Q: From a circular disc of radius R and mass 9 M , a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

(a) $\frac{40}{9} M R^2 $

(b) $ M R^2 $

(c) $ 4 M R^2 $

(d) $\frac{4}{9} M R^2 $

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Sol: The moment of inertia of the disc of mass 9M & radius R about an axis perpendicular to the plane of the disc and passing through its centre will be

$I_1 = \frac{1}{2}(9M)R^2 $

The moment of inertia of the disc of mass M & radius R/3 about an axis perpendicular to the plane of the disc and passing through its centre will be

$I_2 = \frac{1}{2}(M)(R/3)^2 $

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre will be

$I_R = I_1 – I_2 $

$I_R = \frac{9}{2}MR^2 – \frac{1}{18}MR^2 $

$I_R = \frac{40}{9}M R^2 $