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.
Dennis flora said:
4 years ago
No need to find weight of pycnometer, they are given weight of sand directly so we apply 400 for w2-w1.
(1)
Nikhil Dongre said:
5 years ago
Wc = (w2 - w1)/(w2 -w1) - (w3 - w4).
(1)
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)
M.bhavya said:
1 decade ago
w = ((w3-w4)/(w2-w1)*(1/G))-1.
(1)
Ashok Kumar said:
1 year ago
Thank you everyone for giving an explanation of the answer.
Mpyangu Francis said:
4 years ago
Well explained and understood very well. Thanks, everyone.
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.
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.
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.
Santosha kousalya said:
6 years ago
2 * 0.6 - 1 = 0.2 then how it becomes 20%?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers