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(B) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the pivot,$m$ is the m

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$R/\sqrt{2}$

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(B) The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the pivot,$m$ is the mass,and $d$ is the distance from the center of mass to the pivot.For a disc,the moment of inertia about the center is $I_{cm} = \frac{1}{2}mR^2$. By the parallel axis theorem,the moment of inertia about a point at distance $l$ from the center is $I = I_{cm} + ml^2 = \frac{1}{2}mR^2 + ml^2$.Substituting this into the time period formula: $T = 2\pi \sqrt{\frac{\frac{1}{2}mR^2 + ml^2}{mgl}} = 2\pi \sqrt{\frac{R^2/2 + l^2}{gl}} = 2\pi \sqrt{\frac{1}{g} \left( \frac{R^2}{2l} + l \right)}$.To minimize $T$,we must minimize the expression $L = \frac{R^2}{2l} + l$.Taking the derivative with respect to $l$ and setting it to zero: $\frac{dL}{dl} = -\frac{R^2}{2l^2} + 1 = 0$.Solving for $l$: $l^2 = \frac{R^2}{2} \implies l = \frac{R}{\sqrt{2}}$.

$A$ number of holes are drilled along a diameter of a disc of radius $R$. To get the minimum time period of oscillations,the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be:

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