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Time period of a simple pendulum is $4 \mathrm{~s}$ at a place on the earth where the acceleration due to gravity is $\pi^2 \mathrm{~ms}^{-2}$. Then the length of the pendulum in meters is
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The correct answer is:
$4$
We have
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow 4=2 \pi \sqrt{\frac{\ell}{\pi^2}} \Rightarrow 16=4 \pi^2 \cdot \frac{\ell}{\pi^2}$
$\Rightarrow \ell=4 \mathrm{sec}$
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow 4=2 \pi \sqrt{\frac{\ell}{\pi^2}} \Rightarrow 16=4 \pi^2 \cdot \frac{\ell}{\pi^2}$
$\Rightarrow \ell=4 \mathrm{sec}$
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