Q 11-07-175JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
The value of the acceleration due to gravity is $g_1$ at a height $h = \dfrac R2$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_1$ at a depth $d$ below the surface of the earth. The ratio $\left(\dfrac dR\right)$ equals:
Answer: (B) $\dfrac59$
At height $R/2$: $g_1 = \dfrac{g}{(1 + 1/2)^{2}} = \dfrac{4g}{9}$.
At depth $d$: $g\left(1 - \dfrac dR\right) = \dfrac{4g}9 \Rightarrow \dfrac dR = \dfrac59$.
Solution by Sreeraj P, M.Sc Physics