A) \[\sqrt{3}T\]
B) \[T/\sqrt{3}\]
C) \[\frac{\sqrt{3}T}{2}\]
D) T/3
Correct Answer: C
Solution :
Time period of the simple pendulum in the lift \[T=2\pi \sqrt{\frac{l}{ga}}\] Time period when lift starts accelerating upwards with acceleration g/3 \[T=2\pi \sqrt{\frac{l}{g+\frac{g}{3}}}\] \[T=2\pi \times \sqrt{3}\sqrt{\frac{l}{4g}}\] or \[T=2p\sqrt{\frac{l}{g}}\times \frac{\sqrt{3}}{2}\] or \[T=\frac{\sqrt{3}}{2}T\]You need to login to perform this action.
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