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(B) Let the total length of the wire be $L = 1\,m$. Let the length of part $X$ be $\ell_X$ and the length of part $Y$ be $\ell_Y$. Thus,$\ell_X + \ell

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$1: 2$

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(B) Let the total length of the wire be $L = 1\,m$. Let the length of part $X$ be $\ell_X$ and the length of part $Y$ be $\ell_Y$. Thus,$\ell_X + \ell_Y = 1$.The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.For part $X$,$R_X = \rho \frac{\ell_X}{A_X}$. For part $Y$,$R_Y = \rho \frac{\ell_Y}{A_Y}$.When wire $X$ is stretched to length $\ell_W = 2\ell_X$,its volume remains constant $(A_X \ell_X = A_W \ell_W)$.Since $\ell_W = 2\ell_X$,we have $A_W = \frac{A_X}{2}$.The resistance of wire $W$ is $R_W = \rho \frac{\ell_W}{A_W} = \rho \frac{2\ell_X}{A_X/2} = 4 \left( \rho \frac{\ell_X}{A_X} \right) = 4R_X$.Given that $R_W = 2R_Y$,we substitute $R_W = 4R_X$ to get $4R_X = 2R_Y$,which implies $R_Y = 2R_X$.Since both parts $X$ and $Y$ were cut from the same original wire,they have the same cross-sectional area $A$ and resistivity $\rho$. Thus,$R \propto \ell$.Therefore,$\frac{R_X}{R_Y} = \frac{\ell_X}{\ell_Y}$.Substituting $R_Y = 2R_X$,we get $\frac{R_X}{2R_X} = \frac{\ell_X}{\ell_Y} = \frac{1}{2}$.

$A$ $1\,m$ long wire is broken into two unequal parts $X$ and $Y$. The $X$ part of the wire is stretched into another wire $W$. The length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$. Find the ratio of the length of $X$ to the length of $Y$.

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