Q 11-07-209JEE MainMedium
The acceleration due to gravity at a height $\dfrac R2$ above the earth's surface is equal to that at a depth $d$ below the surface, where $R$ is the radius of the earth (assumed uniform). Then $d$ is
Answer: (B) $\dfrac{5R}{9}$
Concept. Above the surface $g_h = \dfrac{g}{(1 + h/R)^2}$ (exact, not the small-height approximation). Inside a uniform earth $g_d = g\left(1 - \dfrac dR\right)$.
Step 1. At height $\frac R2$.
$$g_h = \frac{g}{(3/2)^2} = \frac{4g}{9}$$
Step 2. Equate to depth.
$$g\left(1 - \frac dR\right) = \frac{4g}{9} \;\Rightarrow\; \frac dR = \frac59$$
Answer. $\dfrac{5R}{9}$ (Option B). Using $g_h \approx g(1 - 2h/R)$ would wrongly give $d = R$; the approximation fails for $h = R/2$.
Solution by Sreeraj P, M.Sc Physics