The moment of inertia of a uniform rod of length `2l` and mass `m` about an axis `xy` passing through its centre and inclined at an enable `alpha` is

Options

A. `(ml^(2))/(12)`
B. `(ml^(2))/(12) cos^(2) theta`
C. `(ml^(2))/(12) sin ^(2) theta`
D. `(ml^(2))/(3) sin^(2) theta`

Explanation

When we select a segment along the length of rod at a distance x from the centre then its perpendicular distance from the axis becomes x sin e. Hence sino appears as extra constant term multiplied with the result if we compare this calculation with that when axis is passing through the centre and perpendicular to the length of the rod. So we can see that option (c) is correct.