A ring of diameter $2m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$
Diffcult
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$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mgl}}}=\sqrt{\frac{3 \mathrm{mr}^{2}}{2 \mathrm{mgr}}}$
or $T=2 \pi \sqrt{\frac{3 r}{2 m}}$
or $2 \pi \sqrt{\frac{l}{\mathrm{g}}}=2 \pi \sqrt{\frac{3 \mathrm{r}}{2 \mathrm{g}}}$
$\therefore \quad l=\frac{3}{2} \mathrm{r}=\frac{3}{2} \times 1=1.5 \mathrm{m}$
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