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.
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)
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)
Aryan said:
2 years ago
[{Wt of sample/(difference of big weight)} *(G-1)/G] - 1.
1. wt of sample + pycnometer + water.
2. Wt of water + pycnometer.
Water content = [{wt of sample/(difference of both weights)*((G-1)/G)} - 1].
WC= [{400/(2150-1950)*((2.5-1)/2.5) } - 1].
= {(400/200)*1.5/2.5} - 1.
6/5 - 1 = 0.2.
Since WC in percentage; So, 0.2*100= 20%.
1. wt of sample + pycnometer + water.
2. Wt of water + pycnometer.
Water content = [{wt of sample/(difference of both weights)*((G-1)/G)} - 1].
WC= [{400/(2150-1950)*((2.5-1)/2.5) } - 1].
= {(400/200)*1.5/2.5} - 1.
6/5 - 1 = 0.2.
Since WC in percentage; So, 0.2*100= 20%.
(11)
.Ali Adnan Khalaf said:
5 years ago
W1 =w of pycnometer empty.
W2 = w of pycnometer +soil of the sample.
W3 = w of soil+water+pyconometer.
W4 = w of water+pyconometer.(full of water).
Assume the weight of pycnometer is 100 because , its CONSTANT in w1 w2 w3 and w4
So, w1=100 and w2=100+400 = 500,
w4 and w3 is given;
wc ={(w2-w1)/(w3-w4)*(Gs-1)/Gs}-1.
wc = 20%.
W2 = w of pycnometer +soil of the sample.
W3 = w of soil+water+pyconometer.
W4 = w of water+pyconometer.(full of water).
Assume the weight of pycnometer is 100 because , its CONSTANT in w1 w2 w3 and w4
So, w1=100 and w2=100+400 = 500,
w4 and w3 is given;
wc ={(w2-w1)/(w3-w4)*(Gs-1)/Gs}-1.
wc = 20%.
(7)
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.
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.
Gaurav Panth said:
4 years ago
Let us assume that the weight of the empty pycnometer is 100 gm, i.e M1=100 gm.
According to the question, the weight of sand is 400 gm then, M2=100 + 400 = 500 gm.
Also, M3= 2150 gm and M4=1950 gm.
wc ={(M2-M1)/(M3-M4)*(Gs-1)/Gs}-1.
= 1.2 - 1,
= 0.2 ,
wc = 20%.
According to the question, the weight of sand is 400 gm then, M2=100 + 400 = 500 gm.
Also, M3= 2150 gm and M4=1950 gm.
wc ={(M2-M1)/(M3-M4)*(Gs-1)/Gs}-1.
= 1.2 - 1,
= 0.2 ,
wc = 20%.
(8)
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)
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)
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.
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