The natural moisture content of a fully saturated clay is 32.3%. The specific gravity of soil is 2.65. Calculate the percentage decrease to be expected in the unit volume of clay when its moisture content is reduced by evaporation to 16.4%. The shrinkage limit of clay is 23%.

Volume of soil at natural condition.

$V_{1}$ = $V_{S}$ + $V_{w1}$

w=32 3%

$W_{W1}$ =0.323 $W_{S}$ $=$ $V_{w}=\frac{0.323 W_{s}}{\gamma_{w}}=\frac{0.323 W_{s}}{1}$ $\left[\gamma_{w}=1 \mathrm{~g} / \mathrm{cc}\right]$ Thus $$ V_{1}=V_{s}+0.323 W_{s} $$ Similarly, volume of soil at shrinkage limit $$ V_{d}=V_{s}+0.23 W_{s} $$ If water content of soil is reduced to $16.4 \%$, it is below the shrinkage limit. Hence the volume of soil at this water content will remain same as shrinkage limit. $$ \begin{aligned} \therefore V_{2} &=V_{d} \\ \text { % Volume reduction } &=\frac{V_{1}-V_{2}}{V_{1}} \times 100 \\ &=\frac{\left(V_{s}+0.323 W_{s}\right)-\left(V_{s}+0.23 W_{s}\right)}{V_{s}+0.323 W_{s}} \times 100 \\ &=\frac{9.3 W_{s}}{V_{s}+0.323 W_{s}}=\frac{9.3 W_{s}}{W_{s}\left(\frac{V_{s}}{W_{s}}+0.323\right)} \\ &=\frac{9.3}{\frac{V_{s}}{W_{s}}+0.323}=\frac{9.3}{\left(\frac{V_{s}}{V_{s} \gamma_{s}}\right)+0.323} \\ &=\frac{9.3}{\left(\frac{1}{G \gamma_{w}}\right)+0.323}=\frac{9.3}{\left(\frac{1}{2.65 \times 1}+0.323\right)}=13.28 \% \quad\left[\because G=\frac{\gamma_{s}}{\gamma_{w}}\right] \end{aligned} $$