In a potentiometer arrangement for comparing emf of two cells, it was found that a cell of emf 1.25 V gives a balance point at 0.35 m length of the wire. On replacing the cell by another cell, the balance point shifted to 0.63 m. The emf of the second cell is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2.25 V
In a potentiometer the wire is uniform and carries a steady current, so every unit length has the same potential drop (a constant potential gradient). A cell is balanced when the potential drop across the length l equals its emf, hence \( \varepsilon\propto l \).
\( \displaystyle \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} \)
\( \displaystyle \varepsilon_2 = \varepsilon_1\,\frac{l_2}{l_1} = 1.25\,\mathrm{V}\times\frac{0.63\,\mathrm{m}}{0.35\,\mathrm{m}} \)
\( \varepsilon_2 = 1.25\times1.8\,\mathrm{V} = 2.25\,\mathrm{V} \)
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):
Exercises – …
