question_answer

The time period of a simple pendulum in a lift descending with constant acceleration g is:

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 \[T=2\pi \sqrt{\frac{l}{0}}=\infty \]