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A man measures the time period of a simple pendulum inside a stationary lift and finds it to be 10 sec. If the lift accelerates upwards with an acceleration g4, then find the time period of the pendulum.
  1. 85 sec
  2. 45 sec
  3. 25 sec
  4. 105 sec


Solution

The correct option is B 45 sec
In a stationary lift T=2πlg=10 seconds     ...(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=2πl(g+a)     ...(2)
From Eq. (1) & (2)
TT=gg+a
TT= gg+g4=45
T=2T5=45 seconds

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