Correct Answer - Option 3 : 25 kN/m

^{2}
**Concept:**

We know,

As per triaxial stress, **the relationship between principal stress and cohesion** -

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

For clayey soil subjected unconfined triaxial test, ϕ = 0° and σ_{3} = 0 kN/m^{2}

From (1)

σ_{1} = 2c = q_{u}

⇒** **c = **\(\frac{q_u}{2}\)** ...(2)

Here,

c - Cohesion (kN/m^{2})

q_{u} - Unconfined compressive strength (kN/m^{2})

σ_{3} - Confining stress (kN/m^{2})

σ_{1} - Major principal stress (kN/m^{2})

**Calculation:**

**Given,**

q_{u} = 50 kN/m^{2}

From (2)

c = \(\frac{50}{2}\) = 25 kN/m^{2}

We know,

**As per columb's theory, shear stress is given by** -

τ = c + \(\bar{σ}\) tanϕ

For clayey soil, ϕ = 0°

⇒ **τ = c = 25 kN/m**^{2}