A rod of length L is composed of a uniform length 1/2 L of wood mass is `m_(w)` and a uniform length 1/2 L of brass whose mass is `m_(b)`. The moment of inertia `I` of the rod about an axis perpendicular to the rod and through its centre is equal to

Options

A. `(m_(w)+m_(b))L^(2)//12`
B. `(m_(w)+m_(b))L^(2)//6`
C. `(m_(w)+m_(b))L^(2)//3`
D. `(m_(w)+m_(b))L^(2)//2`

Answer

A

Explanation

Moment of inertia,
`M_(1)=m_(w)xx((L//2)^(2))/(3)=m_(w)=(L^(2))/(12)`
Moment of inertia,
`M_(2)=m_(b)xx((L//2)^(2))/(3)=m_(b)xx(L^(2))/(12)`
Moment of inertia,
`M=M_(1)+M_(2)=m_(w)(L^(2))/(12)+m_(b)xx(L^(2))/(12)=(m_(w)+m_(b))(L^(2))/(12)`.