Tow cells of emf `E_(1)` and `E_(2) ( E_(1) gt E_(2))` are connected individually to a potentiometer and their corresponding balancing length are 625 cm and 500 cm, then the ratio `(E_(1))/(E_(2))` is .

Options

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

Answer

To solve the problem, we need to find the ratio of the EMFs \( \frac{E_1}{E_2} \) using the balancing lengths measured on a potentiometer.

### Step-by-Step Solution:

1. **Understanding the Potentiometer Principle**:
- When two cells are connected to a potentiometer, the EMF of each cell is proportional to the balancing length measured on the potentiometer.
- The relationship can be expressed as:
\[ E_1 = \Phi L_1 \quad \text{and} \quad E_2 = \Phi L_2 \]
where \( \Phi \) is the potential gradient, \( L_1 \) is the balancing length for \( E_1 \), and \( L_2 \) is the balancing length for \( E_2 \).

2. **Setting Up the Ratio**:
- From the above equations, we can derive the ratio of the EMFs:
\[ \frac{E_1}{E_2} = \frac{\Phi L_1}{\Phi L_2} \]
- The \( \Phi \) cancels out, leading to:
\[ \frac{E_1}{E_2} = \frac{L_1}{L_2} \]

3. **Substituting the Values**:
- We know from the problem statement that:
- \( L_1 = 625 \, \text{cm} \)
- \( L_2 = 500 \, \text{cm} \)
- Substituting these values into the ratio gives:
\[ \frac{E_1}{E_2} = \frac{625}{500} \]

4. **Simplifying the Ratio**:
- To simplify \( \frac{625}{500} \):
\[ \frac{625}{500} = \frac{625 \div 125}{500 \div 125} = \frac{5}{4} \]

5. **Final Result**:
- Therefore, the ratio of the EMFs is:
\[ \frac{E_1}{E_2} = \frac{5}{4} \]

### Final Answer:
\[ \frac{E_1}{E_2} = \frac{5}{4} \]