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A simple pendulum of length 1 m is freely suspended from the ceiling of an elevator. The time period of small oscillations as the elevator moves up with an acceleration of $2 m / s^{2}$ is $(use g = 10 m / s^{2})$

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$\frac{π}{\sqrt{5}} s$

$\sqrt{\frac{2}{5}} π s$

$\frac{π}{\sqrt{2}} s$

$\frac{π}{\sqrt{3}} s$

answer is D.

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Detailed Solution

Given, Length of simple pendulum, l = 1m. Acceleration of an elevator, a = 2m/s². Acceleration due to gravity, g = 10m/s² we

know that $T = 2π \sqrt{\frac{l}{g+a}} = 2π \sqrt{\frac{1}{10+2}}$

$= 2π \sqrt{\frac{1}{12}} = π \sqrt{\frac{4}{12}} T = \frac{π}{\sqrt{3}} s$

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