A) \[\frac{T}{\sqrt{3}}\]
B) \[\frac{T}{3}\]
C) \[\frac{\sqrt{3}}{2}T\]
D) \[\sqrt{3T}\]
Correct Answer: C
Solution :
Key Idea: Effective acceleration increases when the lift moves upwards. The time period of a simple pendulum of length \[l\] is \[T=2\,\pi \sqrt{\frac{I}{g}}\] where g is acceleration due to gravity when lift moves up \[g=g+a=g+\frac{g}{3}=\frac{4\,g}{3}\] \[\therefore \] \[\frac{T}{T}=\sqrt{\frac{g}{g}}\] \[\therefore \] \[\frac{T}{T}=\sqrt{\frac{\frac{4}{3}g}{g}}=\frac{2}{\sqrt{3}}\] \[\Rightarrow \] \[T=\frac{\sqrt{3}}{2}T\]
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