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Q.

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3, then the time period of the pendulum is

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a

T3

b

T3

c

32T

d

3 T

answer is C.

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Detailed Solution

For stationary lift T1=2πlgFor ascending lift with acceleration a,  T2=2πlg+a⇒T1T2=g+ag⇒T1T2=g+g3g=43 ⇒T2=32T
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