A potentiometer of length L is used to compare emfs. If the balance lengths for cells of emf ε₁ and ε₂ are l₁ and l₂ respectively, then ε₁ / ε₂ equals
A{'text': 'l₁ + l₂', 'label': 'A'}
B{'text': 'L / (l₁ + l₂)', 'label': 'B'}
C{'text': 'l₁ × l₂', 'label': 'C'}
D{'text': 'l₁ / l₂', 'label': 'D'}
Answer & Solution
Correct answer: D. {'text': 'l₁ / l₂', 'label': 'D'}
1. In a potentiometer, the emf being compared equals K × l, where K is potential gradient and l is balance length.
2. For emf ε₁, ε₁ = K × l₁; for emf ε₂, ε₂ = K × l₂.
3. Dividing the two equations, ε₁ / ε₂ = l₁ / l₂.
4. So the ratio of emfs equals the ratio of balance lengths (K cancels).
_Source: NCERT Class 12 Physics, Ch 3 "Current Electricity", §3 potentiometer_
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