A Daniel cell is balanced on `125 cm` length of a potentiometer wire. Now the cells is short-circuited by a resistance `2 ohm` and the balance is obtained at `100 cm`. The internal resistance of the Dainel cell is

Options

A. `4/3 Omega`
B. `1.5 Omega`
C. `1.25 Omega`
D. `0.5 Omega`

Answer

To find the internal resistance of the Daniel cell, we can follow these steps:

### Step 1: Understand the Problem
We have a Daniel cell that is balanced on a potentiometer wire of length 125 cm. When the cell is short-circuited with a resistance of 2 ohms, the balance is obtained at 100 cm. We need to find the internal resistance of the Daniel cell.

### Step 2: Set Up the Formula
The internal resistance \( r \) of the cell can be calculated using the formula:
\[ r = \left( L_1 - L_2 \right) \times \frac{R}{L_2} \]
where:
- \( L_1 \) = length of the potentiometer wire when the cell is not short-circuited = 125 cm
- \( L_2 \) = length of the potentiometer wire when the cell is short-circuited = 100 cm
- \( R \) = external resistance = 2 ohms

### Step 3: Substitute the Values
Now, we can substitute the values into the formula:
\[ r = \left( 125 \, \text{cm} - 100 \, \text{cm} \right) \times \frac{2 \, \Omega}{100 \, \text{cm}} \]

### 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 value back into the formula:
\[ r = 25 \, \text{cm} \times \frac{2 \, \Omega}{100 \, \text{cm}} \]

### Step 6: Simplify the Expression
Now simplify the expression:
\[ r = 25 \times \frac{2}{100} = \frac{50}{100} = 0.5 \, \Omega \]

### Step 7: Conclusion
Thus, the internal resistance of the Daniel cell is \( 0.5 \, \Omega \).

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

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