Work done in stretching a wire through 1 mm | vedclass

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(D) The work done in stretching a wire is given by $W = \frac{1}{2} k x^2$,where $k = \frac{Y A}{L}$ is the force constant of the wire.Here,$Y$ is You

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$16$

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(D) The work done in stretching a wire is given by $W = \frac{1}{2} k x^2$,where $k = \frac{Y A}{L}$ is the force constant of the wire.Here,$Y$ is Young's modulus,$A = \pi R^2$ is the cross-sectional area,and $L$ is the original length.For the first wire: $W_1 = \frac{1}{2} \left( \frac{Y \pi R^2}{L} ight) x^2 = 2 \ J$ (where $x = 1 \ mm$).For the second wire: $L' = \frac{L}{2}$ and $R' = 2R$.The new force constant is $k' = \frac{Y \pi (2R)^2}{L/2} = \frac{Y \pi (4R^2)}{L/2} = 8 \left( \frac{Y \pi R^2}{L} ight) = 8k$.The work done for the second wire is $W_2 = \frac{1}{2} k' x^2 = \frac{1}{2} (8k) x^2 = 8 \left( \frac{1}{2} k x^2 ight) = 8 W_1$.Substituting $W_1 = 2 \ J$,we get $W_2 = 8 	imes 2 \ J = 16 \ J$.

Work done in stretching a wire through $1 \ mm$ is $2 \ J$. What amount of work will be done for elongating another wire of same material,with half the length and double the radius of cross-section,by $1 \ mm$ (in $J$)?

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