Given:-

ww=$w \times w_{5} $

$w=5%$=0.05

$\gamma=18.44 kn/m^{2}$

G=2.67

$W_{w1}=0.05 w_{s}$

$W_{w2}=0.15 W_{s}$

Quantity of water to be added =$0.15ws-0.05ws$

$\gamma_{d}$=$\frac{\gamma}{1+w}$=$\frac{18.44}{1+0.05}=17.56kn/m^{3}$

$W_{s}=\gamma_{d} \times \gamma$

For $1m^{3}$ volume.

$W_{s}=17.56 \times 1 =17.56 kN$

$W_{w}=0.1\times17.56=17.56kN$

$V_{w}$=$\frac{W_{w}}{\gamma_{w}}=\frac{1.756}{9.81}=0.179m^{3}$ i.e 179 litres.

e=$\frac{G \gamma m _{w}}{\gamma d}-1=\frac{2.67 \ast 9.81}{17.56}-1$

e=0.492

weight of sand filling hole …