A metal wire of diameter D stretches by 0.100 mm when supporting a weight W . If the same-length wire is used to support a weight three times as heavy, what would its diameter have to be (in terms of D ) so it still stretches only 0.100 mm ?

The Young's Modulus is a physical property of any material mostly elastic in nature that determines the ease of deformation of that material with respect to stress and strain applied to it.

Given, that a wire having diameter D is elongated by $∆\mathrm{L}=0.100\mathrm{mm}$due to its weight W on it.

Now, for the initial length ${\mathrm{L}}_{\mathrm{o}}$, Young's Modulus is given by:

$$\begin{gathered} \mathrm{Y}=\frac{\mathrm{F}{\mathrm{L}}_{\mathrm{o}}}{\mathrm{A}\mathrm{L}} \\ =\frac{\mathrm{W}}{{\mathrm{\pi D}}^2}\frac{{\mathrm{L}}_{\mathrm{o}}}{\mathrm{L}} \end{gathered}$$

Since Young's modulus is a property independent of the dimensions of the medium or material, therefore, for the new weight 3W and new diameter D' , Young's Modulus is given by:

role="math" localid="1663824529351" $\mathrm{Y}=\frac{3\mathrm{W}}{\mathrm{\pi D}{\text{'}}^2}=\frac{{\mathrm{L}}_0}{\mathrm{L}}$

Equating both, we get:

role="math" localid="1663824559269" $$\begin{gathered} \frac{\mathrm{W}}{\mathrm{\pi D}{\text{'}}^2}\frac{{\mathrm{I}}_0}{\mathrm{I}}=\frac{3\mathrm{W}}{\mathrm{\tau \tau D}{\text{'}}^2}\frac{{\mathrm{I}}_0}{\mathrm{I}} \\ \mathrm{D}{\text{'}}^2=3{\mathrm{D}}^2 \\ \mathrm{D}\text{'}=\sqrt{3}\mathrm{D} \end{gathered}$$

So, its diameter has to be $\sqrt{3}$times of initial diameter.

Hence, this is the required answer.