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

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\frac{2R}{3}

\frac{3R}{4}

\frac{5 R}{4}

\frac{3R}{2}

answer is D.

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Detailed Solution

Time period of a physical pendulum

$T = 2 π \sqrt{\frac{I_{0}}{mgd}} = 2 π \sqrt{\frac{\frac{1}{2} {mR}^{2} + {mR}^{2}}{mgR}}$

$T = 2 π \sqrt{\frac{3 R}{2 g}} \to (1)$

$T_{simple pendulum} = 2 π \sqrt{\frac{l}{g}} \to (2)$

$equating (1) & (2)$

$2 π \sqrt{\frac{3 R}{2 g}} = 2 π \sqrt{\frac{l}{g}}$

$\sqrt{\frac{3 R}{2 g}} = \sqrt{\frac{l}{g}}$

$\frac{3 R}{2 g} = \frac{l}{g} \Rightarrow \frac{3 Rg}{2 g} = l$

$\frac{3 R}{2} = l$

(or)

$T = 2 π \sqrt{\frac{I}{mgd}} \Rightarrow T = 2 π \sqrt{\frac{\frac{3 {MR}^{2}}{2}}{MgR}} = 2 π \sqrt{\frac{3R}{2g}}$

$\Rightarrow 2 π \sqrt{\frac{3R}{2g}} = 2 π \sqrt{\frac{l}{g}}$

$\Rightarrow l = \frac{3R}{2}$

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