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 along with \(E_1\) so as two cells oppose each other, the balancing length is '\(l_2\)'. The ratio \(E_1:E_2\) is

• A. \((l_1):(l_1+l_2)\)
• B. \((l_1):(l_1-l_2)\)
• C. \((l_1+l_2):(l_1)\)
• D. \((l_1+l_2):(l_1-l_2)\)

Hint:

With the cells opposing, the net emf is E1 - E2, which balances at l2. The emf is proportional to the balancing length.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a potentiometer, the emf of a cell is proportional to the balancing length on the wire: \(E = \phi\,l\), where \(\phi\) is the potential gradient.

Step 2: Key Formula or Approach:
1. Cell \(E_1\) alone: \(E_1 = \phi\,l_1\).
2. Cells in opposition: the net emf is \(E_1 - E_2 = \phi\,l_2\).

Step 3: Detailed Explanation:
Divide the first by the second:
\[ \frac{E_1}{E_1 - E_2} = \frac{l_1}{l_2} \]
Cross-multiply and rearrange:
\[ E_1l_2 = l_1E_1 - l_1E_2 \Rightarrow l_1E_2 = E_1(l_1 - l_2) \]
\[ \frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2} \]
Since \(E_1 > E_2\), the net emf is positive and \(l_2 < l_1\), so the denominator is positive. Options (C) and (D) have \(l_1 + l_2\), which would apply to cells that help each other, where the net emf is \(E_1 + E_2\).

Final Answer:
\(E_1:E_2 = l_1:(l_1 - l_2)\), option (B). \[ \boxed{l_1:(l_1-l_2) \text{ (B)}} \]