Q134 / 160 · AP EAMCET 2023 (16 May Shift 1) · Physics

A steel rod of radius $20 \mathrm{~mm}$ and length of $2 \mathrm{~m}$ is acted upon by a force of $400 \mathrm{kN}$ along the length. The values of stress and strain are respectively $\left(Y_{\text {steel }}=2 \times 10^{11} \mathrm{Nm}^{-2}\right)$

  • A
    $1.96 \times 10^8 \mathrm{Nm}^{-2}, 0.16 \%$
  • Correct answer
    B
    $3.18 \times 10^8 \mathrm{Nm}^{-2}, 0.16 \%$
  • C
    $3.18 \times 10^8 \mathrm{Nm}^{-2}, 0.32 \%$
  • D
    $4 \times 10^8 \mathrm{Nm}^{-2}, 0.2 \%$

Solution

Radius of rod, $\mathrm{r}=20 \mathrm{~mm}=20 \times 10^{-3} \mathrm{~m}$ Length of rod, $L=2 \mathrm{~m}$ Force, $\mathrm{F}=400 \mathrm{~K} \mathrm{~N}=400 \times 10^3 \mathrm{~N}$ $\mathrm{Y}_{\text {steel }}=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ Stress $=\frac{F}{A}=\frac{F}{\pi r^2}$ $=\frac{400 \times 10^3}{3.14 \times\left(20 \times 10^{-3}\right)^2}=3.18 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ Young modulus, $Y=\frac{\text { stress }}{\text { strain }}$ Strain $=\frac{\text { Stress }}{Y}=\frac{3.18 \times 10^8}{2 \times 10^{11}}=0.16 \%$

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

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