A wire of resistance 1 `Omega` is elongated by 10%. The resistance of the elongated wire is

Options

A. `1.1Omega`
B. `11.1Omega`
C. `1.21Omega`
D. `12.1Omega`

Answer

To find the resistance of a wire after it has been elongated by 10%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Resistance**:
The initial resistance of the wire is given as \( R_0 = 1 \, \Omega \).

2. **Determine the Change in Length**:
If the wire is elongated by 10%, the new length \( L' \) can be expressed as:
\[ L' = L + 0.1L = 1.1L \]
where \( L \) is the original length of the wire.

3. **Relate the Area of Cross-section**:
When the length of the wire increases, the volume of the wire remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the original cross-sectional area. After elongation, the new volume can be expressed as:
\[ V = A' \cdot L' = A' \cdot 1.1L \]
Setting the two volumes equal gives:
\[ A \cdot L = A' \cdot 1.1L \]
From this, we can derive the new area \( A' \):
\[ A' = \frac{A}{1.1} \]

4. **Calculate the New Resistance**:
The resistance of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material. The new resistance \( R' \) after elongation can be expressed as:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.1L)}{(A/1.1)} = \frac{\rho L \cdot 1.1^2}{A} \]
Since \( R_0 = \frac{\rho L}{A} = 1 \, \Omega \), we can substitute:
\[ R' = R_0 \cdot 1.1^2 = 1 \cdot 1.21 = 1.21 \, \Omega \]

5. **Final Answer**:
The resistance of the elongated wire is:
\[ R' = 1.21 \, \Omega \]