Search any question & find its solution
Question:
Answered & Verified by Expert
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
$\left(Y_{\text {steel }}=2 \times 10^{11} \mathrm{Nm}^{-2}\right)$
Options:
$\left(Y_{\text {steel }}=2 \times 10^{11} \mathrm{Nm}^{-2}\right)$
Solution:
1666 Upvotes
Verified Answer
The correct answer is:
$3.18 \times 10^8 \mathrm{Nm}^{-2}, 0.16 \%$
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 $=\frac{\text { Stress }}{Y}=\frac{3.18 \times 10^8}{2 \times 10^{11}}=0.16 \%$
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 $=\frac{\text { Stress }}{Y}=\frac{3.18 \times 10^8}{2 \times 10^{11}}=0.16 \%$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.