The moment of inertia of a uniform rod of length $2L$ and mass $m$ about an axis through centre and inclined at an angle $\theta$  to rod is $\frac{m{L}^2}{n}{\sin }^2\theta$,  then find '$n$'.

Consider an element of thickness $dx$ at a distance $x$ from the centre of the rod. Man of that element $=\frac{m}{2L}dx$ Moment of inertia of that element about the required axis is

$\begin{array}{l}dI=dm{r}^2\\ \Rightarrow dI=\frac{m}{2L}{\left(x\sin \theta \right)}^{2}dx\end{array}$

Moment of inertia of the complete rod is obtained by integrating $dI$ from $x=L/2$ to

$x=-L/2.$   then we get  $\frac{M{L}^2}{3}{\sin }^2\theta$