SSC JE Civil Engineering 23-Mar-2021 Shift 1 Paper

Concept:
Dry Unit Weight $\left(\mathrm{\gamma }\mathrm{d}\right):$
It is a measure of the weight of solid particles per unit volume.
${\mathrm{\gamma }}_{\mathrm{d}}={\mathrm{W}}_{\mathrm{s}}∕\mathrm{V}$
Where ${\mathrm{W}}_{\mathrm{s}}=$ weight of solid particles, $\mathrm{V}=$ Volume of solid particles
Now the relationship between the specific gravity of solid particles, void ratio, and dry unit weight
${\mathrm{\gamma }}_{\mathrm{d}}=\frac{\mathrm{G}{\mathrm{\gamma }}_{\mathit{w}}}{\left(1+\mathit{e}\right)}$
where G = specific gravity of solid particles, e = void ratio
Calculation:
GivenData:
The mass of a chunk of moist soil = 20 kg
The dry density of soil $\left({\mathrm{\gamma }}_{\mathrm{d}}\right)=1500\mathrm{kg}∕{\mathrm{m}}^3$
Specific gravity $\left(\mathrm{G}\right)=2.50$
The density of water $\left({\mathrm{\gamma }}_{\mathit{w}}\right)=1000\mathrm{kg}∕{\mathrm{m}}^3$
Using the relationship between the specific gravity of solid particles, void ratio, and dry unit weight
${\mathrm{\gamma }}_{\mathrm{d}}=\frac{\mathrm{G}{\mathrm{\gamma }}_{\mathrm{w}}}{\left(1+\mathrm{e}\right)}$
$1500=\frac{2.50\times 1000}{\left(1+\mathrm{e}\right)}$
$\left(1+\mathrm{e}\right)=\frac{2.50\times 1000}{1500}$
e = 0.6667
The void ratio is 66.67%