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A tangential force of 2100 N is applied...

A tangential force of 2100 N is applied on a surface of area `3xx10^(-6)m^(2)` which is 0.1 m from a fixed face. The force produces a shift of 7 mm of upper surface with respect to bottom. Calculate the modulus of rigidity of the material.

Text Solution

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`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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