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%.

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