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 2 of 3.
Himanshu Dhar said:
5 years ago
@Anand.
How you could get the wt of empty container by subtracting W4 from W3? please mention.
W3 = Wt of pycnometer + Wt Of sand + wt of water to full height.
So W3 - W4 = Wt Of solids.
How you could get the wt of empty container by subtracting W4 from W3? please mention.
W3 = Wt of pycnometer + Wt Of sand + wt of water to full height.
So W3 - W4 = Wt Of solids.
(2)
Angad@@@ said:
6 years ago
How we say that Wt of Picnomecter is 600 gram in 2150-1950 we get 200 which is wt of sand.
(2)
Ranku Biswas said:
6 years ago
W1 =wt of pycnometer.
W2=wt of pycnometer +soil.
W3= wt of soil+water+pyconometer.
W4=wt of water+pyconometer.
Wt of soil (W2-W1) = 400g.
No need to get pycnometer wt.
Wt of soil+pyconometer + water = 2150g.
Wt of pyconometer + water = 1950g,
Water content ={ (W2 - W1/W3 - W4) *(G-1)/G} - 1.
W2=wt of pycnometer +soil.
W3= wt of soil+water+pyconometer.
W4=wt of water+pyconometer.
Wt of soil (W2-W1) = 400g.
No need to get pycnometer wt.
Wt of soil+pyconometer + water = 2150g.
Wt of pyconometer + water = 1950g,
Water content ={ (W2 - W1/W3 - W4) *(G-1)/G} - 1.
Ranku Biswas said:
6 years ago
Weight of (sand + pyconometer + water) - Weight of (pyconometer +water)
After subtraction what we get is the weight of pycnometer.
After subtraction what we get is the weight of pycnometer.
Santosha kousalya said:
6 years ago
2 * 0.6 - 1 = 0.2 then how it becomes 20%?
Fev said:
7 years ago
How about its void ratio and degree of saturation of its bulk density is 2.05 g/cc?
APURBA MUKHERJEE said:
8 years ago
Water content w = ( (M2 - M1) / (M3 - M4) ) ( (G - 1) /G) - 1,
M= weight of sand = 400 gm.
Where, M= M2 - M1
M2= Wt. Of pycnometer + Wt. Of Sand
M1= Wt. Of pycnometer.
M3 = 2150 gm. and
M4 = 1950 gm.
G= 2.5.
Put into the formula, in above mentioned
Then water content w = 20%
Thanks.
M= weight of sand = 400 gm.
Where, M= M2 - M1
M2= Wt. Of pycnometer + Wt. Of Sand
M1= Wt. Of pycnometer.
M3 = 2150 gm. and
M4 = 1950 gm.
G= 2.5.
Put into the formula, in above mentioned
Then water content w = 20%
Thanks.
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)
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.
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)
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