MHTCET 16 Apr 2026 Shift 2 Paper

Concept:The time period of a simple pendulum depends on the acceleration due to gravity, which decreases as we move to a height above the earth's surface.Explanation:The time period of a simple pendulum on the earth's surface is given by:$$T = 2\pi\sqrt{\frac{l}{g}}$$Here, $l$ is the length of the pendulum and $g$ is the acceleration due to gravity on the surface.We know that $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the earth, and $R$ is the earth's radius.When the pendulum is taken to a height $R$ above the surface, its distance from the centre of the earth becomes $R + R = 2R$.The new acceleration due to gravity at this height is:$$g' = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{g}{4}$$The new time period is therefore:$$T' = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}}$$$$T' = 2 \times 2\pi\sqrt{\frac{l}{g}} = 2T$$Given that the new time period is $xT$, we equate:$$xT = 2T$$$$x = 2$$Answer:The value of $x$ is 2, which is option C.