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A load of 4.0kg is suspended from a ceiling through a steel wire of length 20m and radious 2.0mm it is found that the length of the wire increase by 0.031mm as equilibrium is achieved find young's modulus of steel take g=3.1πm/s² |
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Answer» Hii friend, # ANSWER- Y = 2 × 10^12 N/m2 # GIVEN- m=4kg l=20m dl=0.031mm=3.1×10-5m r=2mm=2×10^-3m # Formula- Youngs MODULUS is given by Y = stress/strain Y = (F/A)/(dl/l) Y = (mg/πr^2)/(dl/l) Y = (4×3.1π/π×4×10^-6)/(3.1×10-5/20) Y = 2 × 10^12 N/m2 Youngs modulus = 2 × 10^12 N/m2. Hope that was helpful... |
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