Tow cells of e.m.f `E_(1)` and `E_(2)` are joined in series and the balancing length of the potentiometer wire is 625 cm. If the terminals of `E_(1)` are reversed , the balancing length obtained is 125 cm. Given `E_(2) gt E_(1)`,the ratio `E_(1) : E_(2)` willbe

Options

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

Answer

To find the ratio \( E_1 : E_2 \), we can follow these steps:

### Step 1: Write the equations based on the potentiometer principle
When the two cells \( E_1 \) and \( E_2 \) are connected in series with the same polarity, the total EMF is given by:
\[ E_1 + E_2 = K L_1 \]
where \( L_1 \) is the balancing length of the potentiometer wire when both cells are connected in the same direction and \( K \) is the potential gradient.

When the terminals of \( E_1 \) are reversed, the equation becomes:
\[ E_2 - E_1 = K L_2 \]
where \( L_2 \) is the balancing length when \( E_1 \) is reversed.

### Step 2: Substitute the known values
From the problem, we have:
- \( L_1 = 625 \, \text{cm} \)
- \( L_2 = 125 \, \text{cm} \)

### Step 3: Set up the equations
We can rewrite the equations as:
1. \( E_1 + E_2 = K \cdot 625 \) (Equation 1)
2. \( E_2 - E_1 = K \cdot 125 \) (Equation 2)

### Step 4: Divide the equations
Dividing Equation 1 by Equation 2 gives:
\[ \frac{E_1 + E_2}{E_2 - E_1} = \frac{K \cdot 625}{K \cdot 125} \]
This simplifies to:
\[ \frac{E_1 + E_2}{E_2 - E_1} = \frac{625}{125} = 5 \]

### Step 5: Cross-multiply and rearrange
Cross-multiplying gives:
\[ E_1 + E_2 = 5(E_2 - E_1) \]
Expanding the right side:
\[ E_1 + E_2 = 5E_2 - 5E_1 \]
Rearranging terms:
\[ E_1 + 5E_1 = 5E_2 - E_2 \]
This simplifies to:
\[ 6E_1 = 4E_2 \]

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

### Conclusion
Thus, the ratio \( E_1 : E_2 \) is:
\[ E_1 : E_2 = 2 : 3 \]