From a uniform square plate of side a and mass `m`, a square portion DEFG of side `(a)/(2)` is removed. Then, the moment of inertia of remaining portion about the axis AB is

Options

A. `(7ma^(2))/(16)`
B. `(3ma^(2))/(16)`
C. `(3ma^(2))/(4)`
D. `(9ma^(2))/(16)`

Answer

B

Explanation

`I_(AB)=I_(1)+I_(2)+I_(3)`
`=((m//4)(a//2)^(2))/(3)+((m//4)(a//2)^(2))/(3)`
`+[((m//4)(a//2)^(2))/(12)+(m//4)((3a)/(4))^(2)]`
`=(3)/(16)ma^(2)`