Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,
[tex]A=2\times10^{-6}m^2[/tex]
* The young modulus of the steel is,
[tex]Y=2.1\times10^{11}\text{ Pa}[/tex]
Solution:
The young modulus of the steel in terms of the force and extension is,
[tex]Y=\frac{F\times l}{A\times dl}[/tex]
where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,
[tex]\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}[/tex]
Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.
