In a potentiometer experiment the balancing length with a cell is 560 cm. When an external resistance of `10Omega` is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Problem
In a potentiometer experiment, we have a cell with an initial balancing length of 560 cm. When a 10 Ω resistor is connected in parallel to the cell, the balancing length decreases by 60 cm, resulting in a new balancing length of 500 cm. We need to find the internal resistance of the cell.

### Step 2: Set Up the Equations
Let:
- \( E \) = EMF of the cell
- \( r \) = internal resistance of the cell
- \( V \) = potential drop per unit length of the potentiometer wire

From the initial condition (without the external resistor), the potential difference across the potentiometer wire is given by:
\[ E = V \cdot L_1 \]
Where \( L_1 = 560 \) cm.

From the new condition (with the external resistor), the potential difference across the potentiometer wire is:
\[ E' = V \cdot L_2 \]
Where \( L_2 = 500 \) cm.

### Step 3: Write the Equations
1. For the initial balancing length:
\[ E = V \cdot 560 \quad \text{(Equation 1)} \]

2. For the new balancing length with the external resistor:
The total current \( I \) through the circuit when the 10 Ω resistor is connected in parallel is given by:
\[ I = \frac{E}{r + 10} \]
The potential drop across the 10 Ω resistor (which is the same as the potential difference across points A and B) is:
\[ V_{AB} = I \cdot 10 = \frac{E \cdot 10}{r + 10} \]
This potential drop can also be expressed in terms of the new balancing length:
\[ V_{AB} = V \cdot 500 \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Expressions for \( V_{AB} \)
From Equation 2:
\[ \frac{E \cdot 10}{r + 10} = V \cdot 500 \]

### Step 5: Substitute \( V \) from Equation 1
From Equation 1, we have:
\[ V = \frac{E}{560} \]
Substituting this into the equation gives:
\[ \frac{E \cdot 10}{r + 10} = \frac{E}{560} \cdot 500 \]

### Step 6: Simplify the Equation
Cancelling \( E \) from both sides (assuming \( E \neq 0 \)):
\[ \frac{10}{r + 10} = \frac{500}{560} \]
Simplifying \( \frac{500}{560} \):
\[ \frac{500}{560} = \frac{25}{28} \]
Thus, we have:
\[ \frac{10}{r + 10} = \frac{25}{28} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 10 \cdot 28 = 25 \cdot (r + 10) \]
This simplifies to:
\[ 280 = 25r + 250 \]

### Step 8: Solve for \( r \)
Rearranging gives:
\[ 25r = 280 - 250 \]
\[ 25r = 30 \]
\[ r = \frac{30}{25} = 1.2 \, \Omega \]

### Final Answer
The internal resistance of the cell is \( r = 1.2 \, \Omega \).

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