If the length of simple pendulum is doubled, then its time period becomes

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4 times

$\frac{1}{2} times$

$\sqrt{2} times$

$\frac{1}{\sqrt{2}} times$

answer is C.

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Detailed Solution

$\begin{aligned} T = 2 π \sqrt{\frac{l}{g}} \\ \frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}} \\ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{l_{1}}{l_{2}} \\ T_{2}^{2} = T_{1}^{2} \times \frac{l_{2}}{l_{1}} \\ T_{2}^{2} = T^{2} \times \frac{2 l}{l} (∵ T_{1} = T, l_{2} = 2 l, l_{1} = l) \\ T_{2}^{2} = T^{2} \times 2 \\ T_{2} = (\sqrt{2}) T \end{aligned}$

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