MCQ
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$9.8\, ms ^{-2}$
- ✓$2 \pi^{2}\, ms ^{-2}$
- D$16\, m / s ^{2}$
✓
Answer
Correct option: C.
$2 \pi^{2}\, ms ^{-2}$
c
$T =2 \pi \sqrt{\frac{l}{ g }}$
$T =2 \pi \sqrt{\frac{l}{ g }}$
$2=2 \pi \sqrt{\frac{2}{ g }}$
$\Rightarrow g =2 \pi^{2}$
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