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Sagot :
Answer:
Follows are the solution to the given question:
Explanation:
Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]
solid soil volume =[tex]205 \ cc[/tex]
saturated mass soil = [tex]361 \ g[/tex]
The weight of the soil after drainage is =[tex]295 \ g[/tex]
Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]
Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]
Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]
[tex]= 361- 295 \\\\ = 66 \ cc[/tex]
Soil water retained volume = (draining field weight - dry soil weight)
[tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]
[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]
[tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]
(Its saturated water volume is equal to the volume of voids)
[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]
[tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]
[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]
[tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]
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