In a potentiometer arrangement, a cell of emf 1.5 V gives a balance point at 45.0 cm length of the wire. If the cell is replaced by another cell of emf 2.25 V. Where will the balance point shift to?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 67.5 cm
A potentiometer wire carries a steady current, so the potential drop along it is uniform. At balance no current flows through the cell, and its emf equals the drop across the balancing length: \( E = k\,l \). So the emf is proportional to the balancing length.
\( \displaystyle \frac{E_2}{E_1} = \frac{l_2}{l_1} \)
\( \displaystyle l_2 = l_1 \times \frac{E_2}{E_1} = 45.0\,\mathrm{cm} \times \frac{2.25\,\mathrm{V}}{1.5\,\mathrm{V}} \)
\( \displaystyle l_2 = 45.0 \times 1.5 = 67.5\,\mathrm{cm} \)
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 – “… The jockey is moved along the wire till at a point N1, at a distance l1 from A, there is no deflection in the galvanometer. … From the last two equations
\( \displaystyle \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} \qquad (3.92) \)
This simple mechanism thus allows one to compare the emf’s of any two sources.”