Answer is given below
Explanation:
given data
weight = 6N
temp = 105° C
weight reduced = 5N
solution
weight of container is 1N
SO W = (6-1) = 5
And Wd = 5 - 1 = 4
so
moisture content is
moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex] .......1
moisture content = [tex]\frac{5-4}{4} \times 100[/tex]
moisture content = 25%
and
as we know density of soil soild = 2700 kg/m³
density of water = 1000 kg/m³
and sp gravity of soil = [tex]\frac{2700}{1000}[/tex] = 2.7
so
now we get here bulk unit weight
bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex] ..........2
bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]
bulk unit wt = 19.766 KN/m³
and
so dry unit wt will be
dry unit wt = [tex]\frac{Ysat}{1+w}[/tex] ..............3
dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]
dry unit wt = 15.813 kN/m³