A wire of length 1m has a resistance of 20 at 0^ C. It is uniformly stretched so that its length increases by 21%. Assuming the volume of the wire remains constant, the percentage change in resistance is n % . Alternatively, if the wire is heated [without stretching] through a temperature of 27^ C and if the temperature coefficient of resistance of the material of wire is 0.004K⁻¹ , the percentage change in resistanc

Step-by-step solution

For stretching of the wire, the volume V = A L remains constant. Resistance R = L A = L^2 V R L^2 Given that the length increases by 21 % , the new length is L' = 1.21 L . New resistance R' = R (1.21)^2 = 1.4641 R Percentage change in resistance n = R' - R R 100 = (1.4641 - 1) 100 = 46.41 46.4 For heating the wire, the change in resistance is given by R = R T . Percentage change in resistance m = R R 100 = T 100 Given = 0.004 K⁻¹ and T = 27^ C = 27 K . m = 0.004 27 100 = 10.8 The values of m and n are 10.8 and 46.4