Two cells when connected in series are balanced on `8 m` on a potentiometer. If cells are connected with polarities of one the cellis reversed, they balance on `2 m`. The ratio of e.m.f.'s of the two cellsis

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have two cells with unknown EMFs, E1 and E2. When connected in series, they balance on 8 meters of the potentiometer. When one cell's polarity is reversed, they balance on 2 meters.

### Step 2: Write the equations for the two scenarios
1. **For series connection**: The total EMF (ES) is the sum of the individual EMFs.
\[ E_S = E_1 + E_2 \]
According to the potentiometer principle, we can express this as:
\[ E_S = K \cdot L_S \]
where \(L_S\) is the length of the wire at the balanced point when connected in series, which is 8 m.
Thus,
\[ E_S = K \cdot 8 \]

2. **For parallel connection (one cell reversed)**: The total EMF (EP) is the difference of the individual EMFs.
\[ E_P = E_1 - E_2 \]
Similarly, we express this as:
\[ E_P = K \cdot L_P \]
where \(L_P\) is the length of the wire at the balanced point when connected in parallel, which is 2 m.
Thus,
\[ E_P = K \cdot 2 \]

### Step 3: Set up the equations
From the above, we have:
1. \(E_S = K \cdot 8\)
2. \(E_P = K \cdot 2\)

### Step 4: Relate the EMFs
From the series connection, we have:
\[ E_1 + E_2 = K \cdot 8 \quad \text{(1)} \]
From the parallel connection, we have:
\[ E_1 - E_2 = K \cdot 2 \quad \text{(2)} \]

### Step 5: Solve the equations
Now, we can solve these two equations simultaneously.

From equation (1):
\[ E_1 = K \cdot 8 - E_2 \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ (K \cdot 8 - E_2) - E_2 = K \cdot 2 \]
\[ K \cdot 8 - 2E_2 = K \cdot 2 \]
Rearranging gives:
\[ 2E_2 = K \cdot 8 - K \cdot 2 \]
\[ 2E_2 = K \cdot 6 \]
\[ E_2 = 3K \quad \text{(4)} \]

Substituting equation (4) back into equation (3):
\[ E_1 = K \cdot 8 - 3K \]
\[ E_1 = 5K \quad \text{(5)} \]

### Step 6: Find the ratio of EMFs
Now, we have:
- \(E_1 = 5K\)
- \(E_2 = 3K\)

The ratio of the EMFs is:
\[ \frac{E_1}{E_2} = \frac{5K}{3K} = \frac{5}{3} \]

### Final Answer
The ratio of the EMFs of the two cells is \( \frac{5}{3} \).

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