A composite rod consists of a steel rod of length 25 cm and area 2A and a copper rod of length 50 cm and area A. The composite rod is subjected to an axial load F. If the Young's moduli of steel and copper are in the ratio 2:1, then

$(\text{ stress }{)}_{\mathrm{S}}=\frac{\mathrm{F}}{2\mathrm{A}}, (\text{ stress }{)}_{\mathrm{Cu}}=\frac{\mathrm{F}}{\mathrm{A}}$

Given that $\frac{{\mathrm{Y}}_{\mathrm{S}}}{{\mathrm{Y}}_{\mathrm{Cu}}}=\frac{2}{1}$

$\frac{{\left(\text{ strain }\right)}_{\mathrm{S}}}{(\text{ strain }{)}_{\mathrm{Cu}}}=\frac{(\text{ stress }{)}_{\mathrm{S}}/{\mathrm{Y}}_{\mathrm{S}}}{(\text{ stress }{)}_{\mathrm{Cu}}/{\mathrm{Y}}_{\mathrm{Cu}}}$

$=\frac{(\text{ stress }{)}_{\mathrm{S}}}{(\text{ stress }{)}_{\mathrm{Cu}}}\times \frac{{\mathrm{Y}}_{\mathrm{Cu}}}{{\mathrm{Y}}_{\mathrm{S}}}=\frac{(\mathrm{F}/2\mathrm{A})}{(\mathrm{F}/\mathrm{A})}\times \frac{1}{2}=\frac{1}{4}$

 $$\begin{gathered} \frac{{\mathrm{\Delta L}}_{\mathrm{S}}/{\mathrm{L}}_{\mathrm{S}}}{{\mathrm{\Delta L}}_{\mathrm{Cu}}/{\mathrm{L}}_{\mathrm{Cu}}}=\frac{1}{4}\text{ or }\frac{{\mathrm{L}}_{cu} {\mathrm{\Delta L}}_s}{{\mathrm{L}}_s {\mathrm{\Delta L}}_{\mathrm{cu}}}=\frac{1}{4} \\ \frac{50 {\mathrm{\Delta L}}_s}{25 {\mathrm{\Delta L}}_{\mathrm{cu}}}=\frac{1}{4} \\ \frac{ {\mathrm{\Delta L}}_s}{{\mathrm{\Delta L}}_{\mathrm{cu}}}=\frac{1}{8} \end{gathered}$$

As area of cross section of steel is more than copper. Stress applied to steel is less than copper