A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be
Diffcult
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(d) Time period of a physical pendulum
$T = 2\pi \sqrt {\frac{{{I_0}}}{{mgd}}} = 2\pi \sqrt {\frac{{\left( {\frac{1}{2}m{R^2} + m{R^2}} \right)}}{{mgR}}} $
$ = 2\pi \sqrt {\frac{{3R}}{{2g}}} $…..$(i)$
${T_{simple\;pendulum}} = 2\pi \sqrt {\frac{l}{g}} $…..$(ii)$
Equating $(i)$ and $(ii),$
$T = 2\pi \sqrt {\frac{{{I_0}}}{{mgd}}} = 2\pi \sqrt {\frac{{\left( {\frac{1}{2}m{R^2} + m{R^2}} \right)}}{{mgR}}} $
$ = 2\pi \sqrt {\frac{{3R}}{{2g}}} $…..$(i)$
${T_{simple\;pendulum}} = 2\pi \sqrt {\frac{l}{g}} $…..$(ii)$
Equating $(i)$ and $(ii),$
$l = \frac{3}{2}R.$
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