In order to quadruple the resistance of a uni | vedclass

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(A) Let $l$ be the original length of the wire and $A$ be its original cross-sectional area. The original resistance is $R = \rho \frac{l}{A}$.Let a l

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$1/8$

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(A) Let $l$ be the original length of the wire and $A$ be its original cross-sectional area. The original resistance is $R = \rho \frac{l}{A}$.Let a length $x$ of the wire be stretched to a new length $x' = x + 0.5l$. Since the volume of this part remains constant,$A x = A' x'$,where $A'$ is the new cross-sectional area.Thus,$A' = A \frac{x}{x + 0.5l}$.The new resistance $R'$ is the sum of the resistance of the unstretched part $(l-x)$ and the stretched part $(x+0.5l)$:$R' = \rho \frac{l-x}{A} + \rho \frac{x+0.5l}{A'} = \rho \frac{l-x}{A} + \rho \frac{(x+0.5l)^2}{Ax}$.Given $R' = 4R$,we have $4 \frac{l}{A} = \frac{l-x}{A} + \frac{(x+0.5l)^2}{Ax}$.Multiplying by $Ax$,we get $4lx = x(l-x) + (x+0.5l)^2$.$4lx = lx - x^2 + x^2 + lx + 0.25l^2$.$4lx = 2lx + 0.25l^2$.$2lx = 0.25l^2$.$x/l = 0.25/2 = 1/8$.

In order to quadruple the resistance of a uniform wire,a part of its length $x$ was uniformly stretched till the final length of the entire wire was $1.5$ times the original length $l$. The fraction of the wire that was stretched,$x/l$,is equal to:

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