The moments of inertia of a non-uniform circular disc (of mass M and radius R) about four mutually perpendicular tangents AB, BC, CD, DA are `I_(1), I_(2), I_(3)` and `I_(4)`, respectively (the square ABCD circumscribes the circle). The distance of the center of mass of the disc from its geometrical center is given by -

Options

A. `(1)/(4MR)sqrt((I_(1)-I_(3))^(2)+(I_(2)-I_(4))^(2))`
B. `(1)/(12MR)sqrt((I_(1)-I_(3))^(2)+(I_(2)-I_(4))^(2))`
C. `(1)/(3MR)sqrt((I_(1)-I_(2))^(2)+(I_(2)-I_(4))^(2))`
D. `(1)/(2MR)sqrt((I_(1)+I_(3))^(2)+(I_(2)+I_(4))^(2))`

Answer

A

Explanation

`I_(1)=I_(CM)+M(R-y)^(2)`
I_(3)=I_(CM)+M(R+y)^(2)`
So, `I_(3)-I_(1)=M[(R+y)^(2)-(R-y)^(2)]`
same as
`I_(4)-I_(2)=M[(R+x)^(2)-(R-x)^(2)]`
By distance of CM from centre of disc `=x^(2)+y^(2)=(1)/(4MR)sqrt((I_(1)-I_(3))^(2)+(I_(2)-I_(4))^(2))`