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(B) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.For a second's pendulum,the time period $T$ is constant

Vedclass · Accepted Answer

$4$

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(B) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.For a second's pendulum,the time period $T$ is constant $(T = 2 	ext{ s})$.From the formula,we have $T^2 \propto \frac{l}{g}$,which implies $l \propto g$ for a constant $T$.If the acceleration due to gravity $g'$ on the new planet is $4$ times the acceleration due to gravity $g$ on Earth $(g' = 4g)$,then to keep the time period $T$ constant,the new length $l'$ must satisfy the ratio $\frac{l'}{l} = \frac{g'}{g}$.Substituting the values,we get $\frac{l'}{l} = \frac{4g}{g} = 4$.Therefore,the length of the pendulum must be made $4$ times its original length.

If a second's pendulum is moved to a planet where the acceleration due to gravity is $4$ times that on Earth,the length of the second's pendulum on the planet should be made .......... times to keep the time period constant.

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