See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

The current will be conducted radially outwards from the inner conductor to the outer conductor. The area of cross-section for the conduction of the current is therefore the area of an elementary cylindrical shell which varies with radius. The length of the conducting shell is measured radially from radius a to radius b. Consider an elementary cylindrical shell of radius r and thickness dr. Its area of cross-section (normal to flow of current) = (2r) and its length = dr. Hence the resistance of the elementary cylindrical shell of the medium is dr dr dR 2 r 2 r               The resistance of the medium is obtained by integrating for r from a to b. Hence required resistance b b e a a dr R log r 2 r 2              e b log 2 a          = 7 