Question

A steel rod has a radius $10 \,mm$ and a length of $1.0 \,m$. A force stretches it along its length and produces a strain of $0.32 \%$. Young's modulus of the steel is $2.0 \times 10^{11} Nm ^{-2}$. 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} $
$Y =2 \times 10^{11} Nm ^{2}$
We know
$\frac{F L}{A Y}=\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} $
$F=201 \,kN$