A load W produces an extension of 1 mm in a | vedclass

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(C) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sect

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(C) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since $A = \pi r^2$,we have $l \propto \frac{F}{r^2}$.Given the initial state: $l_1 = 1 \ mm$,$F_1 = W$,and $r_1 = r$.Given the final state: $F_2 = 4W$ and $r_2 = 2r$.Using the proportionality $l \propto \frac{F}{r^2}$,we get the ratio:$\frac{l_2}{l_1} = \frac{F_2}{F_1} 	imes \left( \frac{r_1}{r_2} ight)^2$Substituting the values: $\frac{l_2}{1} = \frac{4W}{W} 	imes \left( \frac{r}{2r} ight)^2 = 4 	imes \left( \frac{1}{2} ight)^2 = 4 	imes \frac{1}{4} = 1$.Therefore,the new extension $l_2 = 1 \ mm$.

$A$ load $W$ produces an extension of $1 \ mm$ in a thread of radius $r$. If the load is increased to $4W$ and the radius is increased to $2r$,while all other factors remain the same,what will be the new extension in $mm$?

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