The figure below shows an infinite-slope model. If \(\tau\) and SS denote the driving shear stress and the resisting shear strength, respectively, then \(\tau\) along the failure plane is
\(\times 10^3\) kg/ms\(^2\) (round off to two decimal places).
[Use: \(d = 2\) m; \(\rho_{soil} = 2.0\) g/cm\(^3\); \(g = 9.8\) m/s\(^2\)]
