1. Use the resistance formula

For a wire, $R = \rho \frac{L}{A}$ where:

• $\rho$ = resistivity of copper
• $L$ = length
• $A$ = cross-sectional area

When the wire is stretched, its volume remains constant (mass unchanged, density unchanged), so: $AL = \text{constant}$

2. New length after stretching

Length is increased by $20\%$, so $L' = 1.2L$

3. Find new area using constant volume

Since $AL = A'L'$ we get $A' = \frac{AL}{L'} = \frac{A}{1.2}$

4. Compute new resistance

$R' = \rho \frac{L'}{A'} = \rho \frac{1.2L}{A/1.2} = \rho \frac{(1.2)^2L}{A} = 1.44R$

So the resistance increases by a factor of $1.44$.

5. Percentage increase

$\%\text{ increase} = \frac{R' - R}{R} \times 100 = \frac{1.44R - R}{R} \times 100$

$= 0.44 \times 100 = 44\%$

Hence, the percentage increase in resistance is: $\boxed{44}$