A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ sec. If the lift accelerates upwards with an acceleration $\frac{g}{4}$, then the period of the pendulum will be
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(c) In stationary lift $T = 2\pi \sqrt {\frac{l}{g}} $
In upward moving lift $T' = 2\pi \sqrt {\frac{l}{{(g + a)}}} $ ($a = $Acceleration of lift)
$ \Rightarrow \frac{{T'}}{T} = \sqrt {\frac{g}{{g + a}}} = \sqrt {\frac{g}{{\left( {g + \frac{g}{4}} \right)}}} = \sqrt {\frac{4}{5}} $
$ \Rightarrow T' = \frac{{2T}}{{\sqrt 5 }}$
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