Correct Answer - Option 1 : 43.3 kPa

**Concept:**

**Unconfined Compression test:**

(i) It is a special case of the triaxial test in which confining pressure is zero. It means only deviator or shear stress is applied.

(ii) The axial stress at failure is called unconfined compressive strength since confining pressure σ_{3} = 0

hence, σ_{1} = σ_{3} + σ_{d} = σ_{d}

Using,

σ_{1} = σ_{3} tan^{2}(45° + ϕ/2) + 2c tan(45° + ϕ/2)

**σ1 = 2c tan(45° + ϕ/2)**

**Calculation:**

Given,

axial stress (σ_{d}) = σ_{1} = 150 kPa

θ = 45° + ϕ/2 = 60°

∵ σ1 = 2c tan(45° + ϕ/2)

150 = 2c tan 60°

c = 43.33 kPa