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A tangentialforce of 2100 N is appliedon a surfaceof area 3xx10^(-6)m^(2) which is0.1 m from a fixed face. The force producesa shift of 7 mm of upper surface with respectto bottom. Calculatethe modulusof rigidity of the material. |
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Answer» Solution :`F=2100N, A=3XX10^(-6) m^(2)`, `L=0.1m, DELTAX=7xx10^(-3)m`. `G=(FL)/( A Deltax)=(2100xx0.1)/(3xx10^(-6)xx7xx10^(-3))=1xx10^(10)NM^(-2)` |
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