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A tangential force 2100 N is applied on a surface area `3xx10^(-6)m^(2)` which is 0.1 m from a fixed face. If the force produces a shift of `7xx10^(-3)` m of the upper surface with respect to the bottom. Then the modulus of rigidity of the material will be,A. `10^(8)N//m^(2)`B. `10^(9)N//m^(2)`C. `10^(10)N//m^(2)`D. `10^(11)N//m^(2)` |
Answer» Correct Answer - C `h=(F)/(A theta)=(Fh)/(Ax) = (2100 xx 0.1)/(3xx10^(-6)xx7xx10^(-3))` `=10^(10)N//m^(2)` |
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