91Ó°ÊÓ

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

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

Where${\mathit{F}}_{\mathbf{âŠ\yen }}$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.

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

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

$A=\frac{F_{âŠ\yen }}{\mathrm{φ}}$

Where $F_{âŠ\yen }$ 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

$F_{âŠ\yen }=350N$

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

Therefore,

$\begin{array}{rcl}A& =& \frac{350N}{4.7×{10}^{8}Pa}\\ & =& 7.45×{10}^{-7}{m}^2\end{array}$

Using equation (2), we can write, that

$d=\sqrt{\frac{4A}{\mathrm{Ï€}}}$

Substituting the values in the above expression, we get,

$\begin{array}{rcl}d& =& \sqrt{\frac{4×\left(7.45×{10}^{-7}{m}^2\right)}{3.14}}\\ & =& 0.97m\end{array}$mm

Hence, the diameter of the wire is 0.97 mm.