For a cell of e.m.f. 2 V , a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a `2Omega` resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is

Options

A. `0.25Omega`
B. `0.50Omega`
C. `0.80Omega`
D. `1.00Omega`

Answer

To find the internal resistance of the cell, we can use the relationship between the lengths of the potentiometer wire and the internal resistance of the cell when a shunt resistor is applied.

### Step-by-Step Solution:

1. **Identify Given Values:**
- E.M.F. of the cell (V) = 2 V
- Length of the potentiometer wire for the original cell (L1) = 50 cm
- Length of the potentiometer wire when shunted (L2) = 40 cm
- Shunt resistor (R) = 2 Ω

2. **Understand the Relation:**
The internal resistance (r) of the cell can be calculated using the formula:
\[ r = R \times \frac{L2 - L1}{L1} \]
where:
- R is the shunt resistance,
- L1 is the length of the wire for the original balance,
- L2 is the length of the wire for the shunted balance.

3. **Substitute the Values:**
Substitute the known values into the formula:
\[ r = 2 \, \Omega \times \frac{40 \, \text{cm} - 50 \, \text{cm}}{50 \, \text{cm}} \]

4. **Calculate the Length Difference:**
Calculate \(L2 - L1\):
\[ L2 - L1 = 40 \, \text{cm} - 50 \, \text{cm} = -10 \, \text{cm} \]

5. **Convert Length to Meters:**
Since we are working in standard units, convert centimeters to meters:
\[ -10 \, \text{cm} = -0.1 \, \text{m} \]

6. **Calculate the Internal Resistance:**
Now substitute this value back into the equation:
\[ r = 2 \, \Omega \times \frac{-0.1 \, \text{m}}{0.5 \, \text{m}} = 2 \, \Omega \times (-0.2) = -0.4 \, \Omega \]

7. **Correct the Sign:**
Since resistance cannot be negative, we take the absolute value:
\[ r = 0.4 \, \Omega \]

### Final Answer:
The internal resistance of the cell is \(0.4 \, \Omega\).