2 R^{2} h-2 R h^{2}=R^{3}$
$R^{2}-R h^{2}-h^{3}=0$
$R^{2}-R h-h^{2}=0$
$h^{2}+R h-R^{2}=0 \Rightarrow h=\frac{-R \pm \sqrt{R^{2}+4 R^{2}}}{2}$
i.e. $h=\frac{-R+\sqrt{5} R}{2}=\left(\frac{\sqrt{5}-1}{2}\right) R$