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Time period of a simple pendulum will be double, if we

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Increase the length 2 times

Decrease the length 4 times

Increase the length 4 times

Decrease the length 2 times

answer is D.

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Detailed Solution

$T \propto \sqrt{l} \frac{T_{2}}{T_{1}} = \sqrt{\frac{l_{2}}{l_{1}}} \frac{T_{2}}{T_{1}} = \sqrt{\frac{4 l_{1}}{l_{1}}} T_{2} = 2 T_{1}$

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