The potentiometer wire of length 400 cm has a resistance of \( 20\,\Omega \). It is connected in series with a resistance of \( 10\,\Omega \) and a cell of emf 9 V having negligible resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. The value of L is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 160 cm
At the balance point the potential drop across the length L of the wire equals the emf of the source. The potential gradient comes from the current in the driver circuit, in which the wire and the \( 10\,\Omega \) resistance are in series with the 9 V cell.
\( \displaystyle I = \frac{9\,\mathrm{V}}{20\,\Omega + 10\,\Omega} = 0.3\,\mathrm{A} \)
\( V_{\mathrm{wire}} = I\times20\,\Omega = 6\,\mathrm{V} \) over 400 cm
\( \displaystyle \text{potential gradient} = \frac{6\,\mathrm{V}}{400\,\mathrm{cm}} = 0.015\,\mathrm{V\,cm^{-1}} \)
\( \displaystyle L = \frac{2.4\,\mathrm{V}}{0.015\,\mathrm{V\,cm^{-1}}} = 160\,\mathrm{cm} \)
Shortcut: The balance length is the same fraction of the wire as the source emf is of the drop across the wire: \( L = \frac{2.4}{6}\times400\,\mathrm{cm} = 160\,\mathrm{cm} \).
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 – “… 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.”