Time period of a simple pendulum in a stationary lift is T. If the lift accelerates with $\frac{\mathrm{g}}{6}$ vertically upwards then the time period will be : (where g=acceleration due to gravity)

Time period of pendulum, ${\mathrm{T}}_{\mathrm{P}}=2\mathrm{\pi }\sqrt{\frac{\text{ string length }}{\text{ downward acceleration }}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{{\mathrm{a}}_{\mathrm{y}}}}$

If the lift is stationary, ${\mathrm{a}}_{\mathrm{y}}=\mathrm{g}$

$\text{ Time period is }\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

If the lift is accelerating with $\frac{\mathrm{g}}{6}$

Time period $\mathrm{T}\text{'}=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{{\displaystyle \frac{7\mathrm{g}}{6}}}}=\sqrt{\frac{6}{7}}\mathrm{T}$