Q65 / 90 · JEE Main 2012 (07 May Online) · Physics

A structural steel rod has a radius of $10 \mathrm{~mm}$ and length of $1.0 \mathrm{~m}$. A $100 \mathrm{kN}$ force stretches it along its length. Young's modulus of structural steel is $2 \times 10^{11} \mathrm{Nm}^{-2}$. The percentage strain is about

  • Correct answer
    A
    $0.16 \%$
  • B
    $0.32 \%$
  • C
    $0.08 \%$
  • D
    $0.24 \%$

Solution

Given: $F=100 \mathrm{kN}=10^5 \mathrm{~N}$ $$ \begin{aligned} & \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2} \\ & \ell_0=1.0 \mathrm{~m} \end{aligned} $$ radius $\mathrm{r}=10 \mathrm{~mm}=10^{-2} \mathrm{~m}$ From formula, $Y=\frac{\text { Stress }}{\text { Strain }}$ $$ \Rightarrow \quad \text { Strain }=\frac{\text { Stress }}{Y}=\frac{F}{A Y} $$ $$ \begin{aligned} & =\frac{10^5}{\pi r^2 Y}=\frac{10^5}{3.14 \times 10^{-4} \times 2 \times 10^{11}} \\ & =\frac{1}{628} \end{aligned} $$ Therefore $\%$ strain $=\frac{1}{628} \times 100=0.16 \%$

Hence, the correct answer is option: 'A'.

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