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(C) $1$. The moment of inertia of the disc about its central axis is $I_{cm} = \frac{1}{2}MR^2$.$2$. Using the Parallel Axis Theorem,the moment of ine

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$(M/2 + 8m)R^2$

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(C) $1$. The moment of inertia of the disc about its central axis is $I_{cm} = \frac{1}{2}MR^2$.$2$. Using the Parallel Axis Theorem,the moment of inertia of the disc about an axis passing through a point on its rim (perpendicular to the plane) is $I_{disc} = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.$3$. Let the four particles be at positions $(R, 0), (0, R), (-R, 0), (0, -R)$. Let the axis pass through the particle at $(R, 0)$.$4$. The distances of the four particles from the axis are: $r_1 = 0$,$r_2 = \sqrt{R^2 + R^2} = R\sqrt{2}$,$r_3 = \sqrt{(2R)^2 + 0} = 2R$,$r_4 = \sqrt{R^2 + R^2} = R\sqrt{2}$.$5$. The moment of inertia of the particles is $I_{particles} = m(r_1^2 + r_2^2 + r_3^2 + r_4^2) = m(0 + 2R^2 + 4R^2 + 2R^2) = 8mR^2$.$6$. Total moment of inertia $I = I_{disc} + I_{particles} = \frac{3}{2}MR^2 + 8mR^2 = (\frac{M}{2} + 8m)R^2$.

Column-$I$	Column-$II$
$(1)$ $SI$ unit of torque	$(a)$ $m$
$(2)$ $SI$ unit of radius of gyration	$(b)$ $N\,m$
	$(c)$ $Js^{-2}$

Four particles each of mass $m$ are lying symmetrically on the rim of a disc of mass $M$ and radius $R$. The moment of inertia of the system about an axis passing through one of the particles and perpendicular to the plane of the disc is:

Similar Questions

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ $SI$ unit of torque $(a)$ $m$

$(2)$ $SI$ unit of radius of gyration $(b)$ $N\,m$

$(c)$ $Js^{-2}$

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