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
5%
10%
15%
20%
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
27 comments Page 3 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)

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%.
(8)

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.
(2)

Tanmoy Karmakar said:   4 years ago
@Dorna.

Well explained, thanks.
(1)

MILI said:   3 years ago
@Anand.

You are right, thanks for explaining.
(3)

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%.
(11)

Ashok Kumar said:   1 year ago
Thank you everyone for giving an explanation of the answer.


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