Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  EAMCET  >>  Physics
0 votes

A tension of $20 \;N$ is applied to a copper wire of cross sectional area $0.01\;cm^2$, Young's Modulus of copper is $1.1 \times 10^{11} \;N/m^2$ and Poisson's ratio is $0.32$. The decrease in cross sectional area of the wire is

\[(a)\;1.16 \times 10^{-6}\;cm^2 \quad (b)\;1.16 \times 10^{-5}\;m^2 \quad (c)\;1.16 \times 10^{-4}\;m^2 \quad (d)\;1.16 \times 10^{-3}\;cm^2 \]
Can you answer this question?

1 Answer

0 votes
$(a)\;1.16 \times 10^{-6}\;cm^2$
answered Nov 7, 2013 by pady_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App