A potentiometer wire is 100 cm long with a constant potential difference maintained across it. Two cells are connected in series, first supporting each other and then opposing each other. The balance points are found at 50 cm and 10 cm from the positive end in the two cases. What is the ratio of the emfs of the two cells?

Introduction / Context:
Potentiometers are precision instruments used to compare emfs of cells and to measure internal resistance without drawing current from the source. This question describes a standard experiment where two cells are connected in series, first in the same direction so that their emfs support each other, and then in opposite directions so that they oppose each other. The balance lengths on the potentiometer wire are given for both cases, and you need to find the ratio of the emfs of the two cells.

Given Data / Assumptions:

• Total potentiometer wire length is 100 cm with uniform potential gradient.
• When two cells are in series aiding, balance point is at 50 cm from the positive end.
• When the same two cells are in series opposing, balance point is at 10 cm from the positive end.
• Let the emfs of the two cells be E1 and E2, with E1 greater than E2 for convenience.
• Potential gradient along the wire is constant and equal in both experiments.

Concept / Approach:
In a potentiometer, emf is proportional to the balancing length, because E equals k times L, where k is the potential gradient and L is the balance length. When cells are connected in series supporting each other, the effective emf is E1 plus E2. When they are connected in series opposing, the effective emf is E1 minus E2, assuming E1 is greater than E2. Using the two balance lengths, we can form two proportionality equations involving the same constant k and then divide them to eliminate k and find the ratio E1 to E2.

Step-by-Step Solution:
Step 1: For the series aiding case, write (E1 + E2) = k * 50. Step 2: For the series opposing case, write (E1 - E2) = k * 10. Step 3: Divide the first equation by the second: (E1 + E2) / (E1 - E2) = (k * 50) / (k * 10) = 50 / 10 = 5. Step 4: So (E1 + E2) / (E1 - E2) = 5, which implies E1 + E2 = 5 * (E1 - E2). Step 5: Expand and rearrange: E1 + E2 = 5E1 - 5E2, so 6E2 = 4E1, giving E1 / E2 = 6 / 4 = 3 / 2.

Verification / Alternative check:
To check the result, assume E2 equals 2 units. Then E1 equals 3 units from the ratio 3 : 2. In the aiding case, total emf is 5 units, proportional to 50 cm, so 1 unit corresponds to 10 cm of length. In the opposing case, effective emf is 1 unit, which should correspond to 10 cm, exactly matching the given data. This consistency confirms that the ratio E1 : E2 equals 3 : 2 and validates the algebraic steps.

Why Other Options Are Wrong:
3 : 4 and 4 : 3 do not satisfy the relation (E1 + E2) / (E1 - E2) = 5 when substituted.
5 : 4 and 5 : 1 also fail to produce the correct ratio of balance lengths and do not agree with the derived equations.
Only 3 : 2 remains consistent with the experimental conditions and the proportionality of emf and balance length.

Common Pitfalls:
A frequent mistake is to mix up which combination of cells (aiding or opposing) corresponds to which length, or to forget that the effective emf in the opposing connection is the difference E1 minus E2, not the sum. Another error is to attempt direct substitution without using the proportionality factor k, leading to confused algebra. Always write expressions for the effective emfs in both cases, set them equal to k times the measured lengths, and then divide to eliminate k before solving for the ratio of emfs.

Final Answer:
The ratio of the emfs of the two cells is 3 : 2.