Q 11-13-092JEE MainJEE Main 2022 (26 Jun, Shift 1)Easy
The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\dfrac g6$ vertically upwards then the time period will be (where $g$ = acceleration due to gravity)
Answer: (C) $\sqrt{\dfrac67}\,T$
Effective gravity $g_{\text{eff}} = g + \dfrac g6 = \dfrac{7g}{6}$, and $T\propto\dfrac1{\sqrt{g_{\text{eff}}}}$:
$$T' = T\sqrt{\frac{g}{7g/6}} = \sqrt{\frac67}\,T$$
Solution by Sreeraj P, M.Sc Physics