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.

Explanation

`F = 2100 N, A = 3 xx 10^(-6) m^2`
`L = 0.1 m , l = 7 xx 10^(-3) m`
`G = (FL)/(Al) = (2100 xx 0.1)/(3 xx 10^(-6) xx 7 xx 10^(-3)) = 1 xx 10^10 Nm^(-2)`