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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 xx 10^(-3) . 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 |
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Answer» `1.6xx10^8 N m^(-2)` Longitudinal strain = `"Lateral strain"/"Poisson.s ratio"` `= (0.01xx10^(-3))/0.4` …(i) Young.s modulus , `Y="Normal stress"/"Longitudinal strain"` (Using (i)) `Y=F/(AXX((0.01xx10^(-3))/0.4))` `=(100xx0.4)/(0.025xx0.01xx10^(-3))N m^(-2) = 1.6xx10^8 N m^(-2)` |
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