Correct Answer - Option 4 : 19.44 KN/m

3
__Concepts:__

The relation between dry density, bulk density and water content is given as:

\(\gamma_d = \dfrac{\gamma_b}{1 +w}\)

Where symbols have the usual meaning

__Calculations:__

**Given: **γ_{b} = 21 kN/m^{3}; w = 8%

Using above relation:

\(\gamma_d = \dfrac{21}{1+0.08}\)

**∴ γ**_{d} = 19.44 kN/m^{3}

**Other Important relations: (Symbols have their usual meaning)**

1. W × G = S × e

2. \(\gamma_b = \dfrac{(G + Se)\gamma_w}{1 + e}\)

3. \(\gamma_d = \dfrac{G\gamma_w}{1 + e}\)

4. \(\rm porosity, n = \dfrac{e}{ 1 + e} \)

5. \(\gamma_s = \dfrac{(G + e)\gamma_w}{1 + e}\)