If time period of a two meter long simple pendulum is 2, s , acceleration due to gravity at place wh

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13.Oscillations

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If the time period of a two meter long simple pendulum is $2\, s$, the acceleration due to gravity at the place where pendulum is executing $S.H.M.$ is

A

$\pi^{2}\, ms ^{-2}$

B

$9.8\, ms ^{-2}$

C

$2 \pi^{2}\, ms ^{-2}$

D

$16\, m / s ^{2}$

(JEE MAIN-2021)

Solution

$T =2 \pi \sqrt{\frac{l}{ g }}$

$2=2 \pi \sqrt{\frac{2}{ g }}$

$\Rightarrow g =2 \pi^{2}$

Standard 11

Physics

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