Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is `I_(0)` as shown in the figure. A cavithy DEF 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 -

Options

A. `(7)/(8)I_(0)`
B. `(15)/(16)I_(0)`
C. `(3)/(4)I_(0)`
D. `(31)/(32)I_(0)`

Answer

B

Explanation

Moment of inertia of triangular lamina about perpendicular to the plane of lamina passing through point of centre of mass
`I_(cm)=I_(0)=(Ma^(2))/(6)`
Mass of this lamina is M and having area is `A=(sqrt(3a^(2)))/(2)`

Mass of the portion which is removed as shown in the figure.
`M.=(M)/(A)xxA.=(M)/((sqrt3a^(2))/(2))xx(sqrt3a^(2))/(8)=(M)/(4)`
`I_("remaining")I_(0)-I_("removed")`
`=(1)/(6)Ma^(2)-(1)/(6)M.((a)/(2))^(2)`
`=(1)/(6)Ma^(2)-(1)/(6)(M)/(4)((a)/(2))^(2)=(1)/(6)Ma^(2)(1-(1)/(16))=(15)/(16)I_(0)`