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
Diffcult
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(a) Let $l$ be the original length of wire and $x$ be its length stretched uniformly such that final length is $1.5\, l$
Then $4R = \rho \frac{{(l - x)}}{A} + \rho \frac{{(0.5l + x)}}{{A'}}$where $A' = \frac{x}{{(0.5l + x)}}A$
$4\rho \frac{l}{A} = \rho \frac{{l - x}}{A} + \rho \frac{{{{(0.5l + x)}^2}}}{{xA}}$
or $4l = l - x + \frac{1}{4}\frac{{{l^2}}}{x} + \frac{{{x^2}}}{x} + \frac{{lx}}{x}$ or $\frac{x}{l} = \frac{1}{8}$
Then $4R = \rho \frac{{(l - x)}}{A} + \rho \frac{{(0.5l + x)}}{{A'}}$where $A' = \frac{x}{{(0.5l + x)}}A$
$4\rho \frac{l}{A} = \rho \frac{{l - x}}{A} + \rho \frac{{{{(0.5l + x)}^2}}}{{xA}}$
or $4l = l - x + \frac{1}{4}\frac{{{l^2}}}{x} + \frac{{{x^2}}}{x} + \frac{{lx}}{x}$ or $\frac{x}{l} = \frac{1}{8}$
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