[tex]79 f t^{3}[/tex] is the volume of the sample when the water content is 10%.
Explanation:
Given Data:
[tex]V_{1}=100\ \mathrm{ft}^{3}[/tex]
First has a natural water content of 25% = [tex]\frac{25}{100}[/tex] = 0.25
Shrinkage limit, [tex]w_{1}=12 \%=\frac{12}{100}=0.12[/tex]
[tex]G_{s}=2.70[/tex]
We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
[tex]V \propto[1+e][/tex]
[tex]\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}[/tex] ------> eq 1
[tex]e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}[/tex]
The above equation is at [tex]S_{r}=1[/tex],
[tex]e_{1}=w_{1} \times G_{s}[/tex]
Applying the given values, we get
[tex]e_{1}=0.25 \times 2.70=0.675[/tex]
Shrinkage limit is lowest water content
[tex]e_{2}=w_{2} \times G_{s}[/tex]
Applying the given values, we get
[tex]e_{2}=0.12 \times 2.70=0.324[/tex]
Applying the found values in eq 1, we get
[tex]\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904[/tex]
[tex]V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}[/tex]
