A $$6\,V$$  battery is connected to the terminals of a $$3\,m$$  long wire of uniform thickness and resistance of $$100\,\Omega .$$  The difference of potential between two points on the wire separated by a distance of $$50\,cm$$  will be

Solution :
Total current drawn from the battery
$$i = \frac{E}{{R + r}} = \frac{6}{{100 + 0}} = 0.06\,A$$
Resistance of $$50\,cm$$  wire is
$$R' = \frac{{\rho l'}}{A} = \left( {\frac{\rho }{A}} \right)l' = \left( {\frac{R}{l}} \right)l'\left( {\because R = \frac{{\rho l}}{A}} \right)$$
$$ = \frac{{100}}{{300}} \times 50$$
So, $$R' = \frac{{50}}{3}\Omega $$
Hence, the potential difference between two points on the wire separated by a distance $${l'}$$ is given by Ohm's law
i.e. $$V = iR' = 0.06 \times \frac{{50}}{3} = 1\,V$$