A potentiometer wire is `100 cm` long hand a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at `50 cm` and `10 cm` from the positive end of the wire in the two cases. The ratio of emf is:

Options

A. `3 : 2`
B. `5 : 1`
C. `5 : 4`
D. `3 : 4`

Explanation

To solve the problem, we need to analyze the situation with the potentiometer wire and the two cells connected in series. Let's break it down step by step.

### Step 1: Understand the Setup
We have a potentiometer wire that is 100 cm long with a constant potential difference across it. Two cells are connected in series:
1. In the first case, they are connected to support each other (E1 + E2).
2. In the second case, they are connected in opposite directions (E1 - E2).

### Step 2: Identify the Balance Points
- When the cells are supporting each other, the balance point is at 50 cm from the positive end of the wire.
- When the cells are opposing each other, the balance point is at 10 cm from the positive end of the wire.

### Step 3: Relate the Balance Points to EMF
Using the principle of the potentiometer, we can relate the lengths to the EMFs of the cells:
- For the first case (supporting), the potential difference across the length of the wire is proportional to the EMF:
\[ E_1 + E_2 \propto 50 \text{ cm} \]
- For the second case (opposing), we have:
\[ E_1 - E_2 \propto 10 \text{ cm} \]

### Step 4: Set Up the Equations
From the balance points, we can write:
1. \( E_1 + E_2 = k \cdot 50 \) (where k is a constant)
2. \( E_1 - E_2 = k \cdot 10 \)

### Step 5: Solve the Equations
Now we can solve these two equations. Let's express them in terms of k:
1. \( E_1 + E_2 = 50k \)
2. \( E_1 - E_2 = 10k \)

We can add these two equations:
\[ (E_1 + E_2) + (E_1 - E_2) = 50k + 10k \]
This simplifies to:
\[ 2E_1 = 60k \quad \Rightarrow \quad E_1 = 30k \]

Now, subtract the second equation from the first:
\[ (E_1 + E_2) - (E_1 - E_2) = 50k - 10k \]
This simplifies to:
\[ 2E_2 = 40k \quad \Rightarrow \quad E_2 = 20k \]

### Step 6: Find the Ratio of EMFs
Now we can find the ratio of the EMFs:
\[ \frac{E_1}{E_2} = \frac{30k}{20k} = \frac{30}{20} = \frac{3}{2} \]

### Conclusion
The ratio of the EMFs \( E_1 \) and \( E_2 \) is \( \frac{3}{2} \).