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When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is $0.01 \times 10^{-3}\;m$. The Poisson's ratio was found to be 0.4. If the area of cross- section of wire is $0.025\;m^2$, its young's modulus is

\[\begin {array} {1 1} (a)\;1.6 \times 10^8 N/m^2 & \quad (b)\;2.5 \times 10^{10} N/m^2 \\ (c)\;1.25 \times 10^{11} N/m^2 & \quad (d)\;16 \times 10^9 N/m^2 \end {array}\]
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$ (a)\;1.6 \times 10^8 N/m^2$
answered Nov 7, 2013 by pady_1

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