The resistance of a wire of length 20 cm is `5Omega`. It is stretched uniformly to a legth of 40 cm. The resistance now becomes:

Options

A. `5Omega`
B. `10Omega`
C. `20Omega`
D. `200Omega`

Answer

To solve the problem, we need to determine the new resistance of a wire after it has been stretched. We start with the known values and apply the relevant formulas step by step.

### Step-by-Step Solution:

1. **Identify Initial Values**:
- Initial length of the wire, \( L_1 = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Initial resistance of the wire, \( R_1 = 5 \, \Omega \)

2. **Determine the New Length**:
- The wire is stretched to a new length, \( L_2 = 40 \, \text{cm} = 0.4 \, \text{m} \)

3. **Understand the Volume Conservation**:
- When the wire is stretched, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A_1 \times L_1 = A_2 \times L_2 \]
- Where \( A_1 \) is the original cross-sectional area and \( A_2 \) is the new cross-sectional area.

4. **Relate Areas Using Volume Conservation**:
- From the volume conservation, we have:
\[ A_1 \times L_1 = A_2 \times L_2 \]
- Substituting the values:
\[ A_1 \times 0.2 = A_2 \times 0.4 \]
- Rearranging gives:
\[ A_2 = \frac{A_1 \times 0.2}{0.4} = \frac{A_1}{2} \]
- This means the new area \( A_2 \) is half of the original area \( A_1 \).

5. **Use the Resistance Formula**:
- The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
- Where \( \rho \) is the resistivity of the material.

6. **Calculate the New Resistance**:
- The new resistance \( R_2 \) can be expressed as:
\[ R_2 = \frac{\rho L_2}{A_2} \]
- Substituting \( L_2 = 0.4 \, \text{m} \) and \( A_2 = \frac{A_1}{2} \):
\[ R_2 = \frac{\rho \times 0.4}{\frac{A_1}{2}} = \frac{0.4 \times 2 \rho}{A_1} = \frac{0.8 \rho}{A_1} \]

7. **Relate New Resistance to Old Resistance**:
- Since \( R_1 = \frac{\rho L_1}{A_1} \):
\[ R_1 = \frac{\rho \times 0.2}{A_1} = 5 \, \Omega \]
- Therefore, \( \rho = \frac{5 \times A_1}{0.2} = 25 A_1 \).

8. **Substituting Back**:
- Now substituting \( \rho \) back into the equation for \( R_2 \):
\[ R_2 = \frac{0.8 \times 25 A_1}{A_1} = 20 \, \Omega \]

### Final Answer:
The new resistance of the wire after being stretched to 40 cm is \( R_2 = 20 \, \Omega \).