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

The time period of a simple pendulum in a stationary lift is $$T$$. When the lift accelerates upward with acceleration $$\frac{g}{6}$$, the effective gravitational acceleration is altered.

Initially, the time period is given by $$T = 2\pi\sqrt{\frac{l}{g}}.$$

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

Therefore, the new time period is $$T' = 2\pi\sqrt{\frac{l}{g_{eff}}} = 2\pi\sqrt{\frac{l}{\frac{7g}{6}}} = 2\pi\sqrt{\frac{6l}{7g}}.$$ Dividing by the original period gives $$\frac{T'}{T} = \frac{2\pi\sqrt{\frac{6l}{7g}}}{2\pi\sqrt{\frac{l}{g}}} = \sqrt{\frac{6}{7}},$$ so that $$T' = \sqrt{\frac{6}{7}}\cdot T.$$

Hence, the correct answer is Option C.