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(A) Let $T_1$ be the time period when the lift moves upwards with acceleration $a$, and $T_2$ be the time period when the lift moves downwards with th

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$6$

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(A) Let $T_1$ be the time period when the lift moves upwards with acceleration $a$, and $T_2$ be the time period when the lift moves downwards with the same acceleration $a$.The effective acceleration due to gravity when moving upwards is $g' = g + a$, and when moving downwards is $g'' = g - a$.The time periods are given by:$T_1 = 2\pi \sqrt{\frac{l}{g+a}}$ $(i)$$T_2 = 2\pi \sqrt{\frac{l}{g-a}}$ $(ii)$Given the ratio $T_1 : T_2 = 1 : 2$, we have:$\frac{T_1}{T_2} = \sqrt{\frac{g-a}{g+a}} = \frac{1}{2}$Squaring both sides:$\frac{g-a}{g+a} = \frac{1}{4}$$4(g - a) = g + a$$4g - 4a = g + a$$3g = 5a$$a = \frac{3g}{5} = \frac{3 	imes 10}{5} = 6 \,ms^{-2}$

$A$ simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio $1: 2$, then the acceleration of the lift is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,ms^{-2}$)

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