Q 11-07-025NEETJEE MainMedium
At what height above the Earth's surface is the acceleration due to gravity equal to its value at a depth of $\dfrac{R}{2}$, where $R$ is the radius of the Earth?
Answer: (A) $(\sqrt{2} - 1)R$
At depth $\dfrac{R}{2}$: $g_d = g\left(1 - \dfrac{1}{2}\right) = \dfrac{g}{2}$.
At height $h$: $g\dfrac{R^2}{(R + h)^2} = \dfrac{g}{2} \Rightarrow R + h = \sqrt{2}R \Rightarrow h = (\sqrt{2} - 1)R \approx 0.41R$.
Solution by Sreeraj P, M.Sc Physics