As shown in the figure,two thin coplanar circ | vedclass

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(D) The moment of inertia of the system is the sum of the moments of inertia of the two discs about the given axis.Let $I_B$ be the moment of inertia

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$5$

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(D) The moment of inertia of the system is the sum of the moments of inertia of the two discs about the given axis.Let $I_B$ be the moment of inertia of disc $B$ about an axis perpendicular to its plane and passing through its centre. $I_B = \frac{1}{2} Mr^2$.Let $I_A$ be the moment of inertia of disc $A$ about the same axis. Since the axis is parallel to the axis passing through the centre of disc $A$ at a distance $d = 2r$,we use the parallel axis theorem: $I_A = I_{cm} + Md^2$.Here,$I_{cm} = \frac{1}{2} Mr^2$ and $d = 2r$.So,$I_A = \frac{1}{2} Mr^2 + M(2r)^2 = \frac{1}{2} Mr^2 + 4Mr^2 = 4.5 Mr^2$.The total moment of inertia $I = I_B + I_A = 0.5 Mr^2 + 4.5 Mr^2 = 5 Mr^2$.

As shown in the figure,two thin coplanar circular discs $A$ and $B$ each of mass $M$ and radius $r$ are attached to form a rigid body. The moment of inertia of this system about an axis perpendicular to the plane of disc $B$ and passing through its centre is (in $Mr^2$)

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