The length of the potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance `10Omega` is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be

Options

A. `20Omega`
B. `40 Omega`
C. `60Omega`
D. `80Omega`

Answer

To find the resistance of the potentiometer wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Length of the potentiometer wire, \( L_1 = 600 \, \text{cm} \)
- Current flowing through the wire, \( I = 40 \, \text{mA} = 40 \times 10^{-3} \, \text{A} \)
- EMF of the cell, \( E = 2 \, \text{V} \)
- Internal resistance of the cell, \( r = 10 \, \Omega \)
- Balance length (null point), \( L_2 = 500 \, \text{cm} \)

2. **Calculate the resistance at the null point (R'):**
- Using Ohm's law, the resistance at the null point can be calculated as:
\[ R' = \frac{E}{I} \]
- Substituting the values:
\[ R' = \frac{2 \, \text{V}}{40 \times 10^{-3} \, \text{A}} = \frac{2}{0.04} = 50 \, \Omega \]

3. **Use the relationship between the resistances and lengths:**
- According to the potentiometer principle:
\[ \frac{R}{R'} = \frac{L_1}{L_2} \]
- Rearranging this gives us:
\[ R = R' \cdot \frac{L_1}{L_2} \]

4. **Substituting the values to find the total resistance (R):**
- Substituting \( R' = 50 \, \Omega \), \( L_1 = 600 \, \text{cm} \), and \( L_2 = 500 \, \text{cm} \):
\[ R = 50 \cdot \frac{600}{500} = 50 \cdot 1.2 = 60 \, \Omega \]

5. **Conclusion:**
- The resistance of the potentiometer wire is \( R = 60 \, \Omega \).

### Final Answer:
The resistance of the potentiometer wire is \( 60 \, \Omega \).

---