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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

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

1 Answer

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

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