Question

If a simple pendulum has length \(l\) and time period \(T\) then a pendulum of length \(2l\) will have a time period of

• A. \(2\pi T\)
• B. \(\dfrac{1}{2\pi}T\)
• C. \(\sqrt{2}T\)
• D. \(\dfrac{1}{\sqrt{2}}T\)

Hint:

For a simple pendulum, the time period varies as the square root of the length: \[ T\propto \sqrt{l}. \] Doubling the length multiplies the time period by \(\sqrt{2}\).

Answer 1

The Correct Option is C

Step 1: Write the time period law.
For a simple pendulum \[ T = 2\pi\sqrt{\frac{l}{g}} \] so the period grows with the square root of the length.

Step 2: Use proportionality.
Since $T \propto \sqrt{l}$, changing the length scales the period by the square root of the same factor.

Step 3: Double the length.
Length goes from $l$ to $2l$, so the period scales by $\sqrt{2}$.

Step 4: Get the new period.
\[ T' = \sqrt{2}\,T \]

Step 5: Answer.
\[ \boxed{\sqrt{2}\,T} \]