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Question

A man measures the period of simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates downwards with acceleration of g4, the period of oscillation will be

A
T×32s
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B
T×23s
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C
T2s
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D
Ts
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Solution

The correct option is B T×23s
Effective acceleration due to gravity when the lift accelerates downwards,
g=gg4=3g4
Time period T=2π(lg) ...... (i)
T =2π(lg) =2π(4l3g)=2π×23(lg) ....(ii)
Now dividing Eq. (ii) by Eq. (i), we get
TT=23
or T=T×23

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