Resistance of rod,

$R=\frac{\rho l}{A}=\frac{\rho l^{2}}{X}$

( $\because$ volume of rod, $X=A l)$

Change in resistance of rod,

$\Delta R=\frac{\rho \cdot 2 l \Delta l}{X}$

Change in potential drop across rod, $\Delta V=i \Delta R=\frac{i \cdot \rho \cdot 2 l \Delta l-2 i \rho l^{2} \Delta l}{X}=\frac{2 i \rho l^{2}}{X} \cdot \varepsilon$ or $\quad \Delta V \propto \varepsilon$

As change in potential drop is directly proportional to strain, voltage as a function of strain is as shown below;