Question:
When cell of e.m.f. ' \( E_1 \) ' is connected to potentiometer wire, the balancing length is ' \( l_1 \) '. Another cell of e.m.f. ' \( E_2 \) ' (\( E_1>E_2 \)) is connected so that two cells oppose each other, the balancing length is ' \( l_2 \) '. The ratio \( E_1 : E_2 \) is
When cell of e.m.f. ' \( E_1 \) ' is connected to potentiometer wire, the balancing length is ' \( l_1 \) '. Another cell of e.m.f. ' \( E_2 \) ' (\( E_1>E_2 \)) is connected so that two cells oppose each other, the balancing length is ' \( l_2 \) '. The ratio \( E_1 : E_2 \) is
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In opposition method: $E_1 - E_2 = k l_{opp}$ and $E_1 + E_2 = k l_{sum}$. Here $E_1$ alone is compared to opposition.
Updated On: Apr 26, 2026
- \( \frac{l_1}{l_1+l_2} \)
- \( \frac{l_1}{l_1-l_2} \)
- \( \frac{l_1-l_2}{l_1} \)
- \( \frac{l_1+l_2}{l_1-l_2} \)
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The Correct Option is B
Solution and Explanation
Step 1: Principles
$E \propto l \implies E = kl$.
Case 1: $E_1 = k l_1$.
Case 2 (Opposition): $E_1 - E_2 = k l_2$.
Step 2: Divide Equations
$\frac{E_1 - E_2}{E_1} = \frac{l_2}{l_1} \implies 1 - \frac{E_2}{E_1} = \frac{l_2}{l_1}$.
Step 3: Ratio
$\frac{E_2}{E_1} = 1 - \frac{l_2}{l_1} = \frac{l_1 - l_2}{l_1} \implies \frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.
Final Answer: (B)
$E \propto l \implies E = kl$.
Case 1: $E_1 = k l_1$.
Case 2 (Opposition): $E_1 - E_2 = k l_2$.
Step 2: Divide Equations
$\frac{E_1 - E_2}{E_1} = \frac{l_2}{l_1} \implies 1 - \frac{E_2}{E_1} = \frac{l_2}{l_1}$.
Step 3: Ratio
$\frac{E_2}{E_1} = 1 - \frac{l_2}{l_1} = \frac{l_1 - l_2}{l_1} \implies \frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$.
Final Answer: (B)
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