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(A) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material $(Y)$ and the length $(L)$ are

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$8$

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(A) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material $(Y)$ and the length $(L)$ are constant,we have the proportionality $l \propto \frac{F}{r^2}$.Let the initial force be $F_1 = F$,initial radius be $r_1 = r$,and initial elongation be $l_1 = 1 \ mm$.For the second wire,the force is $F_2 = 2F$ and the radius is $r_2 = \frac{r}{2}$.Using the ratio: $\frac{l_2}{l_1} = \frac{F_2}{F_1} 	imes \left( \frac{r_1}{r_2} ight)^2$.Substituting the values: $\frac{l_2}{1} = \frac{2F}{F} 	imes \left( \frac{r}{r/2} ight)^2 = 2 	imes (2)^2 = 2 	imes 4 = 8$.Therefore,the elongation $l_2 = 8 \ mm$.

$A$ wire extends by $1 \ mm$ when a force is applied. If double the force is applied to another wire of the same material and length but half the radius of the cross-section,what will be the elongation of the wire in $mm$?

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