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A long cylindrical conductor of radius a and length l is made of a material whose resistivity depends only on the distance r from the axis of the conductor as ρ = α r 2 , where α is a constant. The resistance of the conductor across the two ends will be

Options

  1. AR = 4 α l π a 3
  2. BR = 2 α l π a 4
  3. CR = 4 α π l a 3
  4. DR = 2 l π α a 3

Correct answer

B. R = 2 α l π a 4

Step-by-step solution

We consider a long co-axial hollow cylinder of radius r and thickness dr. The cross-sectional area of considered element is dA = 2​πrdr ∴ Electric resistance between ends of conductor of considered element is dR = ρ l dA = ρ l 2 π rdr = α l ⁡ 2 π r 3 dr [as resistivity is ρ = α r 2 ] The system may be assumed as a parallel combination of a number of such types of elementary resistance. ∴ 1 R = ∫ 1 dR = 2 π α l ⁡ ∫ 0 a r 3 dr or 1 R

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