Question

A composite wire consists of a steel Wire of length 1 5 and a co uniform cross-sectional area of $2.5\times {10}^{-5}{m}^{-5}$.It is loaded with a mass of 200kg. Find the extension produced. Young's modulus of copper is $2.5\times {10}^{11}{Nm}^{-2}$ and that of steel $2.0\times {10}^{11}{Nm}^{-2}$

• A. 4.156 mm.
• B. 2.156 mm.
• C. 2.256 mm.
• D. 3.156 mm.

Answer 1

Answer: (B)

2.156 mm.

$\begin{array}{c}\text{Length of steel wire}1.5m\text{Length of copper wire}=2m\\ \text{Tension in}\text{composite wire}=mg\text{.}\end{array}$

$\begin{array}{c}\text{given load mass =}200kg\\ \text{Tensile stress}=\frac{\text{Load Force}}{\text{Area of cross section}}\end{array}$

$\text{load}\mathrm{mass}=200kg$

$\begin{array}{c}\text{Given area of Cross section}=2.5\times {10}^{-5}\\ \begin{array}{c}\text{Stress}=\frac{2000}{2.5\times {10}^{-5}}N/m^2\\ \text{Let}\Delta l_1,\Delta l_2\text{be elongations of Steel, copper wires}\\ \text{respectively.}\end{array}\end{array}$

$\begin{array}{c}{Y}_{\text{Steel}}=\frac{\text{Stress}}{\text{strain}}\Rightarrow 2\times {10}^{11}=\frac{2000}{2.5\times {10}^{-5}\times \frac{\Delta l_1}{1.5}}\\ \Rightarrow \Delta l_1=1.52mm\\ {\text{Y}}_{copper}=\frac{\text{Stress}}{\text{Strain}}\Rightarrow 2.5\times {10}^{11}=\frac{2000}{2.5\times {10}^{-5}\times \frac{\Delta l_2}{2}}\\ \Rightarrow \Delta l_2=0.64mm\end{array}$

$\begin{array}{c}\text{So net elongation,}\Delta l=\Delta l_1+\Delta l_2\\ \Delta l=2.16mm\end{array}$