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(D) This is a case of a physical pendulum.The time period of a physical pendulum is given by $T = 2 \pi \sqrt{\frac{I}{M g L}}$,where $I$ is the momen

Vedclass · Accepted Answer

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

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(D) This is a case of a physical pendulum.The time period of a physical pendulum is given by $T = 2 \pi \sqrt{\frac{I}{M g L}}$,where $I$ is the moment of inertia about the point of suspension and $L$ is the distance between the center of mass and the point of suspension.For a uniform disc of mass $M$ and radius $R$,the moment of inertia about its center of mass is $I_{	ext{cm}} = \frac{1}{2} M R^2$.Using the parallel axis theorem,the moment of inertia about a point on its periphery is $I = I_{	ext{cm}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.The distance from the center of mass to the point of suspension is $L = R$.Substituting these values into the formula for the time period:$T = 2 \pi \sqrt{\frac{\frac{3}{2} M R^2}{M g R}}$$T = 2 \pi \sqrt{\frac{3 R}{2 g}}$Thus,the correct option is $(d)$.

$A$ uniform disc of mass $M$ and radius $R$ is suspended in a vertical plane from a point on its periphery. Its time period of oscillation is ........

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