If the work done in stretching a wire by 1 m | vedclass

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Question

(A) The work done in stretching a wire is given by the formula: $w = \frac{1}{2} \frac{AY}{L} (\Delta \ell)^2$. Here,$A$ is the cross-sectional area,$

Vedclass · Accepted Answer

$16$

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(A) The work done in stretching a wire is given by the formula: $w = \frac{1}{2} \frac{AY}{L} (\Delta \ell)^2$. Here,$A$ is the cross-sectional area,$Y$ is Young's modulus,$L$ is the original length,and $\Delta \ell$ is the extension. Since $A = \pi r^2$,we can write $w \propto \frac{r^2}{L}$ for a constant extension $\Delta \ell$ and material $Y$. Given $w_1 = 2 \ J$,$r_2 = 2r_1$,and $L_2 = L_1/2$. Taking the ratio: $\frac{w_2}{w_1} = \left( \frac{r_2}{r_1} ight)^2 	imes \left( \frac{L_1}{L_2} ight) = (2)^2 	imes \left( \frac{L_1}{L_1/2} ight) = 4 	imes 2 = 8$. Therefore,$w_2 = 8 	imes w_1 = 8 	imes 2 \ J = 16 \ J$.

If the work done in stretching a wire by $1 \ mm$ is $2 \ J$,the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $1 \ mm$ is $.... \ J$.

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