A current of 1.0 mA is flowing through a potentiometer wire of length 4 m and of resistance \( 40\,\Omega \). The potential gradient of the potentiometer wire will be:
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 of the wire. By Ohm's law the drop across the whole wire is \( IR \).
\( I = 1.0\,\mathrm{mA} = 1.0\times10^{-3}\,\mathrm{A} \)
\( V = IR = 1.0\times10^{-3}\,\mathrm{A} \times 40\,\Omega = 0.04\,\mathrm{V} \)
\( \displaystyle k = \frac{V}{L} = \frac{0.04\,\mathrm{V}}{4\,\mathrm{m}} = 10^{-2}\,\mathrm{V/m} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Potentiometer – “… A current I flows through the wire which can be varied by a variable resistance (rheostat, R) in the circuit. Since the wire is uniform, the potential difference between A and any point at a distance l from A is
\( \displaystyle \varepsilon(l) = \phi\, l \qquad (3.89) \)
where φ is the potential drop per unit length.”