The resistivity of a potentiometer wire is \( 10^{-7}\,\Omega\,\mathrm{m} \) and its cross-section area is \( 10^{-6}\,\mathrm{m^{2}} \). When a current of 0.1 A flows through the wire, its potential gradient is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{-2}\,\mathrm{V/m} \)
The potential gradient of a potentiometer wire is the potential drop per unit length, \( k = V/l \). For a wire of length l the drop is \( V = IR \) with \( R = \rho l/A \), so the length cancels:
\( \displaystyle k = \frac{IR}{l} = \frac{I\rho}{A} \)
\( \displaystyle k = \frac{0.1\times10^{-7}}{10^{-6}}\,\mathrm{V/m} = 10^{-2}\,\mathrm{V/m} \)