A) \[\sqrt{2}T\]
B) \[\frac{T}{\sqrt{2}}\]
C) \[\frac{\sqrt{3}}{2}T\]
D) \[\frac{T}{3}\]
E) \[\frac{2}{3}T\]
Correct Answer: C
Solution :
When lift accelerates upwards, then effective acceleration on the pendulum \[{{g}_{eff}}=g+\frac{g}{3}=\frac{4g}{3}\] \[\therefore \]Time period\[T=2\pi \sqrt{\frac{l}{{{g}_{eff}}}}=2\pi \sqrt{\frac{l}{4g/3}}\] \[=\frac{\sqrt{3}}{2}.2\pi \sqrt{\frac{l}{g}}\] \[=\frac{\sqrt{3}}{2}T\]You need to login to perform this action.
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