A rod of length L has non-uniform linear mass density given by  $\rho \left(x\right)=a+b{\left(\frac{x}{L}\right)}^2$, where a and b are constants and  $0\le x\le L$. The value of x for the centre of mass of the rod is at

$\begin{array}{l}Given : \frac{dm}{dx}=a+b\frac{x^2}{L^2} \text{}\\ \Rightarrow dm=\left(a+b\frac{x^2}{L^2}\right)dx \\ Now, x_{cm}=\frac{\int xdm}{\int dm}=\frac{\underset{0}{\overset{L}{{\displaystyle \int }}}\left(ax+\frac{b{x}^3}{L^2}\right)dx}{\underset{0}{\overset{L}{{\displaystyle \int }}}\left(a+\frac{b{x}^2}{L^2}\right)dx} \\ \Rightarrow x_{cm}={\frac{\frac{a{x}^2}{2}+\frac{b{x}^4}{4L^2}}{ax+\frac{b{x}^3}{3L^2}}}_0^L\text{}\\ \Rightarrow x_{cm}=\frac{\frac{a{L}^2}{2L}+\frac{b{L}^4}{4L^2}}{aL+\frac{b{L}^3}{3L^2}} \text{}\\ \Rightarrow x_{cm}=\frac{\frac{2a{L}^2+b{L}^2}{4}}{\frac{3aL+bL}{3}} =\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L \end{array}$