125 cm of potentiometer wire balances the emf. of a cell and 100 cm of the wire is required for balance, if the poles of the cell are joined by a `2 Omega` resistor. Then the internal resistance of the cell is

Options

A. `0.25 Omega`
B. `0.5 Omega`
C. `0.75 Omega`
D. `1.25 Omega`

Explanation

To find the internal resistance of the cell using the given information, we can follow these steps:

### Step 1: Understand the given information
We have a potentiometer wire of length 125 cm that balances the EMF of a cell. When a 2 Ω resistor is connected across the cell, the length of the wire required for balance is 100 cm.

### Step 2: Set up the formula
The formula to find the internal resistance (r) of the cell when using a potentiometer is given by:

\[ r = \frac{(L_1 - L_2)}{L_2} \times R \]

Where:
- \(L_1\) = total length of the potentiometer wire (125 cm)
- \(L_2\) = length of the wire used for balance (100 cm)
- \(R\) = external resistance connected (2 Ω)

### Step 3: Substitute the values into the formula
Substituting the values into the formula:

\[ r = \frac{(125 \, \text{cm} - 100 \, \text{cm})}{100 \, \text{cm}} \times 2 \, \Omega \]

### Step 4: Calculate the difference in lengths
Calculate \(L_1 - L_2\):

\[ L_1 - L_2 = 125 \, \text{cm} - 100 \, \text{cm} = 25 \, \text{cm} \]

### Step 5: Substitute the difference back into the formula
Now substitute this back into the formula:

\[ r = \frac{25 \, \text{cm}}{100 \, \text{cm}} \times 2 \, \Omega \]

### Step 6: Simplify the fraction
Simplifying the fraction:

\[ r = \frac{25}{100} \times 2 \, \Omega = 0.25 \times 2 \, \Omega \]

### Step 7: Calculate the internal resistance
Now calculate the internal resistance:

\[ r = 0.5 \, \Omega \]

### Final Answer
The internal resistance of the cell is \(0.5 \, \Omega\).

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