A wire has a resistance of `10Omega`. It is stretched by `1//10` of its original length. Then its resistance will be

Options

A. `9Omega`
B. `10Omega`
C. `11Omega`
D. `12.1Omega`

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial resistance of the wire, \( R_1 = 10 \, \Omega \).
- The wire is stretched by \( \frac{1}{10} \) of its original length.

2. **Determine the Change in Length**:
- Let the original length of the wire be \( L_1 \).
- The new length after stretching, \( L_2 = L_1 + \frac{1}{10}L_1 = \frac{11}{10}L_1 \).

3. **Understand the Relationship Between Resistance and Length**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
- When the wire is stretched, its volume remains constant. Thus, if the length increases, the cross-sectional area decreases.

4. **Volume Conservation**:
- The volume \( V \) of the wire is given by:
\[ V = A_1 L_1 = A_2 L_2 \]
- From this, we can express the new area \( A_2 \) in terms of the original area \( A_1 \):
\[ A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1 L_1}{\frac{11}{10}L_1} = \frac{10}{11} A_1 \]

5. **Calculate the New Resistance**:
- The new resistance \( R_2 \) can now be expressed as:
\[ R_2 = \rho \frac{L_2}{A_2} \]
- Substituting \( L_2 \) and \( A_2 \):
\[ R_2 = \rho \frac{\frac{11}{10}L_1}{\frac{10}{11}A_1} = \rho \frac{11^2}{10^2} \frac{L_1}{A_1} = \frac{121}{100} R_1 \]
- Since \( R_1 = 10 \, \Omega \):
\[ R_2 = \frac{121}{100} \times 10 = 12.1 \, \Omega \]

6. **Conclusion**:
- The new resistance of the wire after stretching is \( R_2 = 12.1 \, \Omega \).

### Final Answer:
The resistance after stretching the wire will be \( 12.1 \, \Omega \).