Discussion
If a simple pendulum has length l and time period T then a pendulum of length 2l will have a time period of
Answer: C
The new time period is \(\sqrt{2}T\).
For a simple pendulum the period depends on the square root of the length:
\[ T = 2\pi\sqrt{\frac{l}{g}} \]
\[ \begin{aligned} T' &= 2\pi\sqrt{\frac{2l}{g}} \\ &= \sqrt{2}\cdot 2\pi\sqrt{\frac{l}{g}} = \sqrt{2}T \end{aligned} \]
Doubling the length multiplies the period by \(\sqrt{2}\), about 1.41. The options \(2\pi T\) and \(\frac{1}{2\pi}T\) wrongly bring the constant \(2\pi\) into the ratio, where it cancels. \(\frac{1}{\sqrt{2}}T\) inverts the relation: a longer pendulum swings more slowly, not faster.
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