A disc of radius R and mass M is pivoted at t | vedclass

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(D) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I_0}{Mgd}}$,where $I_0$ is the moment of inertia about the pivot point an

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$\frac{3}{2}R$

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(D) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I_0}{Mgd}}$,where $I_0$ is the moment of inertia about the pivot point and $d$ is the distance from the pivot to the center of mass.For a disc pivoted at its rim,the moment of inertia about the center is $I_{cm} = \frac{1}{2}MR^2$. Using the parallel axis theorem,the moment of inertia about the rim is $I_0 = I_{cm} + Md^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Here,$d = R$. Substituting these values into the formula:$T = 2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}} = 2\pi \sqrt{\frac{3R}{2g}}$.The time period of a simple pendulum of length $l$ is $T = 2\pi \sqrt{\frac{l}{g}}$.Equating the two time periods:$2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{3R}{2g}}$.Squaring both sides and simplifying,we get $l = \frac{3}{2}R$.

$A$ disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If a 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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