Question:
If the time period of a two meter long simple pendulum is $2 \mathrm{~s}$, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
Correct Option: , 3
Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
$2=2 \pi \sqrt{\frac{2}{g}}$
$\Rightarrow \mathrm{g}=2 \pi^{2}$
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