A ring is suspended from a point $S$ on its rim as shown in the figure. When displaced from equilibrium, it oscillates with time period of $1\,second.$ The radius of the ring is ..... $m$ (take $g = \pi ^2$ )
AIEEE 2012, Diffcult
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We know that the time period of a physical pendulum is given by $T=2 \pi \sqrt{\frac{I_{\text {support }}}{m g l_{\text {cm }}}}$
$I_{\text {support }}=m R^{2}+m R^{2}=2 m R^{2}$
$T=2 \pi \sqrt{\frac{\left(2 m R^{2}\right)}{m g R}} =2 \pi \sqrt{\frac{2 R}{g}}$
$\Rightarrow R=\frac{T^{2} g}{8 \pi^{2}}$
A seconds pendulum is a pendulum whose period is precisely $two\, seconds; one \,second$ for a swing in one direction and $one\, second$ for the return swing. Ie. $T=2 \varepsilon,$ so we have
$R=\frac{2^{2} \times \pi^{2}}{8 \pi^{2}}=0.5 \mathrm{m}$
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