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(A) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is

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It increases.

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(A) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.Since the acceleration due to gravity on the Moon $(g_m)$ is approximately $1/6$th of the acceleration due to gravity on Earth $(g_e)$,we have $g_m According to the relation $T \propto \frac{1}{\sqrt{g}}$,as the value of $g$ decreases,the time period $T$ increases.Therefore,when a simple pendulum is taken from the Earth's surface to the Moon's surface,its time period increases.

What happens to the time period of a simple pendulum when it is taken to the moon's surface from the earth's surface?

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