Search any question & find its solution
Question:
Answered & Verified by Expert
An iron rod of length $2 \mathrm{~m}$ and cross-sectional area of $50 \mathrm{~mm}^{2}$ stretched by $0.5 \mathrm{~mm}$, when a mass of $250 \mathrm{~kg}$ is hung from its lower end. Young's modulus of iron rod is
Options:
Solution:
2373 Upvotes
Verified Answer
The correct answer is:
$19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
$Y=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}$
$=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
$=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
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.