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
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)`
5 views
1 Answers
Correct Answer - C
`h=(F)/(A theta)=(Fh)/(Ax) = (2100 xx 0.1)/(3xx10^(-6)xx7xx10^(-3))`
`=10^(10)N//m^(2)`
5 views
Answered