Question

Two cells of e.m.f. \(E_1\) and \(E_2\) (\(E_1 > E_2\)) are connected as shown in figure.

When a potentiometer is connected between points A and B the balancing length of potentiometer wire is \(412\) cm. When same potentiometer wire is connected between points A and C the balancing length is \(103\) cm. The ratio \(E_1:E_2\) is

• A. \(6:1\)
• B. \(4:1\)
• C. \(4:3\)
• D. \(3:4\)

Hint:

The cells oppose each other, so A to C measures E1 minus E2.

Answer 1

The Correct Option is C

Step 1: Potentiometer principle:
The balancing length is proportional to the emf (potential difference) being measured, with the same wire gradient in both cases: $E \propto l$.

Step 2: Set up with the gradient k:
$E_1 = k\times412$ and $E_1 - E_2 = k\times103$.

Step 3: Subtract:
$E_2 = k(412 - 103) = 309k$. Then $\frac{E_1}{E_2} = \frac{412}{309} = \frac43$.

Final Answer:
E1 to E2 is 4 to 3, option (C). \[ \boxed{4:3} \]