The resistance of a wire is `2Omega` if it is drawn in such a way that it experiences a longitudinal strain `200%` its new resistance is

Options

A. `4Omega`
B. `8Omega`
C. `16Omega`
D. `18Omega`

Explanation

To solve the problem of finding the new resistance of a wire after it is stretched, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The initial resistance \( R \) of the wire is given as \( 2 \, \Omega \).
- The longitudinal strain is \( 200\% \).

2. **Calculate the New Length**:
- Longitudinal strain of \( 200\% \) means the wire is stretched to \( 2 \) times its original length.
- If the original length of the wire is \( L \), the new length \( L' \) will be:
\[ L' = L + 2L = 3L \]

3. **Volume Conservation**:
- The volume of the wire remains constant during stretching. Therefore, we can use the relationship:
\[ A \cdot L = A' \cdot L' \]
- Where \( A \) is the original cross-sectional area, and \( A' \) is the new cross-sectional area.

4. **Express the New Cross-Sectional Area**:
- Substituting \( L' = 3L \) into the volume conservation equation:
\[ A \cdot L = A' \cdot (3L) \]
- Rearranging gives:
\[ A' = \frac{A}{3} \]

5. **Calculate the New Resistance**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
- The new resistance \( R' \) can be calculated using:
\[ R' = \rho \frac{L'}{A'} \]
- Substituting \( L' = 3L \) and \( A' = \frac{A}{3} \):
\[ R' = \rho \frac{3L}{\frac{A}{3}} = \rho \frac{3L \cdot 3}{A} = 9 \cdot \rho \frac{L}{A} \]
- Since \( \rho \frac{L}{A} = R \):
\[ R' = 9R \]
- Given \( R = 2 \, \Omega \):
\[ R' = 9 \cdot 2 = 18 \, \Omega \]

6. **Conclusion**:
- The new resistance of the wire after it is stretched is \( 18 \, \Omega \).