Moment of inertia of an equilateral triangular lamina A B C , about the axis passing through its centre O and perpendicular to its plane is I 0 as shown in the figure. A cavity D E F is cut out from the lamina, where D , E , F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is:

Step-by-step solution

Given, A B = B C = A C = l Moment of inertia of a triangular lamina A B C I 0 = k m l 2 D E = E F = D F = 1 2 , A B = l 2 ∴ Moment of inertia of ∆ D E F I D E F = k m 4 l 2 2 I D E F = k 16   m l 2 . I D E F = I 0 16 Moment of inertia of the remaining part I r e m a i n = I 0 - I 0 16 = 15 I 0 16