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Potentiometer

Mediumphysics

A potentiometer wire balances the EMF of one cell at a length of 240 cm and that of a second cell at 160 cm under identical conditions. What is the ratio of the two EMFs?

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About This Question

Subject
physics
Chapter
current electricity
Topic
potentiometer
Difficulty
Medium
Year
2025
Tags
potentiometerEMF comparisonbalancing lengthpotential gradientnull method

Solution

Correct Answer:

3 : 2

A potentiometer works on the principle that, for a steady current through a uniform wire, the potential drop is directly proportional to the length of wire. At the balance point no current is drawn from the cell under test, so the cell's true EMF is compared without any internal-resistance error, giving (\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}). Substituting the balancing lengths (l_1 = 240) cm and (l_2 = 160) cm gives (\frac{\varepsilon_1}{\varepsilon_2} = \frac{240}{160} = \frac{3}{2}), i.e. 3 : 2. The ratio 2 : 3 simply inverts the correct proportion. The ratio 4 : 3 comes from misreducing 240 : 160. The ratio 8 : 5 arises from an arithmetic slip in simplification. This is the NCERT comparison-of-EMF method using a potentiometer. A plausibility check confirms it: the cell balanced over the longer wire length must have the greater EMF, so the ratio must exceed unity, and 3 : 2 satisfies this expectation neatly. The genuine strength of the potentiometer over an ordinary voltmeter is that, by drawing zero current at balance, it imposes no loading on the cell, so the comparison reflects the true EMFs rather than terminal voltages diminished by internal-resistance drops, which is exactly why it is the preferred laboratory instrument for accurate EMF comparison.

This medium difficulty physics question is from the chapter current electricity, covering the topic of potentiometer. It appeared in the 2025 exam.

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