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The hydraulic head that would produce a quick condition in a sand stratum of thickness 1.5 m,specific gravity 2.67 and voids ratio 0.67 is equal to
- a)1.0m
- b)1.5m
- c)2.0m
- d)3m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The hydraulic head that would produce a quick condition in a sand stra...
hydraulic gradient= (G-1)/(1+e)
G=2.67,e=0.67
we get, hydraulic gradient=1
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The hydraulic head that would produce a quick condition in a sand stra...
Given data:
Thickness of sand stratum (h) = 1.5 m
Specific gravity (G) = 2.67
Voids ratio (e) = 0.67
We need to determine the hydraulic head (H) that would produce a quick condition in the sand stratum.
Quick condition occurs in a soil when the upward seepage velocity is equal to the critical seepage velocity. The critical seepage velocity is given by:
Vc = (k/γ) [(1+e) ln(1+e)-e]
where k is the hydraulic conductivity, γ is the unit weight of water, and e is the voids ratio.
For a quick condition, the hydraulic gradient (i) should be equal to or greater than the critical hydraulic gradient (ic) given by:
ic = Vc/H
where H is the hydraulic head.
From the given data, we can calculate the hydraulic conductivity (k) using the relation:
k = q/n
where q is the discharge per unit area and n is the coefficient of permeability.
Since no values are given for q and n, we cannot calculate k.
However, we can use the fact that a quick condition occurs when the hydraulic gradient is equal to the critical hydraulic gradient. For a sand stratum, the critical hydraulic gradient is approximately 1.0. Therefore, to achieve a quick condition, the hydraulic gradient should be 1.0 or greater.
The hydraulic gradient is given by:
i = H/h
where h is the thickness of the sand stratum.
Substituting the given values, we get:
i = H/h = H/1.5
To achieve a quick condition, i should be 1.0 or greater. Therefore,
H/1.5 ≥ 1.0
H ≥ 1.5
Therefore, the hydraulic head that would produce a quick condition in the sand stratum is 1.5 m, which is option (B).
Thickness of sand stratum (h) = 1.5 m
Specific gravity (G) = 2.67
Voids ratio (e) = 0.67
We need to determine the hydraulic head (H) that would produce a quick condition in the sand stratum.
Quick condition occurs in a soil when the upward seepage velocity is equal to the critical seepage velocity. The critical seepage velocity is given by:
Vc = (k/γ) [(1+e) ln(1+e)-e]
where k is the hydraulic conductivity, γ is the unit weight of water, and e is the voids ratio.
For a quick condition, the hydraulic gradient (i) should be equal to or greater than the critical hydraulic gradient (ic) given by:
ic = Vc/H
where H is the hydraulic head.
From the given data, we can calculate the hydraulic conductivity (k) using the relation:
k = q/n
where q is the discharge per unit area and n is the coefficient of permeability.
Since no values are given for q and n, we cannot calculate k.
However, we can use the fact that a quick condition occurs when the hydraulic gradient is equal to the critical hydraulic gradient. For a sand stratum, the critical hydraulic gradient is approximately 1.0. Therefore, to achieve a quick condition, the hydraulic gradient should be 1.0 or greater.
The hydraulic gradient is given by:
i = H/h
where h is the thickness of the sand stratum.
Substituting the given values, we get:
i = H/h = H/1.5
To achieve a quick condition, i should be 1.0 or greater. Therefore,
H/1.5 ≥ 1.0
H ≥ 1.5
Therefore, the hydraulic head that would produce a quick condition in the sand stratum is 1.5 m, which is option (B).
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The hydraulic head that would produce a quick condition in a sand stra...
Correct option must be A i.e. 1
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The hydraulic head that would produce a quick condition in a sand stratum of thickness 1.5 m,specific gravity 2.67 and voids ratio 0.67 is equal toa)1.0mb)1.5mc)2.0md)3mCorrect answer is option 'B'. Can you explain this answer?
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