Q 11-07-082JEE MainJEE Main 2023 (31 Jan, Shift 1)Medium
At a certain depth $d$ below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3R$ above earth surface, where $R$ is radius of earth (Take $R=6400\ \text{km}$). The depth $d$ is equal to
Answer: (D) $4800\ \text{km}$
At height $3R$: $g_h=\dfrac{g}{(4)^2}=\dfrac{g}{16}$.
At depth $d$: $g\left(1-\dfrac dR\right)=4\times\dfrac{g}{16}=\dfrac g4\Rightarrow d=\dfrac{3R}{4}=4800\ \text{km}$.
Solution by Sreeraj P, M.Sc Physics