Q 11-07-170JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
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)
Answer: (C) $\dfrac{\sqrt5 R - R}{2}$
Equate $g$ at height $h$ and at depth $h$ (exact expressions, since $h$ is not small):
$$\frac{g}{(1+h/R)^{2}} = g\left(1 - \frac{h}{R}\right)$$
With $x = h/R$: $(1+x)^{2}(1-x) = 1 \Rightarrow x - x^{2} - x^{3} = 0 \Rightarrow x^{2} + x - 1 = 0$.
So $x = \dfrac{\sqrt5 - 1}{2}$ and $h = \dfrac{\sqrt5R - R}{2}$.
Solution by Sreeraj P, M.Sc Physics