Search any question & find its solution
Question:
Answered & Verified by Expert
\(A 3 \mathrm{~m}\) long steel wire is stretched to increase its length by \(0.3 \mathrm{~cm}\). Poisson's ratio for steel is 0.26 . The lateral strain produced in the wire is
Options:
Solution:
2939 Upvotes
Verified Answer
The correct answer is:
\(0.26 \times 10^{-3}\)
Length of steel wire, \(l=3 \mathrm{~m}\)
Increased length, \(\Delta l=0.3 \mathrm{~cm}=3 \times 10^{-3} \mathrm{~m}\)
Poisson's ratio \(=0.26\)
\(\Rightarrow \quad \frac{\text { Lateral strain }}{\text { Longitudinal strain }}=0.26\)
\(\Rightarrow\) Lateral strain \(=0.26 \times\) longitudinal strain
\(\begin{aligned}
& =0.26 \times \frac{\Delta l}{l} \\
& =0.26 \times \frac{3 \times 10^{-3}}{3}=0.26 \times 10^{-3}
\end{aligned}\)
Increased length, \(\Delta l=0.3 \mathrm{~cm}=3 \times 10^{-3} \mathrm{~m}\)
Poisson's ratio \(=0.26\)
\(\Rightarrow \quad \frac{\text { Lateral strain }}{\text { Longitudinal strain }}=0.26\)
\(\Rightarrow\) Lateral strain \(=0.26 \times\) longitudinal strain
\(\begin{aligned}
& =0.26 \times \frac{\Delta l}{l} \\
& =0.26 \times \frac{3 \times 10^{-3}}{3}=0.26 \times 10^{-3}
\end{aligned}\)
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.