Question

A steel rod of length $1 \,m$ and radius $10\, mm$ is stretched by a force $100\, kN$ along its length. The stress produced in the rod is $\left(Y_{Steel}=2\times10^{11}N m^{-2}\right)$, the percentage strain produced in the rod is

• A. $0.04 \%$
• B. $0.08 \%$
• C. $0.16\%$
• D. $0.24\%$

Answer 1

Answer: (C)

Elongation,
$\Delta L$ $=\frac{\left(F /A\right)L}{Y}$ $=\frac{\left(3.18\times10^{8} N m^{-2}\right)\left(1 m\right)}{2\times10^{11}N m^{-2}}$
$=1.59\times10^{-3}m$ $=1.59 \,mm$
Strain produced in the rod is
Strain $=\frac{\Delta L}{L}$ $=\frac{1.59\times10^{-3} m}{1 m}$ $=1.59\times10^{-3}$ $=0.16 \%$