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A `0.05 m` cube has its upper face displaced by `0.2 cm` by a tangential force of `8N`. Calculate the shearing strain, shearing stress and modulus of rigidity of the material of the cube. |
Answer» `L=5 cm =5 xx 10^(-2) m, Delta L=0.2 cm =0.2 xx 10^(-2) m`, `F=8N` , Shearing strain `=(Delta L)/(L) = (0.2)/(5) = 0.04`, Shearing stress = `(F)/(L xx L) = (8)/((5 xx 10^(-2))^(2))` `= 3200 N//m^(2)` Modulus of rigidity, `G = ("shearing stress")/("shearing strain") =(3200)/(0.04)` `=80000 N//m^(2)` `=8 xx 10^(4)N//m^(2)`. |
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