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: 1
Solution:
(1)
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{g}}}$
$\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{g}}$
$\mathrm{g}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{T}^{2}}$
$=\frac{4 \pi^{2} \times 2}{(2)^{2}}=2 \pi^{2} \mathrm{~ms}^{-2}$
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