A wire is stretched by 0.01 m by a certain f | vedclass

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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 force $F$ and Young's modulus $Y$ are co

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$0.005$

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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 force $F$ and Young's modulus $Y$ are constant,we have $l \propto \frac{L}{r^2}$.Let the original length be $L_1 = L$ and radius be $r_1 = r$. The elongation is $l_1 = 0.01 \ m$.For the second wire,the length $L_2 = 2L$ and the diameter is doubled,so the radius $r_2 = 2r$.Using the proportionality: $\frac{l_2}{l_1} = \frac{L_2}{L_1} 	imes \left( \frac{r_1}{r_2} ight)^2$.Substituting the values: $\frac{l_2}{l_1} = \left( \frac{2L}{L} ight) 	imes \left( \frac{r}{2r} ight)^2 = 2 	imes \left( \frac{1}{2} ight)^2 = 2 	imes \frac{1}{4} = \frac{1}{2}$.Therefore,$l_2 = \frac{l_1}{2} = \frac{0.01 \ m}{2} = 0.005 \ m$.

$A$ wire is stretched by $0.01 \ m$ by a certain force $F$. Another wire of the same material whose diameter and length are double that of the original wire is stretched by the same force. Then its elongation will be (in $m$)

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