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A metal cube of side 0.20 m is subjected to a shearing force of 4000 N. The top surface is displaced through 0.50 cm with respect to the bottom. Calculate the shear modulus of elasticity of the metal. |
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Answer» Solution :Here , `L = 0.20 m, F = 4000 N, X = 0.50 cm = 0.005m and "Area" A = L^2 = 0.04 m^2` THEREFORE, Shear modulus `eta_(R ) = F/A xx L/x = (4000)/(0.4) xx (0.20)/(0.005) = 4 xx 10^(6) N m^(-2)`. |
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