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The differemce b/w maximum and minimum void ratio of a sand sample is 0.30. if the relative density of this
sample is 66.6% at a void ratio of 0.40, then the void ratio of this sample at its lossest state will be
sample is 66.6% at a void ratio of 0.40, then the void ratio of this sample at its lossest state will be
Correct answer is between '0.50,0.64'. Can you explain this answer?
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The differemce b/w maximum and minimum void ratio of a sand sample is ...
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Calculation:
Given, maximum void ratio (e_max) - minimum void ratio (e_min) = 0.30
Relative density (Dr) = 66.6%
Void ratio at Dr = 0.40 (e1) = 0.40
From the relation between relative density, void ratio and maximum void ratio:
Dr = (emax - e)/(emax - emin) x 100
Substituting the given values, we get:
66.6 = (emax - 0.40)/(emax - e_min) x 100
Solving for emax - e_min, we get:
emax - e_min = 0.60
Adding e_min to both sides, we get:
emax = e_min + 0.60
Substituting this value in the given relation between relative density, void ratio and maximum void ratio, we get:
66.6 = (e_min + 0.60 - 0.40)/(e_min + 0.60 - e_min) x 100
Simplifying, we get:
66.6 = 60/(e_min + 0.60)
Solving for e_min, we get:
e_min = 0.82
Substituting this value in the equation emax = e_min + 0.60, we get:
emax = 1.42
Therefore, the void ratio of this sample at its lowest state will be between 0.50 and 0.64.
Given, maximum void ratio (e_max) - minimum void ratio (e_min) = 0.30
Relative density (Dr) = 66.6%
Void ratio at Dr = 0.40 (e1) = 0.40
From the relation between relative density, void ratio and maximum void ratio:
Dr = (emax - e)/(emax - emin) x 100
Substituting the given values, we get:
66.6 = (emax - 0.40)/(emax - e_min) x 100
Solving for emax - e_min, we get:
emax - e_min = 0.60
Adding e_min to both sides, we get:
emax = e_min + 0.60
Substituting this value in the given relation between relative density, void ratio and maximum void ratio, we get:
66.6 = (e_min + 0.60 - 0.40)/(e_min + 0.60 - e_min) x 100
Simplifying, we get:
66.6 = 60/(e_min + 0.60)
Solving for e_min, we get:
e_min = 0.82
Substituting this value in the equation emax = e_min + 0.60, we get:
emax = 1.42
Therefore, the void ratio of this sample at its lowest state will be between 0.50 and 0.64.
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The differemce b/w maximum and minimum void ratio of a sand sample is 0.30. if the relative density of thissample is 66.6% at a void ratio of 0.40, then the void ratio of this sample at its lossest state will beCorrect answer is between '0.50,0.64'. Can you explain this answer?
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