Time period of a simple pendulum in a stationary lift

For a pendulum of length $l$, the time period is $T = 2\pi \sqrt{\frac{l}{g}}.$

When the lift accelerates upward

If the lift accelerates vertically upward with acceleration $a = \dfrac{g}{6}$, then the effective gravitational acceleration becomes $g_{\text{eff}} = g + a = g + \frac{g}{6} = \frac{7g}{6}.$

New time period

The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{7g/6}}.$

Rewrite in terms of $T$: $T' = 2\pi \sqrt{\frac{6l}{7g}} = \sqrt{\frac{6}{7}}\left(2\pi \sqrt{\frac{l}{g}}\right).$

Therefore, $T' = \sqrt{\frac{6}{7}}\,T.$

Match with options

$\sqrt{\frac{6}{7}}T$ corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

Hence, the derived answer agrees with the stored answer.