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.
Ashok Kumar said:
1 year ago
Thank you everyone for giving an explanation of the answer.
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)
MILI said:
3 years ago
@Anand.
You are right, thanks for explaining.
You are right, thanks for explaining.
(3)
Tanmoy Karmakar said:
4 years ago
@Dorna.
Well explained, thanks.
Well explained, thanks.
(1)
Dorna said:
4 years ago
@Ranku, @Suresh.
You are correct w2-w1= 400 is what we need, no need to find a separate weight of the pycnometer.
You are correct w2-w1= 400 is what we need, no need to find a separate weight of the pycnometer.
(2)
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)
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)
Mpyangu Francis said:
4 years ago
Well explained and understood very well. Thanks, everyone.
Nikhil Dongre said:
5 years ago
Wc = (w2 - w1)/(w2 -w1) - (w3 - w4).
(1)
.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)
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