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: Convert lengths to emfs:
$E_1 = \phi l_1$ and $E_1 - E_2 = \phi l_2$. Subtract the second from the first: $E_2 = \phi(l_1 - l_2)$.

Step 2: Take the ratio:
$\dfrac{E_1}{E_2} = \dfrac{\phi l_1}{\phi(l_1-l_2)} = \dfrac{l_1}{l_1 - l_2}$.

Step 3: Reasonableness:
For $E_2$ small, $l_2\to l_1$ and the ratio becomes large, which is right. For $E_2\to E_1$, $l_2\to0$ and the ratio tends to 1, which is also right.

Final Answer:
Option (B). \[ \boxed{l_1:(l_1-l_2) \text{ (B)}} \]