Q 11-13-024NEETJEE MainMedium
A simple pendulum has period $T$ in a stationary lift. If the lift accelerates upward at $\dfrac{g}{3}$, the period becomes
Answer: (B) $\dfrac{\sqrt{3}}{2}T$
Effective gravity in the lift: $g_{eff} = g + \dfrac{g}{3} = \dfrac{4g}{3}$.
$$T' = T\sqrt{\frac{g}{g_{eff}}} = T\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}T$$
Solution by Sreeraj P, M.Sc Physics