All sides of an equilateral triangle are diameter of three identical uniform semi- circular rings each of mass m. Plane of each ring is perpendicular to the plane of paper. Then moment of inertia of this system of three semicircular rings about an axis through centroid of triangle and perpendicular to plane of paper is :

Options

A. `(5 ma^(2))/24`
B. `(5ma^(2))/16`
C. `(5ma^(2))/8`
D. `(5ma^(2))/6`

Explanation

`I'=I_(cm)+m(a/(2sqrt(3)))^(2)=(5ma^(2))/24=I_(cm)=(m(a/2)^(2))/2=(ma^(2))/8`
`I=3I'=(5ma^(2))/8`