A man measures the time period of a simple pendulum inside a stationary lift and finds it to be 10sec. If the lift accelerates upwards with an acceleration g4, then find the time period of the pendulum.
A
8√5sec
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B
4√5sec
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C
10√5sec
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D
2√5sec
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Solution
The correct option is B4√5sec In a stationary lift T=2π√lg=10seconds...(1)
For a lift accelerated upwards, the effective value of acceleration due to gravity: g′=g+a
Replacing g by g′ in Eq.(1), T′=2π√l(g+a)...(2)
From Eq. (1) & (2) T′T=√gg+a T′T=
⎷gg+g4=√45 ∴T′=2T√5=4√5seconds