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(D) The extension $\Delta l$ of an elastic spring is governed by Hooke's Law,$F = k \Delta l$,where $k$ is the spring constant.$1$. $\Delta l = \frac{

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(D) The extension $\Delta l$ of an elastic spring is governed by Hooke's Law,$F = k \Delta l$,where $k$ is the spring constant.$1$. $\Delta l = \frac{F}{k} = \frac{mg}{k}$,so the extension is directly proportional to the applied weight $(mg)$.$2$. The spring constant $k$ for a wire is given by $k = \frac{YA}{l}$,where $Y$ is Young's modulus,$A$ is the cross-sectional area,and $l$ is the original length.$3$. Substituting $k$ into the extension formula: $\Delta l = \frac{mg l}{YA}$.$4$. Thus,$\Delta l$ depends on the weight $(mg)$,the original length $(l)$,the cross-sectional area $(A)$,and the material properties ($Y$,which defines the type of spring).

When a weight is suspended from the end of an elastic spring,its increased length depends upon what?

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