Acceleration Due to Gravity question from MHT CET 2026 (18 Apr, Shift 1)

Answer: (a)

1. Seconds pendulum: $T = 2\pi\sqrt{l/g}$ fixed, so $l \propto g$.
2. $\frac{l_h}{l_s} = \frac{g_h}{g} = \frac{R^2}{(R+h)^2} \Rightarrow \frac{\sqrt{l_h}}{\sqrt{l_s}} = \frac{R}{R+h}$
3. $R\sqrt{l_s} = (R+h)\sqrt{l_h} \Rightarrow R\left(\sqrt{l_s} - \sqrt{l_h}\right) = h\sqrt{l_h}$
4. So the answer is (a) $R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$.