The cells of e.m.f. \( E_1 \) and \( E_2 \) are connected as shown in figure and \( E_2 \gt E_1 \). When a potentiometer is connected between point X and Y, the balancing length of potentiometer wire is 150 cm. On connecting the same potentiometer between X and Z, the balancing length is 50 cm. The ratio of emf \( E_1 \) and \( E_2 \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 : 4
In a potentiometer the potential difference balanced is proportional to the balancing length, so the potential differences across XY and XZ are in the ratio of their balancing lengths. The two cells are joined in opposition, so between X and Y the potentiometer measures \( E_1 \), and between X and Z it measures the difference of the emfs, here \( E_2 - E_1 \).
\( E_1\propto150\,\mathrm{cm},\qquad E_2 - E_1\propto50\,\mathrm{cm} \)
\( \displaystyle \frac{E_1}{E_2-E_1} = \frac{150}{50} = 3 \Rightarrow E_2 = \frac{4}{3}E_1 \)
\( \displaystyle \frac{E_1}{E_2} = \frac{3}{4} \)
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.”
Potentiometer – “… From the last two equations
\( \displaystyle \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} \qquad (3.92) \)”