MixedJEE Physics · Original learning card15 original chapter questions

Radial Conduction in Cylinders and Spheres

Integrating Fourier's law over concentric surfaces gives R_cyl = ln(r2/r1)/(2 pi k L) for a cylindrical shell and R_sph = (1/(4 pi k))(1/r1-1/r2) for a spherical shell.

Why this shows up in the exam

Pipe insulation · Spherical thermal shields · Radial heat loss from shells

Learn the idea

Radial conduction needs geometry-specific resistance because heat-flow area changes with radius. Heat spreading through a shell crosses larger area as radius increases, so a curved shell cannot be replaced by a flat slab unless a justified thin-shell approximation is stated.

🧠 Memory hook: Curved flow area changes with radius, so integrate.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • R_cyl = ln(r2/r1)/(2 pi k L) — radial resistance of a long cylindrical shell with negligible end effects
  • R_sph = (1/(4 pi k))(1/r1-1/r2) — radial resistance of a spherical shell
  • H = Delta T/R_radial — steady radial heat current after selecting the correct geometry

How to approach it

  1. 1Identify cylindrical, spherical, or thin-shell geometry
  2. 2Choose radii and boundary temperatures
  3. 3Use the proper radial resistance and test the thin-shell limit

Common slip-ups that cost marks

  • •Using L/kA with one arbitrary shell area
  • •Confusing axial and radial cylinder flow
  • •Dropping the logarithm for a thick cylindrical shell

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 15

A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.

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