Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 50)
50.
The weight of a pycnometer containing 400 g sand and water full to the top is 2150 g. The weight of pycnometer full of clean water is 1950 g. If specific gravity of the soil is 2.5, the water content is
Discussion:
27 comments Page 1 of 3.
Ankush said:
1 decade ago
W1 = 200g.
W2 = 600g.
W3 = 2150g.
W4 = 1950g.
G=2.5.
w = (((W3-W4)/W2-W1))*(1-1/G))-1.
Solve and you will get the answer.
W2 = 600g.
W3 = 2150g.
W4 = 1950g.
G=2.5.
w = (((W3-W4)/W2-W1))*(1-1/G))-1.
Solve and you will get the answer.
M.bhavya said:
1 decade ago
w = ((w3-w4)/(w2-w1)*(1/G))-1.
(1)
Deepanshu said:
1 decade ago
W = (((w2-w1)/(w3-w4))*((G-1)/G)))-1.
W = {((600-200)/(2150-1950)*((2.5-1)/2.5)))-1.
W = 2*0.6-1.
W = .2*100.
= 20.
W = {((600-200)/(2150-1950)*((2.5-1)/2.5)))-1.
W = 2*0.6-1.
W = .2*100.
= 20.
Hema said:
1 decade ago
How w1 = 200g? Where you get this value?
AFZAL said:
1 decade ago
@Hema,
It is mentioned in the question that sand is 400gm so he has just assumed that what of pycnometer + sand is 600gm, so automatically you get what of pycnometer as 200gm.
It is mentioned in the question that sand is 400gm so he has just assumed that what of pycnometer + sand is 600gm, so automatically you get what of pycnometer as 200gm.
Anand Engineer said:
1 decade ago
Wt. of Pycnometer + Wt. of Sand + Wt. of Water = 2150 g........(1).
Wt. of Pycnometer + Wt. of Water = 1950 g.......(2).
Subtracting (1) from (2) we get = 200 g.....Wt. of empty Pycnometer.
Therefore, W1 = 200 g, W2 = W1(200 g) + Wt. of Sand(400 g) = 600 g
W3 = 2150 g, W4 = 1950 g.
Replace this values in the above mentioned equation and you get the answer.
Wt. of Pycnometer + Wt. of Water = 1950 g.......(2).
Subtracting (1) from (2) we get = 200 g.....Wt. of empty Pycnometer.
Therefore, W1 = 200 g, W2 = W1(200 g) + Wt. of Sand(400 g) = 600 g
W3 = 2150 g, W4 = 1950 g.
Replace this values in the above mentioned equation and you get the answer.
(1)
KRISHNA said:
1 decade ago
(w3-w4)/(w2-w2)*(1/G).
Suresh.veerla said:
9 years ago
As it is mention in question the weight of a pycnometer containing 400 g sand.
So,
W2 - W1 = 400.
W = ((w2 - w1) / (w3 - w4)) * ((G - 1)/G) - 1.
= ((400)/(2150 - 1950)) * ((2.5 - 1)/2.5) - 1.
= 0.2.
= 20%.
So,
W2 - W1 = 400.
W = ((w2 - w1) / (w3 - w4)) * ((G - 1)/G) - 1.
= ((400)/(2150 - 1950)) * ((2.5 - 1)/2.5) - 1.
= 0.2.
= 20%.
(2)
KRISHAN DUTT YADAV said:
9 years ago
Water content w = ( (M1 - M2) / (M3 - M4) ) ( (G - 1) /G) - 1,
Where M1 - M2 = 400GM,
M3 = 2150,
M4 = 1950.
Where M1 - M2 = 400GM,
M3 = 2150,
M4 = 1950.
Anomi said:
8 years ago
Wt. of Pycnometer + Wt. of Sand + Wt. of Water = 2150 g --->(1).
Wt. of Pycnometer + Wt. of Water = 1950 g --->(2).
Subtracting (1) from (2) we get = 200 g.....Wt. of empty Pycnometer.
Therefore, W1 = 200 g, W2 = W1(200 g) + Wt. of Sand(400 g) = 600 g.
W3 = 2150 g, W4 = 1950 g.
Replace this values in the above-mentioned equation and you get the answer.
Wt. of Pycnometer + Wt. of Water = 1950 g --->(2).
Subtracting (1) from (2) we get = 200 g.....Wt. of empty Pycnometer.
Therefore, W1 = 200 g, W2 = W1(200 g) + Wt. of Sand(400 g) = 600 g.
W3 = 2150 g, W4 = 1950 g.
Replace this values in the above-mentioned equation and you get the answer.
(1)
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