Question

A composite rod consists of a steel rod of length $25\text{cm}$ and area $2A$ and a copper rod of length $50\text{cm}$ 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

• A. The extension produced in copper rod will be more.
• B. The extension in copper and steel parts will be in the ratio 1 : 2.
• C. The stress applied to copper rod will be more.
• D. No extension will be produced in the steel rod.

Answer 1

Answer: (A), (C)

The correct options are
A The extension produced in copper rod will be more.
C The stress applied to copper rod will be more.

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

Given that $\frac{Y_S}{Y_{Cu}}=\frac{2}{1}$
$\frac{\text{(strain)}}{{\text{(strain)}}_{Cu}}=\frac{\frac{{\text{(stress)}}_S}{Y_S}}{\frac{{\text{(stress)}}_{Cu}}{Y_{Cu}}}$

$\frac{\text{(strain)}}{{\text{(strain)}}_{Cu}}=\frac{Y_{Cu}}{Y_S}\times \frac{\left(\frac{F}{2A}\right)}{\left(\frac{F}{A}\right)}=\frac{1}{4}$

$\frac{\frac{\Delta L_S}{L_S}}{\frac{\Delta L_{Cu}}{L_{Cu}}}=\frac{1}{4}$

$\frac{25}{50}\times \frac{\Delta L_{Cu}}{\Delta L_S}=4$ or $\frac{\Delta L_{Cu}}{\Delta L_S}=\frac{8}{1}$
So, options (a)and (c) are correct.