When rod is stretched, its length increases. Potential drop across the rod also increases due to increase in

resistance of rod.

Resistance of rod, $R=\frac{\rho l}{A}=\frac{\rho l^2}{X}$

($\because$ volume of rod, $X=Al$)

Change in resistance of rod, $\mathrm{\Delta }R=\frac{\rho .2l\mathrm{\Delta }l}{X}$

Change in potential drop across rod, 

$\mathrm{\Delta }V=i\mathrm{\Delta }R=\frac{i.\rho .2l\mathrm{\Delta }l}{X}=\frac{2i\rho l^2}{X}.\frac{\mathrm{\Delta }l}{l}=\frac{2i\rho l^2}{X}.\epsilon$

or $\mathrm{\Delta }V\propto \epsilon$

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