Question:
The density of a non-uniform rod of length 1 m is given by $\rho(x)=a\left(1+b x^{2}\right)$ where $\mathrm{a}$ and $\mathrm{b}$ are constant and $0 \leq \mathrm{x} \leq 1$. The centre of mass of the rod will be at
(a) $\frac{3(2+b)}{4(3+b)}$
(b) $\frac{4(2+b)}{3(3+b)}$
(c) $\frac{3(3+b)}{4(2+b)}$
(d) $\frac{4(3+b)}{3(2+b)}$
Solution:
The correct answer is (a)
$\frac{3(2+b)}{4(3+b)}$