A) 2 g
B) 3 g
C) 4 g
D) g
Correct Answer: B
Solution :
Time period of simple pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}}\] ... (i) When the lift is moving up with an acceleration a, then time period becomes \[T'=2\pi \sqrt{\frac{l}{g+a}}\] Here, \[T'=\frac{T}{2}\] \[\Rightarrow \] \[\frac{T}{2}=2\pi \sqrt{\frac{l}{g+a}}\] ?. (ii) Dividing Eq. (ii) by Eq. (i), we get \[a=3g\]You need to login to perform this action.
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