A cell of emf 10 V and internal resistance \( 3\,\Omega \) is connected to a uniform wire of length 500 cm and resistance \( 3\,\Omega \). The drop of potential with length in the wire is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10\,\mathrm{mV/cm} \)
The cell drives a current through its internal resistance and the wire in series. The potential drop along a uniform wire is the same for every centimetre, so it equals the total drop across the wire divided by its length.
\( \displaystyle I = \frac{\varepsilon}{R + r} = \frac{10}{3 + 3} = \frac{5}{3}\,\mathrm{A} \)
\( \displaystyle V_{\mathrm{wire}} = I R = \frac{5}{3} \times 3 = 5\,\mathrm{V} \)
\( \displaystyle \text{Drop per cm} = \frac{5\,\mathrm{V}}{500\,\mathrm{cm}} = 0.01\,\mathrm{V/cm} = 10\,\mathrm{mV/cm} \)