91Ó°ÊÓ

Given that a brass wire is to withstand a tensile force of 350 N without breaking.

Breaking stress of wire = $\frac{{\mathbf{F}}_{\mathbf{âŠ\yen }}}{\mathbf{A}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(1)

Where${\mathbf{F}}_{\mathbf{âŠ\yen }}$is the tensile force applied to an object. A is the area over which force is exerted.

Let d be the diameter of the wire.

$\mathbf{A}\mathbf{=}\mathbf{Ï„Ï„}\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(2)

Using equations (1) and (2), we can write,

$\mathrm{A}=\frac{{\mathrm{F}}_{âŠ\yen }}{\mathrm{φ}}$

Where ${\mathrm{F}}_{âŠ\yen }$ is the tensile force applied to an object, A is the area over which force is exerted, and$\mathrm{φ}$ is breaking stress of brass

${\mathrm{F}}_{âŠ\yen }=350\mathrm{N}$

Breaking stress of brass is $\mathrm{φ}=4.7×{10}^8\mathrm{Pa}$

Therefore,

$\begin{array}{r}A=\frac{350\mathrm{N}}{4.7×{10}^8\mathrm{Pa}}\\ =7.45×{10}^{−7}{\mathrm{m}}^2\end{array}$

Using equation (2), we can write, that

$\mathrm{d}=\sqrt{\frac{4\mathrm{A}}{\mathrm{Ï„Ï„}}}$

Substituting the values in the above expression, we get,

$\mathrm{d}=\sqrt{\frac{4×\left(7.45×{10}^{-7}{\mathrm{m}}^2\right)}{3.14}}$mm

= 0.97 m

Hence, the diameter of the wire is 0.97 mm.