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.

Options

A. `2.5 Omega`
B. `1 Omega`
C. `1.5 Omega`
D. `2 Omega`

Answer

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

### Step 1: Understand the initial condition
In the first scenario, no current passes through the galvanometer when the terminals of the cell are connected across `0.52 m` of the potentiometer wire. This means that the electromotive force (E) of the cell is balanced by the potential drop across the potentiometer wire.

Let:
- \( V \) = potential drop per unit length of the potentiometer wire
- \( L_1 = 0.52 \, m \)

From this, we can write the equation:
\[ E = V \cdot L_1 \]
Substituting \( L_1 \):
\[ E = V \cdot 0.52 \]

### Step 2: Understand the second condition
In the second scenario, the cell is shunted by a resistance of `5Ω`, and balance is obtained when the cell is connected across `0.4 m` of the wire.

Let:
- \( L_2 = 0.4 \, m \)

The current through the circuit when the resistance is connected can be calculated as:
\[ I = \frac{E}{R + r} \]
where \( R = 5 \, \Omega \) (the shunt resistance) and \( r \) is the internal resistance of the cell.

The potential drop across the shunt resistance can be expressed as:
\[ V \cdot L_2 = I \cdot R \]
Substituting for \( I \):
\[ V \cdot L_2 = \left(\frac{E}{R + r}\right) \cdot R \]
Substituting \( L_2 \):
\[ V \cdot 0.4 = \left(\frac{E}{5 + r}\right) \cdot 5 \]

### Step 3: Set up the equations
Now we have two equations:
1. \( E = V \cdot 0.52 \)
2. \( V \cdot 0.4 = \frac{5E}{5 + r} \)

### Step 4: Substitute the first equation into the second
Substituting \( E \) from the first equation into the second:
\[ V \cdot 0.4 = \frac{5(V \cdot 0.52)}{5 + r} \]

### Step 5: Simplify the equation
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ 0.4 = \frac{5 \cdot 0.52}{5 + r} \]
Cross-multiplying gives:
\[ 0.4(5 + r) = 5 \cdot 0.52 \]
Expanding the left side:
\[ 2 + 0.4r = 2.6 \]

### Step 6: Solve for \( r \)
Rearranging gives:
\[ 0.4r = 2.6 - 2 \]
\[ 0.4r = 0.6 \]
Dividing both sides by \( 0.4 \):
\[ r = \frac{0.6}{0.4} = 1.5 \, \Omega \]

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

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