CGPET 2021 Solved Paper

Gravitational acceleration at a depth $d$ below the surface of earth or at a distance $r < R$ from the centre of earth is given as $g' = g \left(1 - \frac{d}{R}\right)$ .....(i) But $d + r = R \Rightarrow d = R - r$ .....(ii) From eqs.(i) and (ii), we have $g' = g \left(1 - \frac{R - r}{R}\right) = g \left(1 - \frac{R}{R} + \frac{r}{R}\right) = g \left(1 - 1 + \frac{r}{R}\right)$ ⇒ $g' = \frac{g r}{R}$ ∴ $g' \propto r$ ....(iii) Again, gravitational acceleration at a height $h$ from the surface of earth or at distance $r > R$ from the centre of earth is given as $g'' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$ $= \frac{g}{\left(1 + \frac{r - R}{R}\right)^2} \quad \left[\because r = h + R \Rightarrow h = r - R\right]$ $= \frac{g}{\left(\frac{r}{R}\right)^2} = \frac{g R^2}{r^2}$ ∴ $g^n \propto \frac{1}{r^2}$ ....(iv) Hence, from Eqs. (iii) and (iv), we conclude that the correct variation of $g$ with distance $r$ is shown in option (d).