A disc of radius R is made to oscillate about | vedclass

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(A) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgl}}$.Here,$I$ is the moment of inertia about the axis of rotation,$m

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$2\pi \sqrt{\frac{3R}{2g}}$

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(A) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgl}}$.Here,$I$ is the moment of inertia about the axis of rotation,$m$ is the mass of the disc,and $l$ is the distance from the center of mass to the axis of rotation.For a disc oscillating about an axis passing through its periphery,the distance $l = R$.The moment of inertia of a disc about its center is $I_{cm} = \frac{1}{2}mR^2$.Using the parallel axis theorem,the moment of inertia about the periphery is $I = I_{cm} + mR^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$.Substituting these values into the formula:$T = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}}$.

$A$ disc of radius $R$ is made to oscillate about a horizontal axis passing through its periphery. Its time period would be

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