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

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(A) The work done in stretching a wire is given by $W = \frac{1}{2} kx^2$,where $k = \frac{YA}{L}$.Substituting $k$,we get $W = \frac{1}{2} \left( \fr

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$16 \ J$

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(A) The work done in stretching a wire is given by $W = \frac{1}{2} kx^2$,where $k = \frac{YA}{L}$.Substituting $k$,we get $W = \frac{1}{2} \left( \frac{YA}{L} ight) x^2$.Since the material $(Y)$ and the extension $(x)$ are the same for both wires,$W \propto \frac{A}{L}$.Since $A = \pi r^2$,we have $W \propto \frac{r^2}{L}$.Let $r_1 = r$ and $L_1 = L$. Then $r_2 = 2r$ and $L_2 = \frac{L}{2}$.Taking the ratio: $\frac{W_2}{W_1} = \left( \frac{r_2}{r_1} ight)^2 \left( \frac{L_1}{L_2} ight) = (2)^2 \left( \frac{L}{L/2} ight) = 4 	imes 2 = 8$.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 the same material but with double the radius of cross-section and half the length by $1 \ mm$ is:

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