The Young's modulus of a steel wire of length 6m and...

The external force or tension $\left(F\right)$ acting on the wire is its own weight. It is given by

$\begin{cases}F & = & mg \\ = & 4kg \times \dfrac{10}{4}m{s}^{- 2} \\ = & 10N\end{cases}$

The formula to calculate the Young's modulus of the material of the wire is given by

$\begin{cases}Y & = & \dfrac{\dfrac{F}{A}}{\dfrac{∆ L}{L}} \\ = & \dfrac{FL}{∆ LA} . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )\end{cases}$

where, $A$ is the cross-sectional area, $L$ is the original length and $∆ L$ is the elongation of the wire.

Substitute the given values of the known parameters into equation (1) and solve to calculate the required elongation of the wire.

$\begin{cases}2 \times {10}^{11}N / {m}^{2} & = & \dfrac{10N \times 6m}{∆ L \times 3{mm}^{2} \times \dfrac{1{m}^{2}}{{10}^{6}{mm}^{2}}} \\ ∆ L & = & \dfrac{10N \times 6m}{2 \times {10}^{11}N / {m}^{2} \times 3{mm}^{2} \times \dfrac{1{m}^{2}}{{10}^{6}{mm}^{2}}} \\ = & 0 . 0001m \times \dfrac{{10}^{3}mm}{1m} \\ = & 0 . 1mm\end{cases}$