A rod of length $L$ has non-uniform linear mass density
given by $\rho(x)=a+b\left(\frac{x}{\mathrm{~L}}\right)^{2}$, where $a$ and $b$ are constants
and $0 \leq x \leq \mathrm{L}$. The value of $x$ for the centre of mass of the rod is at:
Correct Option: , 2
(2) Given,
Linear mass density, $\rho(x)=a+b\left(\frac{x}{L}\right)^{2}$
$X_{C M}=\frac{\int x d m}{\int d m}$ 
$\int d m=\int_{0}^{L} \rho(x) d x$
$=\int_{0}^{L}\left[a+b\left(\frac{x}{L}\right)^{2}\right] d x=a L+\frac{b L}{3}$
$\int_{0}^{L} x d m=\int_{0}^{L}\left(a x+\frac{b x^{3}}{L^{2}}\right) d x=\left(\frac{a L^{2}}{2}+\frac{b L^{2}}{4}\right)$
$\therefore \quad X_{\mathrm{CM}}=\frac{\left(\frac{a L^{2}}{2}+\frac{b L^{2}}{4}\right)}{a L+\frac{b L}{3}}$
$\Rightarrow X_{\mathrm{CM}}=\frac{3 L}{4}\left(\frac{2 a+b}{3 a+b}\right)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.