Answer:
246.3%
the complete question is found in the attached document
Step-by-step explanation:
1st step:
using 1936 data,
w1= 356% = 3.56(356/100) , H= 79 feet
specific gravity = 2.65
Sₓ= 100%= 1
initial void ratio(e₀)= (w1 x specific gravity)/Sₓ
=3.56 x 2.65/1 = 9.434
2nd step
using 1996 data
ΔH= 22ft
ΔH/H = Δe/(1 + e₀)
22/79 = Δe/(1+9.434)
0.278=Δe/10.434
Δe= 0.278 x 10.434
Δe= 2.905
Δe= e₀ - eₓ
eₓ= e₀-Δe
eₓ= 9.434 - 2.905
eₓ= 6.529
3rd step
calculating water content in 1996
eₓ =6.529, specific gravity= 2.65, Sₓ= 100%
W2 X 2.65 = 1 x 6.529
w2 = 6.529/2.65 = 2.463 = 246.3%
Following are the calculating to the given question with the steps:
Step 1:
Analysis of the 1936 data,
[tex]\to w_1= 356\% = (\frac{356}{100})= 3.56\\\\ \to H= 79\ ft\\\\\to G_s= 2.65\\\\\to S_r= 100\%= \frac{100}{100}= 1\\\\[/tex]
Calculating the initial void ratio
[tex]= (\frac{w_1 \times G_s}{S_r}) \\\\ =\frac{3.56 \times 2.65}{1}\\\\= 9.434\\[/tex]
Step:2
Analysis of the 1996 data:
[tex]\to \Delta H= 22\ ft\\\\\therefore \\\\ \to \frac{\Delta H}{H} = \frac{\Delta e}{(1 + e_0)}\\\\\to \frac{22}{79} = \frac{\Delta e}{(1+9.434)}\\\\\to \Delta e =2.9057\\\\\therefore \\\\\to e_0=e_f=\Delta e\\\\\to 9.434-e_f=2.9057\\\\\to e_f=6.5283\\\\[/tex]
Step 3:
Calculating water content in 1996:
[tex]\to e_f =6.5283\\\\\to G_s=2.65\\\\ \to S_r= 100\%=\frac{100}{100}= 1\\\\\therefore \\\\\to W_2 \times G_s =s_n \times e_f\\\\ \to w_2 \times 2.65 = 1 \times 6.5283\\\\\to w_2 = \frac{6.5283 \times 1}{2.65} = 2.4635 = 246.35\%\\\\[/tex]
Therefore, the final answer is "246.35%".
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