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2 answers:
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Answer:
Explanation:
xx=33 vvalues
Answer:
Given that
LHS of above given equation have dimension .
Now find the dimension of RHS
Dimension of P = .
Dimension of d= .
Dimension of μ = .
Dimension of L= .
So
It means that both sides have same dimensions.
Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e = =
Specific gravity =
% Saturation S = ×
=
×
water content w = =
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w =
Recall; mass = Density × volume
w =
From eqn. (2) =
∴
putting eqn. (6) into (5)
w =
Again, from eqn (1)
substituting into eqn. (7)
∴
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given,
∴
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil ( )
Given, , S = 50% = 0.5
%S = ×
=
×
0.50 =
mass of water =