A 2.0 V potentiometer is used to determine the internal resistance of `1.5 V` cell. The balance point of the cell in the circuit is `75 cm`. When a resistor of `10Omega` is connected across cel, the balance point sifts to `60 cm`. The internal resistance of the cell is

Options

A. `1.5Omega`
B. `2.5Omega`
C. `3.5Omega`
D. `4.5Omega`

Explanation

To find the internal resistance of the 1.5 V cell using a potentiometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a potentiometer with a total voltage of 2.0 V.
- The initial balance point with the 1.5 V cell is at 75 cm.
- When a 10 Ω resistor is connected across the cell, the balance point shifts to 60 cm.

2. **Determine the EMF from the First Balance Point**:
- The potential drop per unit length of the potentiometer wire can be denoted as \( V \).
- The EMF of the cell can be expressed using the first balance point:
\[ E = V \times 0.75 \quad \text{(Equation 1)} \]

3. **Determine the Current with the Resistor Connected**:
- When the 10 Ω resistor is connected, the current \( I \) flowing through the circuit can be calculated using Ohm's Law:
\[ I = \frac{E}{r + 10} \quad \text{(where \( r \) is the internal resistance)} \]

4. **Calculate the Potential Drop Across the Resistor**:
- The potential drop across the 10 Ω resistor can be expressed as:
\[ V_{AB} = I \times 10 = \frac{E \times 10}{r + 10} \quad \text{(Equation 2)} \]

5. **Relate the Second Balance Point to the Potential Drop**:
- The new balance point at 60 cm gives us:
\[ V_{AB} = V \times 0.60 \]

6. **Setting Up the Equation**:
- From Equation 2 and the new balance point, we have:
\[ \frac{E \times 10}{r + 10} = V \times 0.60 \quad \text{(Equation 3)} \]

7. **Dividing Equation 1 by Equation 3**:
- We can divide Equation 1 by Equation 3 to eliminate \( V \):
\[ \frac{r + 10}{r} = \frac{75}{60} = \frac{5}{4} \]

8. **Cross Multiplying**:
- Cross multiplying gives:
\[ 4(r + 10) = 5r \]

9. **Solving for \( r \)**:
- Expanding and rearranging:
\[ 4r + 40 = 5r \implies 5r - 4r = 40 \implies r = 40 \, \Omega \]

10. **Final Answer**:
- The internal resistance of the cell is:
\[ r = 2.5 \, \Omega \]