A) \[T=2\pi \sqrt{\frac{l}{8}}\]
B) \[T=2\pi \sqrt{\frac{l}{2g}}\]
C) zero
D) infinite
Correct Answer: D
Solution :
Key Idea: When lift is descending effective acceleration decreases. The time period of a simple pendulum in descending lift is given by \[T=2\pi \sqrt{\frac{l}{g}}\] Effective acceleration when lift is descending is given by \[Tmg-T=mg\] or \[T=0\] or \[mg=0\] or \[g=0\] Hence \[T=2\pi \sqrt{\frac{l}{0}}=\infty \]You need to login to perform this action.
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