In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across `0.52 m` of the potentiometer wire. If the cell is shunted by a resistance of `5Omega` balance is obtained when the cell connected across` 0.4m` of the wire. Find the internal resistance of the cell.

Explanation

To find the internal resistance of the cell in the given potentiometer experiment, we will follow these steps:

### Step 1: Understand the Problem
In the experiment, we have two scenarios:
1. When no current passes through the galvanometer, the length of the potentiometer wire is \( l_1 = 0.52 \, \text{m} \).
2. When the cell is shunted with a resistance of \( R = 5 \, \Omega \), the balance is obtained at \( l_2 = 0.4 \, \text{m} \).

### Step 2: Use the Formula for Internal Resistance
The formula for internal resistance \( r \) of the cell can be derived from the relationship between the lengths of the potentiometer wire and the resistances involved:
\[ r = R \left( \frac{l_1}{l_2} - 1 \right) \]
where:
- \( R \) is the shunt resistance,
- \( l_1 \) is the length of the wire when no current flows through the galvanometer,
- \( l_2 \) is the length of the wire when the shunt is connected.

### Step 3: Substitute the Values
Now, we will substitute the known values into the formula:
- \( R = 5 \, \Omega \)
- \( l_1 = 0.52 \, \text{m} \)
- \( l_2 = 0.4 \, \text{m} \)

Substituting these values into the formula:
\[ r = 5 \left( \frac{0.52}{0.4} - 1 \right) \]

### Step 4: Calculate the Ratio
First, calculate the ratio \( \frac{0.52}{0.4} \):
\[ \frac{0.52}{0.4} = 1.3 \]

### Step 5: Substitute Back and Solve
Now substitute this back into the equation:
\[ r = 5 \left( 1.3 - 1 \right) = 5 \times 0.3 = 1.5 \, \Omega \]

### Final Answer
The internal resistance of the cell is \( r = 1.5 \, \Omega \).

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