MHTCET 18 Apr 2026 Shift 1 Paper

Concept:For a seconds pendulum, the time period is fixed ($T=2$ s) at both locations, so the length is directly proportional to the local acceleration due to gravity, i.e. $l \propto g$.Explanation:The time period of a simple pendulum is given by:$$T = 2\pi \sqrt{\frac{l}{g}}$$Since $T$ is the same for a seconds pendulum at both places, we get $l \propto g$.On the surface of the Earth, the acceleration due to gravity is:$$g_s = \frac{GM}{R^2}$$At an altitude $h$ above the surface, the acceleration due to gravity is:$$g_h = \frac{GM}{(R+h)^2}$$Therefore, the ratio of lengths equals the ratio of gravitational accelerations:$$\frac{l_s}{l_h} = \frac{g_s}{g_h} = \frac{(R+h)^2}{R^2}$$Taking square roots on both sides:$$\frac{\sqrt{l_s}}{\sqrt{l_h}} = \frac{R+h}{R}$$Cross-multiplying gives:$$R\sqrt{l_s} = (R+h)\sqrt{l_h}$$Expanding and rearranging:$$R\left(\sqrt{l_s}-\sqrt{l_h}\right) = h\sqrt{l_h}$$Hence, the radius of the Earth is:$$R = \frac{h\sqrt{l_h}}{\sqrt{l_s}-\sqrt{l_h}}$$Answer:The correct option is Option A: $\displaystyle \frac{h \sqrt{l_{h}}}{\sqrt{l_{s}}-\sqrt{l_{h}}}$.