The period of simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5\, g$, the period will be
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(d) $\frac{{T'}}{T} = \sqrt {\frac{g}{{g' + a}}} = \sqrt {\frac{g}{{g + 5g}}} = \sqrt {\frac{1}{6}} $
==> $T' = \frac{T}{{\sqrt 6 }}$
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