Question

If the length of a conductor is increased by 20% and cross-sectional area is decreased by 4% ,then calculate the percentage change in the resistance of the conductor.

Answer 1

The resistance of a conductor depends on its length, cross-sectional area, and resistivity.
The resistivity of a material is a constant that depends only on the material and its temperature.
The resistance of the conductor is given by:

$R$ = $\frac{ρL}{A}$

where ρ is the resistivity of the material, $L$ is the length of the conductor, and A is the cross-sectional area of the conductor.
If the length of the conductor is increased by 20% and the cross-sectional area is decreased by 4%, we have:
$L' = 1.2 L$ (20% increase in length)
$A' = 0.96 \;A$ (4% decrease in cross-sectional area)
The resistance of the conductor with the new length and cross-sectional area is:

R' = $\frac{ρL'}{A'}$ = $\frac{ρ1.2L}{0.96A}$

The percentage change in resistance is given by:

$\frac {R'-R}{R}$ $\times$ $100\%$

Substituting the expressions for R and R', we get:

$\frac{(R' - R)}{ R} * 100$ = ($\frac{ρ1.2L}{0.96A}$) - $\frac{ρL}{A}$) / ($\frac{ρL}{A}$) * 100%

Simplifying this expression, we get:

$\frac{(R' - R)}{ R} \times 100$ = $25\%$

Therefore, the percentage change in resistance is 25%.

$\text{Hence, the correct answer is 25\%.}$