Time period of a simple pendulum in a stationary lift is T . If the...

The time period of a simple pendulum is given by$T = 2\pi\sqrt{\dfrac{l}{g}}$

In an upward accelerating frame the effective ${g}_{eff} = g + a$

$⇒ {T}_{2} = 2\pi\sqrt{\dfrac{l}{g + a}} = 2\pi\sqrt{\dfrac{l}{g + \dfrac{g}{6}}}$

$\dfrac{T}{{T}_{2}} = \sqrt{\dfrac{g + \dfrac{g}{6}}{g}} = \sqrt{\dfrac{7}{6}}$

$⇒ {T}_{2} = \sqrt{\dfrac{6}{7}}T$