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MHT CET 2023 (10 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2023 (10 May, Shift 1)

Q64MHT CET 2023 · 10 MAY · SHIFT 1Medium

Acceleration Due to Gravity

There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be

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Answer: (a)

$g_p = \frac{G(2M)}{(2R)^2} = \frac{g}{2}$
$T = 2\pi\sqrt{\frac{l}{g}} \propto \frac{1}{\sqrt{g}} \Rightarrow T_p = 2\sqrt{2}\text{ s}$ (seconds pendulum has $T = 2\text{ s}$ on earth).
So the answer is (a) $2\sqrt{2}\text{ s}$ .

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