Question

When a uniform wire of resistance R is stretched uniformly to increase its length by 10%, the new resistance value would?

Answer 1

When a uniform wire is stretched, its resistance changes due to the changes in its length and cross-sectional area. Let’s analyze the problem step by step to determine the new resistance.

1. **Original Resistance**: The original resistance \( R \) of the wire can be expressed as:
   \[
   R = \rho \frac{L}{A}
   \]
   where \( \rho \) is the resistivity of the material, \( L \) is the original length of the wire, and \( A \) is the original cross-sectional area.

2. **Stretching the Wire**: The wire is stretched to increase its length by 10%. This means the new length \( L' \) is:
   \[
   L' = L + 0.10L = 1.10L
   \]

3. **Change in Cross-Sectional Area**: When the wire is stretched, its volume remains constant. The volume \( V \) of the wire is:
   \[
   V = A \times L
   \]
   After stretching, the new volume \( V' \) must be the same as the original volume:
   \[
   V' = A' \times L' = A \times L
   \]
   Solving for the new cross-sectional area \( A' \):
   \[
   A' = \frac{A \times L}{L'} = \frac{A \times L}{1.10L} = \frac{A}{1.10}
   \]

4. **New Resistance**: The new resistance \( R' \) can be calculated using the new length \( L' \) and the new cross-sectional area \( A' \):
   \[
   R' = \rho \frac{L'}{A'}
   \]
   Substituting \( L' \) and \( A' \):
   \[
   R' = \rho \frac{1.10L}{\frac{A}{1.10}}
   \]
   Simplify the expression:
   \[
   R' = \rho \frac{1.10L \times 1.10}{A} = \rho \frac{(1.10)^2 L}{A}
   \]
   \[
   R' = (1.10)^2 \cdot \frac{\rho L}{A} = (1.10)^2 \cdot R
   \]
   \[
   R' = 1.21 \cdot R
   \]

So, the new resistance value \( R' \) is \( 1.21R \), which means the resistance increases by 21% when the length of the wire is increased by 10% due to stretching.

Answer 2

When a uniform wire is stretched uniformly, its resistance changes due to the change in its length and cross-sectional area. Resistance \( R \) of a wire is given by:

\[
R = \rho \frac{L}{A}
\]

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material (which stays constant when stretching),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

Now, let's break it down step by step:

### 1. Increase in Length
If the wire is stretched by 10%, the new length \( L' \) is:

\[
L' = 1.1L
\]

### 2. Change in Cross-Sectional Area
Since the volume of the wire remains constant, the product of length and cross-sectional area before and after stretching must be the same. Therefore:

\[
L \cdot A = L' \cdot A'
\]

Thus, the new cross-sectional area \( A' \) is:

\[
A' = \frac{L \cdot A}{L'} = \frac{A}{1.1}
\]

### 3. New Resistance
The new resistance \( R' \) is given by:

\[
R' = \rho \frac{L'}{A'}
\]

Substitute the values of \( L' \) and \( A' \):

\[
R' = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1^2L}{A}
\]

Since \( R = \rho \frac{L}{A} \), we have:

\[
R' = 1.1^2 R = 1.21 R
\]

### Conclusion:
The new resistance of the wire after it is stretched uniformly by 10% is **1.21 times the original resistance**.