Correct Answer - Option 1 : 617.5 kPa

__Concept:__

Using the relation

\({\sigma _1} = {\sigma _3}{\tan ^2}\left( {45^\circ + \frac{\phi }{2}} \right) + 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)\)

Where,

σ1 = maximum principal stress

σ3 = all round cell pressure

\({\sigma _1} = {\sigma _d} + {\sigma _3}\)

σd = deviator stress at failure

ϕ = Angle of internal friction of the soil

c = cohesion of the soil

__Calculation:__

ϕ = 36°, C = 12 kPa

σ3 = 200 kN/m2

σ1 = σ3 + σd

σ1 = 200 + σd

\({\sigma _1} = 200{\tan ^2}\left( {45^\circ + \frac{{36^\circ }}{2}} \right) + 2\times 12\ \times \tan \left( {45^\circ + \frac{36^\circ }{2}} \right)\)

σ1 = 817.47 kN/m2

**Deviator stress at failure**

σd = 817.47 – 200

**σd = 617.47 kN/m2**