In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of `2Omega`, the balancing length becomes 120 cm.The internal resistance of the cell is

Options

A. `0.5 Omega`
B. `1 Omega`
C. `2Omega`
D. `4 Omega`

Explanation

To find the internal resistance of the cell in the potentiometer experiment, we can use the formula derived from the balancing lengths before and after shunting the cell. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial balancing length (L1) = 240 cm
- New balancing length after shunting (L2) = 120 cm
- Shunt resistance (R) = 2 Ω

### Step 2: Write the formula for internal resistance (r)
The formula for the internal resistance of the cell when using a potentiometer is given by:

\[ r = \frac{L1 - L2}{L2} \times R \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values we have into the formula:

\[ r = \frac{240 \, \text{cm} - 120 \, \text{cm}}{120 \, \text{cm}} \times 2 \, \Omega \]

### Step 4: Calculate the numerator
Calculate \(L1 - L2\):

\[ L1 - L2 = 240 \, \text{cm} - 120 \, \text{cm} = 120 \, \text{cm} \]

### Step 5: Substitute the result back into the equation
Now substitute this result back into the equation:

\[ r = \frac{120 \, \text{cm}}{120 \, \text{cm}} \times 2 \, \Omega \]

### Step 6: Simplify the equation
Now simplify the fraction:

\[ r = 1 \times 2 \, \Omega = 2 \, \Omega \]

### Step 7: State the final answer
Thus, the internal resistance of the cell is:

\[ \boxed{2 \, \Omega} \]

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