A wire of length L and radius r is rigidly fi | vedclass

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(A) The formula for the extension $l$ in a wire is given by $l = \frac{FL}{AY}$,where $A = \pi r^2$ is the cross-sectional area and $Y$ is the Young's

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$l$

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(A) The formula for the extension $l$ in a wire is given by $l = \frac{FL}{AY}$,where $A = \pi r^2$ is the cross-sectional area and $Y$ is the Young's modulus.Thus,$l = \frac{FL}{\pi r^2 Y}$.Since the material is the same,$Y$ is constant. Therefore,$l \propto \frac{FL}{r^2}$.For the first wire: $l_1 = l$,$F_1 = F$,$L_1 = L$,$r_1 = r$.For the second wire: $F_2 = 2F$,$L_2 = 2L$,$r_2 = 2r$.Taking the ratio: $\frac{l_2}{l_1} = \left( \frac{F_2}{F_1} ight) 	imes \left( \frac{L_2}{L_1} ight) 	imes \left( \frac{r_1}{r_2} ight)^2$.Substituting the values: $\frac{l_2}{l} = \left( \frac{2F}{F} ight) 	imes \left( \frac{2L}{L} ight) 	imes \left( \frac{r}{2r} ight)^2 = 2 	imes 2 	imes \left( \frac{1}{4} ight) = 1$.Therefore,$l_2 = l$.

$A$ wire of length $L$ and radius $r$ is rigidly fixed at one end. On stretching the other end of the wire with a force $F$,the increase in its length is $l$. If another wire of same material but of length $2L$ and radius $2r$ is stretched with a force of $2F$,the increase in its length will be

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