The force constant of a wire is k and that of | vedclass

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(B) The work done in stretching a wire (or spring) by a distance $x$ is given by the formula $W = \frac{1}{2}kx^2$.Here,$k$ is the force constant and

Vedclass · Accepted Answer

$W_2 = 2W_1$

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(B) The work done in stretching a wire (or spring) by a distance $x$ is given by the formula $W = \frac{1}{2}kx^2$.Here,$k$ is the force constant and $x$ is the extension.For the first wire,the work done is $W_1 = \frac{1}{2}k_1x^2$,where $k_1 = k$.For the second wire,the work done is $W_2 = \frac{1}{2}k_2x^2$,where $k_2 = 2k$.Since both wires are stretched by the same distance $x$,we can see that $W \propto k$.Therefore,the ratio of the work done is $\frac{W_2}{W_1} = \frac{k_2}{k_1} = \frac{2k}{k} = 2$.This implies $W_2 = 2W_1$.

The force constant of a wire is $k$ and that of another wire is $2k$. When both the wires are stretched through the same distance,then the work done is:

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