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A copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is ...... (Assume that the volume of the wire remains constant.) |
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Answer» increase by 0.2 % Suppose the LENGTH of the wire is l and area of cross-section is A. The RESISTANCE of a wire, = R = `(rho l)/(A)` ` therefore R = (rho l^(2))/(Al) = (rho l^(2))/(V) `.... (1) `therefore (dR)/(dl)= (rho)/(V) (2l) ` `therefore dR = (2 rho)/(V) l dl "" ` .... (2) Taking ratio of equations (2) and (1), `(dR)/(R) = ((rho)/(V) 2ldl)/((rho)/(V)l^(2))` `therefore (dR)/(R) = 2 (dl)/(l)` Percentage change = `(dR)/(R) xx 100 %` = 2 `((dl)/(l) ) xx 100 %` = `2 (0.1 %) = 0.2 %` Thus, the resistance of the wire increases by 0.2 % . |
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