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(D) When the lift is stationary,the time period of the simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.When the lift moves upwards with an

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$\sqrt{\frac{2}{3}} T$

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(D) When the lift is stationary,the time period of the simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.When the lift moves upwards with an acceleration $a = g/2$,a pseudo force acts downwards on the pendulum bob.The effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{2} = \frac{3g}{2}$.The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{L}{g_{eff}}}$.Substituting $g_{eff} = \frac{3g}{2}$,we get $T' = 2 \pi \sqrt{\frac{L}{3g/2}} = 2 \pi \sqrt{\frac{2L}{3g}}$.Comparing this with the original time period $T$,we have $T' = \sqrt{\frac{2}{3}} 	imes (2 \pi \sqrt{\frac{L}{g}}) = \sqrt{\frac{2}{3}} T$.

Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2$,the time period of the pendulum will be

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