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) $\dfrac{\sqrt{3}}{2}T$
Concept. In an upward-accelerating lift $g_{\text{eff}} = g + a$; $T \propto \dfrac{1}{\sqrt{g_{\text{eff}}}}$.
Step 1. Effective gravity. $g + \dfrac{g}{3} = \dfrac{4g}{3}$.
Step 2. Period. $T' = T\sqrt{\dfrac34} = \dfrac{\sqrt3}{2}T$.
Answer. $\dfrac{\sqrt3}{2}T$ (Option B).
Solution by Sreeraj P, M.Sc Physics