Correct Answer - Option 4 : 59.59 kN/m

^{2}
**Concept:**

**Relationship between C and principal stresses at failure:**

Mohr coulomb failure criteria can be expressed in terms of the relationship between the principal stresses σ_{1} and σ_{3}

\({σ _1} = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{ϕ }{2}} \right) + 2C\tan \left( {{{45}^o} + \frac{ϕ }{2}} \right)\)

Where, σ_{1} = Total principal stress, σ_{3} = Lateral princiapl stress

C = Cohesion, ϕ = angle of shearing or friction angle

**Calculation:**

Given,

C = 35 kN/m^{2}, σ_{1} = 300 kN/m^{2}, ϕ = 30°

∵ We know that, \({σ _1} = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{ϕ }{2}} \right) + 2C\tan \left( {{{45}^o} + \frac{ϕ }{2}} \right)\)

⇒ \(300 = {σ _3}{\tan ^2}\left( {{{45}^o} + \frac{{{{30}^o}}}{2}} \right) + 2 \times 35 \times \tan \left( {{{45}^o} + \frac{{{{30}^o}}}{2}} \right)\)

σ_{3} = 59.585 kN/m^{2}