A light rod of length `l` has two masses `m_1` and `m_2` attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is.

Options

A. `sqrt(m_(1)m_(2)l^(2))`
B. `(m_(1) m_(2))/(m_(1) + m_(2)) l^(2)`
C. `(m_(1) + m_(2))/(m_(1) m_(2)) l^(2)`
D. `(m_(1) + m_(2)) l^(2)`

Answer

B

Explanation

`I= m_(1)r_(1)^(2) + m_(2)r_(2)^(2)`, Here, `r_(1)= (m_(2)l)/((m_(1) + m_(2))) and r_(2)= (m_(1)l)/((m_(1) + m_(2)))`
`rArr I= m_(1) ((m_(2))/(m_(1) + m_(2))l)^(2) + m_(2) ((m_(1))/(m_(1) + m_(2))l)^(2)`
`therefore I= (m_(1)m_(2)l^(2))/((m_(1) + m_(2)))`