Question

A copper wire ($Y={10}^{11}N/m^2$) of length $8$ m and a steel wire $(Y=2{\times }^{11}N/m^2)$ of length $4$ m, each of $0.5c{m}^2$ cross-section are fastened end to end and streatched with a tension of $500$ N.

• A. Elongation in copper wire is $0.8$ mm
• B. Elongation in steel is $1/4^{th}$ the elongation in copper wire
• C. Total elongation is $1$ mm
• D. All of the above

Answer 1

The correct option is D All of the above

$\begin{array}{c}\text{For steel}:\\ Y=2\times {10}^{11}N/m^2\\ l=4m\\ A=0.5{cm}^2\end{array}$

$\begin{array}{c}\text{for copper}:\\ Y={10}^{11}\times N/m^2\\ l=8m\\ A=0.5{cm}^2\end{array}$

$T=500N$

$\begin{array}{c}\Delta l_c=\frac{F{l}_c}{A{Y}_c}=\frac{500\times 8}{0.5\times {10}^{-4}\times {10}^{11}}=8\times {10}^{-4}m=0.8mm\\ \Delta l_S=\frac{F{l}_S}{A{Y}_S}=\frac{500\times 4}{0.5\times {10}^{-4}\times 2\times {10}^{11}}=2\times {10}^{-4}m=0.2mm\end{array}$

$\Delta l_c+\Delta l_s=0.8+0.2=1mm$

$\text{Hence, option (D) is correct}$