A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance $100 \Omega$. The…
A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance $100 \Omega$. The difference of potential between two points 50 cm apart on the wire is
1 V
2 V
1.5 V
3 V
Solution
$\begin{aligned}
& \mathrm{R}=\frac{\rho l}{\mathrm{~A}} \\
& \frac{100}{3}=\frac{\rho}{A}
\end{aligned}$
$\because \quad$ Total resistance for 50 cm wire is
$\begin{aligned}
& \mathrm{R}^{\prime}=\frac{\rho}{\mathrm{A}} l=\frac{100}{3} \times\left(50 \times 10^{-2}\right)=\frac{50}{3} \Omega ...(i)\\
& \mathrm{I} \doteq \frac{\mathrm{~V}}{\mathrm{R}}=\frac{6}{100} \mathrm{~A}...(ii)
\end{aligned}$
$\therefore \quad$ The potential difference between two points 50 cm apart is
$V=I R^{\prime}=\frac{6}{100} \times \frac{50}{3}=1 \mathrm{~V}$
...[From (i) and (ii)]