Why can't we use Δ R/R ≈ Δ ell/ell - Δ A/A = 24%?

The differential approximation dR/R = dell/ell - dA/A is only valid for very small changes (typically ≤ 5%). A 20% change is too large for first-order approximations.

JEE MainPhysics

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+4 marks−1 if incorrectNumericalPrevious-year question

Calculating Percentage Change in Electrical Resistance from Length and Area Variations

The length of a metallic wire is increased by 20% and its area of cross section is reduced by 4%. The percentage change in resistance of the metallic wire is ________.

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Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why can't we use

Δ R / R \approx Δ ℓ / ℓ - Δ A / A = 24 %

The differential approximation $d R / R = d ℓ / ℓ - d A / A$ is only valid for very small changes (typically $\leq 5 %$ ). A $20 %$ change is too large for first-order approximations.

Answer

Accepted answer: 25 %

The percentage change in resistance of the metallic wire is 25%.

Why each option works or fails

Step-by-step solution

Given: Initial resistance $R = ρ \frac{ℓ}{A}$ , length increased by $20 %$ (new length $ℓ^{'} = 1.20 ℓ$ ), area decreased by $4 %$ (new area $A^{'} = 0.96 A$ ).
Find: The percentage change in resistance $\frac{R^{'} - R}{R} \times 100 %$ .
Visualise: A cylindrical metallic wire stretched or processed such that length becomes $1.2 ℓ$ and cross-sectional area narrows to $0.96 A$ , keeping resistivity $ρ$ constant.
Strategise: Use the formula for electrical resistance $R = ρ \frac{ℓ}{A}$ . Substitute $ℓ^{'} = 1.20 ℓ$ and $A^{'} = 0.96 A$ to find the ratio $R^{'} / R$ , then compute percentage increase as $(R^{'} / R - 1) \times 100$ .
Execute: $R^{'} = ρ \frac{1.20 ℓ}{0.96 A} = \frac{1.20}{0.96} R = 1.25 R$ . The percentage change is $(1.25 - 1) \times 100 = 25 %$ .
Verify: Length increased ( $+ 20 %$ ) and area decreased ( $- 4 %$ ); both effects increase resistance, so the result must be greater than $+ 20 %$ . $25 %$ is positive and fully consistent.

Calculating Percentage Change in Electrical Resistance from Length and Area Variations

What feels right?

The percentage change in resistance of the metallic wire is 25%.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

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