Question

A $6\, V$ battery is connected to the terminals of a three metre 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

• A. 2 V
• B. 3 V
• C. 1 V
• D. 1.5 V

Answer 1

Answer: (C)

Total current drawn from the battery
$i=\frac{E}{R+r}=\frac{6}{100+0}=0.06 A$
Resistance of $50 cm$ wire is $R^{\prime}=\frac{\rho l^{\prime}}{A}=\left(\frac{\rho}{A}\right) l^{\prime}$
$=\left(\frac{R}{l}\right) l^{\prime}\left(\because R=\frac{\rho l}{A}\right) $
$=\frac{100}{300} \times 50$
So, $R^{\prime}=\frac{50}{30} \Omega$
Hence, the potential between two point on the wire spareted by a distance $l$ is
$V=i R=0.06 \times \frac{50}{3}=1 \,V$