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,
![A=2\times10^{-6}m^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes10%5E%7B-6%7Dm%5E2)
* The young modulus of the steel is,
![Y=2.1\times10^{11}\text{ Pa}](https://tex.z-dn.net/?f=Y%3D2.1%5Ctimes10%5E%7B11%7D%5Ctext%7B%20Pa%7D)
Solution:
The young modulus of the steel in terms of the force and extension is,
![Y=\frac{F\times l}{A\times dl}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7BF%5Ctimes%20l%7D%7BA%5Ctimes%20dl%7D)
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,
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202.1%5Ctimes10%5E%7B11%7D%3D%5Cfrac%7BF%5Ctimes4%7D%7B2%5Ctimes10%5E%7B-6%7D%5Ctimes0.3%5Ctimes10%5E%7B-3%7D%7D%20%5C%5C%20F%3D0.315%5Ctimes10%5E2%5Ctext%7B%20N%7D%20%5C%5C%20F%3D31.5%5Ctext%7B%20N%7D%20%5Cend%7Bgathered%7D)
Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.