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 obatined at `50 cm` and `10 cm` from the positive end of the wire in the two cases. The ratio of emfs is:

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
We have a potentiometer wire of length \(100 \, \text{cm}\) with a constant potential difference across it. Two cells are connected in two different configurations: first in series supporting each other and then in opposite directions.

### Step 2: Define the First Case
In the first case, when the cells are connected in series supporting each other, the total EMF \(E\) is given by:
\[ E = E_1 + E_2 \]
The balance point is obtained at \(50 \, \text{cm}\) from the positive end of the wire. According to the potentiometer principle, the EMF is proportional to the length of the wire:
\[ E \propto L_1 \quad \text{(where } L_1 = 50 \, \text{cm}\text{)} \]
Thus, we can write:
\[ E_1 + E_2 \propto 50 \quad \text{(Equation 1)} \]

### Step 3: Define the Second Case
In the second case, when the cells are connected in opposite directions, the total EMF \(E\) is given by:
\[ E = E_1 - E_2 \]
The balance point is obtained at \(10 \, \text{cm}\) from the positive end of the wire:
\[ E \propto L_2 \quad \text{(where } L_2 = 10 \, \text{cm}\text{)} \]
Thus, we can write:
\[ E_1 - E_2 \propto 10 \quad \text{(Equation 2)} \]

### Step 4: Set Up the Ratios
From Equations 1 and 2, we can express the relationships as:
\[ \frac{E_1 + E_2}{E_1 - E_2} = \frac{50}{10} \]
This simplifies to:
\[ \frac{E_1 + E_2}{E_1 - E_2} = 5 \]

### Step 5: Cross Multiply and Rearrange
Cross-multiplying gives:
\[ E_1 + E_2 = 5(E_1 - E_2) \]
Expanding this, we have:
\[ E_1 + E_2 = 5E_1 - 5E_2 \]
Rearranging terms leads to:
\[ E_1 + E_2 + 5E_2 = 5E_1 \]
This simplifies to:
\[ 6E_2 = 4E_1 \]

### Step 6: Solve for the Ratio of EMFs
Dividing both sides by \(E_2\) and \(E_1\) gives:
\[ \frac{E_1}{E_2} = \frac{6}{4} = \frac{3}{2} \]

### Conclusion
Thus, the ratio of the EMFs \(E_1\) and \(E_2\) is:
\[ \frac{E_1}{E_2} = \frac{3}{2} \]