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NCERT gives you R = ρl/A for a wire of uniform cross-section. Then a JEE-style question hands you a conductor shaped like a truncated cone, wider at one end, and the formula has no single A to use. Averaging the radius looks reasonable and gives the wrong answer. This post shows how to set up the integral, works a cone and a wedge with numbers, and shows how far off the average is.
Where is this in the syllabus?
Chapter 3 of the NCERT Class 12 Physics textbook (2026-27 reprint) builds R = ρl/A (Eq. 3.9) from two facts: resistance grows in proportion to length and falls in inverse proportion to area. The JEE (Main) 2026 syllabus lists "electrical resistance" and "electrical resistivity and conductivity", and the CBSE Physics curriculum for 2026-27 has the same topics. NCERT does not work a tapered conductor; it is an extension that uses integration from the JEE mathematics syllabus. The NEET (UG) 2026 syllabus lists resistivity too, but an integral of this kind is a JEE-style question. For the microscopic side of resistance, see our guide to drift velocity and resistivity.
How do you set up the integral?
Use the four-step slicing method.
Pick a slice. Take a thin disc of thickness dx at distance x from the narrow end.
Write its resistance. Over a short length, the area is nearly constant, so dR = ρ dx / A(x).
Decide how slices combine. The current flows along the length, so every slice carries the same current, one after another. They are in series, and series resistances add: R = ∫ ρ dx / A(x).
Write A(x) and integrate. For a cone, the radius grows linearly with x.
The series step is the one to say out loud. If the current flowed across the slices instead, such as from the inner to the outer surface of a tube, the slices would be in parallel, and you would add conductances instead.
This set-up step is where most errors happen. In interviews with students in a calculus-based physics course, Nguyen and Rebello (2011) found that "most of the students could recognize the need for an integral" but "failed to set up the desired integral", and traced this to difficulty understanding the small term and the idea of adding up small contributions.
Worked example 1: a truncated cone
A copper conductor is 0.50 m long, with radius a = 1.0 mm at one end and b = 2.0 mm at the other. Take ρ = 1.68 × 10⁻⁸ Ω m, the value for copper in HyperPhysics' resistivity table. Find its resistance.
Radius at x: r = a + (b − a)x/L. Then
R = ∫ from 0 to L of ρ dx / (πr²).
Change variable: dr = (b − a)/L × dx, so
R = (ρL / (π(b − a))) × ∫ from a to b of dr/r² = (ρL / (π(b − a))) × (1/a − 1/b) = ρL / (πab).
Numbers: R = 1.68 × 10⁻⁸ × 0.50 / (π × 1.0 × 10⁻³ × 2.0 × 10⁻³) = 1.34 × 10⁻³ Ω, or 1.34 mΩ.
Limiting check. If a = b, the formula becomes ρL/(πa²), the uniform-wire result. Always run this check.
Compare the shortcuts. Using the average radius, 1.5 mm, gives 1.19 mΩ, about 11% too low. Using the average of the two end areas gives 1.07 mΩ, 20% too low. Both fail because the narrow end contributes most of the resistance, and averages give it too little weight. The exact result uses the product ab, as if the radius were the geometric mean √(ab).
Worked example 2: a wedge-shaped strip
A nichrome strip is 10 cm long and 1.0 cm wide, and its thickness grows linearly from 1.0 mm to 3.0 mm. Take ρ = 1.00 × 10⁻⁶ Ω m, from the same table. Find its resistance.
Here A(x) = w × t(x), with t growing linearly. The same steps give
R = ρL ln(t₂/t₁) / (w(t₂ − t₁)).
Numbers: R = 1.00 × 10⁻⁶ × 0.10 × ln 3 / (0.010 × 0.0020) = 5.49 × 10⁻³ Ω, or 5.49 mΩ. The average thickness, 2.0 mm, would give 5.00 mΩ, about 9% low.
Notice the different function: a cone gives 1/r², which integrates to a power, while a wedge gives 1/t, which integrates to a logarithm. Write A(x) carefully before you integrate. We checked both answers by adding up 200,000 thin slices numerically; they match the formulas.
One honest caveat: the method assumes the current spreads evenly across each slice. That is a good approximation for a gentle taper, which is what these problems describe.
Which mistakes cost marks?
Averaging the radius or area. It underestimates the resistance of any taper.
Adding slices in parallel. Slices along the current are in series.
Writing r = b − a in the denominator. The radius changes with x; write r(x) first.
Unit slips. Convert millimetres to metres before squaring.
A practice routine
Rederive R = ρL/(πab) from a blank page, saying what each small term means.
Redo example 1 with b = 3.0 mm. Predict first: the answer should fall to two-thirds, about 0.89 mΩ.
Find the resistance of a uniform wire whose resistivity grows linearly from ρ₀ to 2ρ₀ along its length. The answer is 1.5ρ₀L/A.
Then try building integrals in other topics with our calculus for JEE physics guide.
If your set-up keeps going wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, derive the cone formula without notes, then check it with the limit a = b.
