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.
Correct Answer: 25
Approach Solution - 1
\(R=\frac{Pl}{A}\)
\(R'=\frac{P\ 1.2\ l}{0.96A}=\frac{1.25\ pl}{A}\)
R' = 1.25 R
So, the percentage change in resistance is 25%
Approach Solution -2
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:
\(\frac {R'-R}{R}\) * 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:
\(\frac{(R' - R)}{ R} * 100%\) = 25%
Therefore, the percentage change in resistance is 25%.
Hence, the correct answer is 25%.
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