In order to quadruple the resistance of a uniform wire, a part of its length was uniformly stretched till the final length of the entire wire was 1.5 times the original length, the part of the wire was fraction equal to
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Let l be the original length of wire and x be its length stretched uniformly such that final length is 1.5 l
Then $4R = \rho \frac{(l - x)}{A} + \rho \frac{(0.5 + x)}{A'}$ where $A' = \frac{x}{(0.5 + x)} A$
∴ ∴ $4 \rho \frac{l}{A} = \rho \frac{l - x}{A} + \rho \frac{(0.5 l + x)^2}{x A}$
or $4l = l - x + \frac{1}{4x^2} + \frac{x^2}{x} + \frac{|x|}{x}$ or $\underline{x} = \underline{1}$ $\underline{l} \quad 8$
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