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A 20 Kg load is suspended by a wire of crosssection 0.4 mm2. The stress produced in N/m2ISis

Answer»

0.4mm^2=0.4*10^-6m^2hence stress=mg/A=20*10/0.4*10^-6=200/4*10^-7=50*10^7N/m¢^2



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