Question

A steel rod has a radius $10mm$ and a length of $1.0m$. A force stretches it along its length and produces a strain of $0.32\%$. Young's modulus of the steel is $2.0\times 10^{11}N{m}^{-2}$. What is the magnitude of the force stretching the rod?

• A. 100.5 kN
• B. 201 kN
• C. 78 kN
• D. 150 kN

Answer 1

Answer: (B)

Strain $=0.32\%$
$\Rightarrow \frac{\Delta L}{L}\times 100=0.32$
$\Rightarrow \frac{\Delta L}{L}=\frac{0.32}{100}$
$A=\pi r^2=3.14\times {\left(\frac{10}{1000}\right)}^2$
$Y=2\times 10^{11}N{m}^2$
We know
$\frac{FL}{AY}=\Delta L$
$F=\left(\frac{\Delta L}{L}\right)\times A\times Y$
Substituting values
$F=\frac{0.32}{100}\times 3.14\times {\left(\frac{10}{1000}\right)}^2\times 2\times 10^{11}$
$F=201kN$