The e.m.f. of a standard cell balances across 150 cm length of a wire of potentiometer. When a resistance of `2Omega` is connected as a shunt with the cell, the balance point is obtained at 100cm . The internal resistance of the cell is

Options

A. 0.1ohm
B. 1ohm
C. 2ohm
D. 0.5ohm

Answer

To find the internal resistance of the cell, we can use the information provided in the question and apply the relevant formula. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Original length of the potentiometer wire (L1) = 150 cm
- Length of the wire when the shunt is connected (L2) = 100 cm
- Resistance of the shunt (R) = 2 Ω

2. **Understand the Relationship:**
- The internal resistance (r) of the cell can be calculated using the formula:
\[ r = \frac{L1 - L2}{L2} \times R \]
- Here, \(L1 - L2\) represents the change in length of the potentiometer wire when the shunt is connected.

3. **Substitute the Values into the Formula:**
- Calculate \(L1 - L2\):
\[ L1 - L2 = 150 \, \text{cm} - 100 \, \text{cm} = 50 \, \text{cm} \]
- Now substitute the values into the formula:
\[ r = \frac{50 \, \text{cm}}{100 \, \text{cm}} \times 2 \, \Omega \]

4. **Simplify the Expression:**
- The fraction \(\frac{50}{100}\) simplifies to \(\frac{1}{2}\):
\[ r = \frac{1}{2} \times 2 \, \Omega = 1 \, \Omega \]

5. **Conclusion:**
- The internal resistance of the cell is:
\[ r = 1 \, \Omega \]

### Final Answer:
The internal resistance of the cell is **1 Ω**.

---