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

We first note the general expression for the position of the centre of mass of a continuous one-dimensional object:

$$x_{\text{cm}}=\frac{1}{M}\int_{0}^{l}x\,\mathrm d m$$

Here the linear mass density is given as $$\rho(x)=a+b\left(\dfrac{x}{l}\right)^2$$, so for an infinitesimal element of length $$\mathrm d x$$ at position $$x$$ we have

$$\mathrm d m=\rho(x)\,\mathrm d x=\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x.$$

We therefore need two separate integrals: one for the total mass $$M$$ and one for the numerator of $$x_{\text{cm}}$$.

Total mass

$$\begin{aligned} M &=\int_{0}^{l}\rho(x)\,\mathrm d x \\ &=\int_{0}^{l}\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x \\ &=\int_{0}^{l}a\,\mathrm d x+\int_{0}^{l}b\left(\frac{x}{l}\right)^2\mathrm d x. \end{aligned}$$

We integrate term by term:

$$\int_{0}^{l}a\,\mathrm d x=a\bigl[x\bigr]_{0}^{l}=a\,l,$$

and

$$\int_{0}^{l}b\left(\frac{x}{l}\right)^2\mathrm d x =b\int_{0}^{l}\frac{x^{2}}{l^{2}}\mathrm d x =\frac{b}{l^{2}}\left[\frac{x^{3}}{3}\right]_{0}^{l} =\frac{b}{l^{2}}\cdot\frac{l^{3}}{3} =\frac{b\,l}{3}.$$

Adding these contributions,

$$M=a\,l+\frac{b\,l}{3}=l\left(a+\frac{b}{3}\right).$$

Numerator for $$x_{\text{cm}}$$

We next evaluate $$\displaystyle\int_{0}^{l}x\,\rho(x)\,\mathrm d x$$:

$$\begin{aligned} \int_{0}^{l}x\,\rho(x)\,\mathrm d x &=\int_{0}^{l}x\left[a+b\left(\frac{x}{l}\right)^2\right]\mathrm d x \\ &=\int_{0}^{l}a\,x\,\mathrm d x+\int_{0}^{l}b\,x\left(\frac{x}{l}\right)^2\mathrm d x. \end{aligned}$$

For the first part,

$$\int_{0}^{l}a\,x\,\mathrm d x =a\left[\frac{x^{2}}{2}\right]_{0}^{l} =a\cdot\frac{l^{2}}{2}=\frac{a\,l^{2}}{2}.$$

For the second part,

$$\int_{0}^{l}b\,x\left(\frac{x}{l}\right)^2\mathrm d x =b\int_{0}^{l}\frac{x^{3}}{l^{2}}\mathrm d x =\frac{b}{l^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{l} =\frac{b}{l^{2}}\cdot\frac{l^{4}}{4} =\frac{b\,l^{2}}{4}.$$

Hence the numerator is

$$\int_{0}^{l}x\,\rho(x)\,\mathrm d x =\frac{a\,l^{2}}{2}+\frac{b\,l^{2}}{4} =l^{2}\left(\frac{a}{2}+\frac{b}{4}\right) =l^{2}\left(\frac{2a+b}{4}\right).$$

Centre of mass position

Substituting the numerator and the total mass into the formula, we get

$$x_{\text{cm}}=\frac{l^{2}\left(\dfrac{2a+b}{4}\right)}{\,l\left(a+\dfrac{b}{3}\right)}.$$ We simplify step by step:

First cancel one factor of $$l$$:

$$x_{\text{cm}}=l\,\frac{\,2a+b\,}{4\left(a+\dfrac{b}{3}\right)}.$$

Next write the denominator over a common denominator of three:

$$a+\frac{b}{3}=\frac{3a+b}{3}.$$

So

$$x_{\text{cm}}=l\,\frac{2a+b}{4}\cdot\frac{3}{3a+b} =l\,\frac{3(2a+b)}{4(3a+b)}.$$

Finally, pull out the numerical factor:

$$x_{\text{cm}}=\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)l.$$

This matches option B exactly. Hence, the correct answer is Option B.