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Numerical · Q44

Q.A steel wire having cross sectional area 1.2 mm² is stretched by a force of 120 N. If a lateral strain of 1.455×10⁻⁴ is produced in the wire, calculate the Poisson's ratio. [Given: Ysteel = 2×10¹¹ N/m²]

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Step 1. Given: A=1.2 mm2=1.2×10−6 m2A = 1.2\ \text{mm}^2 = 1.2\times10^{-6}\ \text{m}^2, F=120F = 120 N, lateral strain =1.455×10−4= 1.455\times10^{-4}, Y=2×1011 N/m2Y = 2\times10^{11}\ \text{N/m}^2.

Step 2. Longitudinal stress =F/A=120/(1.2×10−6)=1×108 N/m2= F/A = 120/(1.2\times10^{-6}) = 1\times10^{8}\ \text{N/m}^2.

Step 3. Longitudinal (linear) strain =stress/Y=(1×108)/(2×1011)=5×10−4= \text{stress}/Y = (1\times10^8)/(2\times10^{11}) = 5\times10^{-4}. …

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