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A simple pendulum of length 1 is made to oscillate with amplitude of 45 degrees. The acceleration due to gravity is $\mathrm{g} .$ Let $T_{0}=2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be
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The correct answer is:
Slightly more then $\mathrm{T}_{0}$
$T=2 \pi \sqrt{\frac{\ell}{g}}\left(1+\frac{\theta_{0}^{2}}{16}\right)$
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