Q 11-07-068JEE MainJEE Main 2024 (29 Jan, Shift 1)Medium
At what distance above and below the surface of the earth a body will have same weight? (Take radius of earth as $R$)
Answer: (D) $\dfrac{\sqrt5R - R}{2}$
At height $h$: $g_h = \dfrac{gR^2}{(R+h)^2}$. At depth $h$: $g_d = g\left(1 - \dfrac hR\right)$.
Setting them equal with $x = h/R$:
$$\frac{1}{(1+x)^2} = 1 - x \;\Rightarrow\; (1-x)(1+x)^2 = 1 \;\Rightarrow\; x - x^2 - x^3 = 0$$
$$x^2 + x - 1 = 0 \;\Rightarrow\; x = \frac{\sqrt5 - 1}{2}$$
So $h = \dfrac{\sqrt5R - R}{2}$.
Solution by Sreeraj P, M.Sc Physics