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(D) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$

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$3.2$

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(D) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the volume $V = A 	imes L$ is constant,we can write $A = \frac{V}{L}$.Substituting this into the elongation formula: $\Delta L = \frac{FL}{(V/L)Y} = \frac{FL^2}{VY}$.Since $F$,$V$,and $Y$ are constant for both wires,we have $\Delta L \propto L^2$.Given $\Delta L_1 = 2\, mm$ for $L_1 = 2\, m$,and we need to find $\Delta L_2$ for $L_2 = 8\, m$.Using the ratio: $\frac{\Delta L_2}{\Delta L_1} = \left( \frac{L_2}{L_1} ight)^2 = \left( \frac{8}{2} ight)^2 = 4^2 = 16$.Therefore,$\Delta L_2 = 16 	imes \Delta L_1 = 16 	imes 2\, mm = 32\, mm$.Converting to centimeters: $32\, mm = 3.2\, cm$.

$A$ wire of length $2\, m$ is made from $10\, cm^3$ of copper. $A$ force $F$ is applied so that its length increases by $2\, mm$. Another wire of length $8\, m$ is made from the same volume of copper. If the same force $F$ is applied to it,its length will increase by ......... $cm$.

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