Question

A copper wire and a steel wire of the same diameter and length 1 m and 2 m respectively are connected end and a force is applied which stretches their combined length by 1 cm. How much each wire is elongated respectively. Y of copper = $1.2\times {10}^{10}$ $M{m}^{-2}$ and Y of steel +$2.0\times {10}^{10}N{m}^{-2}$

• A. 0.45 cm, 0.55 cm
• B. 0.55 cm, 0.45 cm
• C. 0.045 cm , 0.55 cm
• D. 0.45 cm, 0.055 cm

Answer 1

The correct option is C 0.045 cm , 0.55 cm

$\begin{array}{c}\text{Y of copper}=1.2\times {10}^{10}N/m^2\\ \text{Yof steel}=2\times {10}^{10}N/m^2\\ \text{L of copper}=tm\\ \text{L of steel}=2m\\ \text{both have same diameder}=d\\ \Delta L_{\text{net}}=1cm\end{array}$

$\text{Let}F\text{force is applied,}$

$\begin{array}{c}\Delta L_{\text{net}}=\Delta L_1+\Delta L_2\\ \text{we know}\\ Y=\frac{FL}{A\Delta L}\end{array}$

$\Delta L=\frac{FL}{AY}\Delta L\alpha \frac{L}{Y}\begin{array}{c}(\text{since}F\text{and}A\text{are same}\\ \text{For both})\end{array}$

$\begin{array}{cc}\text{For steel , For copper}& \\ \Delta L_1=\frac{x{L}_1}{Y_1}& \Delta L_2=x\frac{L_2}{Y_2}\end{array}$

$\begin{array}{cc}\mathrm{1x}{10}^{-2}& =\frac{2\times x}{2\times {10}^{10}}+\frac{x\times 1}{1.2\times {10}^{10}}\\ x& =\frac{6}{11}\times {10}^8\end{array}$

$\begin{array}{cc}\Delta L_{net}=& 1cm\\ & \{\begin{array}{c}\{x\text{is proportionality}\\ \text{constant})\end{array}\\ \therefore \Delta L_1=\frac{6}{11}\times \frac{2}{2\times {10}^{10}\times {10}^8}=0.055cm& \\ \Delta L_2=\frac{6}{11}\times {10}^8\times \frac{1}{1.2\times {10}^{10}}=0.045cm& \end{array}$