Two thin discs, each of mass M and radius r meter, are attached as shown in the figure, to form a rigid body. The rotational inertia of this body about an axis perpendicular
to the plane of disc B passing through its center is

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$4 M r^{2}$

$2 M r^{2}$

$3 M r^{2}$

$5 M r^{2}$

answer is D.

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Detailed Solution

Moment of inertia of discs A and B about the axis through their center of mass and perpendicular to the plane will be $I_{A A} = I_{B B} = \frac{1}{2} M r^{2}$

Now, moment of inertia of disc A about an axis through B, by theorem of parallel axis will be

$\begin{array}{l} I_{A B} = I_{A A} + M (2 r)^{2} = \frac{1}{2} M r^{2} + 4 M r^{2} = \frac{9}{2} M r^{2} \\ So I = I_{B B} + I_{A B} = \frac{1}{2} M r^{2} + \frac{9}{2} M r^{2} = 5 M r^{2} \end{array}$