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

Option 1 - 9.8 ms-2

Option 2 - 16 m/s2

Option 3 - 2π2 ms-2

Option 4 - π2 ms-2

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 3

Detailed Solution:

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

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