A massive rod of length 1m placed straight on a smooth horizontal. surface is pulled longitudinally by a force F of 10 N as shown in the figure.

The tension in rod varies as T=10$\sqrt{x}$. The linear mass density of rod 

$\text{ Here, }T=ma,10\sqrt{x}=ma$

$$\begin{gathered} \Rightarrow m=\frac{10}{a}\sqrt{x} \\ \text{ or } \frac{dm}{dx}=\frac{10}{a}\frac{d\left(\sqrt{x}\right)}{dx} \\ or \frac{dm}{dx}=\frac{10}{a}\frac{1}{2\sqrt{x}} \\ or \frac{dm}{dx}=\frac{5}{a\sqrt{x}} \\ or dm=\frac{5}{a\sqrt{x}}dx \\ or \\ \lambda dx=\frac{5}{a\sqrt{x}}dx \\ \lambda =\frac{5}{a\sqrt{x}} \\ \Rightarrow \lambda \propto \frac{1}{\sqrt{x}} \end{gathered}$$

ALTERNATE METHOD :

  T = 10$\sqrt{X}$ = m a

but   m = $\rho x$  where $\rho$ is linear density.

so     $\rho x a = 10 \sqrt{x}$

$\Rightarrow \rho \alpha \frac{1}{\sqrt{x}}$