Solution:

When a wire is stretched, its length increases, and its cross-sectional area decreases. The resistance of a wire is given by the formula:

$$R = \rho \frac{L}{A}$$

Where:

• $R$ is the resistance,
• $\rho$ is the resistivity of the material (constant),
• $L$ is the length of the wire,
• $A$ is the cross-sectional area of the wire.

When the wire is stretched:

1. Length increases by 10%: The new length $L'$ is given by:

   $$L' = L + 0.1L = 1.1L$$

2. Volume remains constant: The volume of the wire before and after stretching remains the same. Volume is the product of length and area:

   $$\text{Volume before} = L \times A$$ $$\text{Volume after} = L' \times A' = 1.1L \times A'$$Since the volume remains constant:

   $$L \times A = 1.1L \times A'$$Solving for $A'$, the new cross-sectional area:

   $$A' = \frac{A}{1.1}$$

Step 2: New Resistance

The new resistance $R'$ of the stretched wire is given by:

$$R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1L \times 1.1}{A} = \rho \frac{1.21L}{A}$$

So, the new resistance $R'$ is 1.21 times the original resistance $R$.

Step 3: Conclusion

The resistance increases by 21% when the wire is stretched by 10%.