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From a circular disc of radius R and mass 9M, a small disc of mass M and radius $\frac{R}{3}$ 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
$\frac{40}{9}MR^2$
B
$MR^2$
C
$4MR^2$
D
$\frac{4}{9}MR^2$
Detailed Solution
Moment of inertia of a disc about its axis: $I = \frac{1}{2}(\text{mass})(\text{radius})^2$
Complete disc: $I_1 = \frac{1}{2}(9M)R^2 = \frac{9MR^2}{2}$
Removed disc (same axis, since it is concentric): $I_2 = \frac{1}{2}M\left(\frac{R}{3}\right)^2 = \frac{MR^2}{18}$
Remaining disc: $I = I_1 - I_2 = \frac{9MR^2}{2} - \frac{MR^2}{18} = \frac{81MR^2 - MR^2}{18}$
$I = \frac{80MR^2}{18} = \frac{40}{9}MR^2$
Complete disc: $I_1 = \frac{1}{2}(9M)R^2 = \frac{9MR^2}{2}$
Removed disc (same axis, since it is concentric): $I_2 = \frac{1}{2}M\left(\frac{R}{3}\right)^2 = \frac{MR^2}{18}$
Remaining disc: $I = I_1 - I_2 = \frac{9MR^2}{2} - \frac{MR^2}{18} = \frac{81MR^2 - MR^2}{18}$
$I = \frac{80MR^2}{18} = \frac{40}{9}MR^2$