Question

In a potentiometer experiment, when two cells of emfs \( E_1 \) and \( E_2 \) (\( E_2>E_1 \)) are connected in series, the balancing length is 160 cm. If one of the cells is reversed, the balancing length decreases by 75%. If \( E_1 = 1.2 \, V \), then \( E_2 \) is

• A. 2 V
• B. 2.4 V
• C. 1.8 V
• D. 1.5 V

Hint:

Use the ratio of balancing lengths to find unknown emf when cells are connected in series aiding and opposing.

Answer 1

The Correct Option is A

Step 1: Given data
- \(E_1 = 1.2\, V\), - Balancing length for series aiding = \(l = 160\, cm\), - Balancing length for series opposing = \(l' = 0.25 \times 160 = 40\, cm\) (since decreases by 75%). Step 2: Relation between emf and balancing length
\[ E_{total} = k \times l, \quad E'_{total} = k \times l', \] where \(k\) is potential gradient.
Step 3: Setup equations
For series aiding: \[ E_1 + E_2 = k \times 160 \] For series opposing: \[ E_2 - E_1 = k \times 40 \] Step 4: Solve for \(E_2\)
Divide equations: \[ \frac{E_1 + E_2}{E_2 - E_1} = \frac{160}{40} = 4 \] \[ E_1 + E_2 = 4 E_2 - 4 E_1 \implies 3 E_2 = 5 E_1 \implies E_2 = \frac{5}{3} \times 1.2 = 2\, V \] Step 5: Conclusion
The emf \( E_2 \) is 2 V.