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

MOI of system
`I= I_(omega) + I_(b)`

`I= [(m_(w) ((L)/(2))^(2))/(12) + m_(w) ((L)/(4))^(2)] + [(m_(b) ((L)/(2))^(2))/(12) + m_(b) ((L)/(4))^(2)]`
`I= (L^(2))/(12) [M_(w) + m_(b)]`