Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 1)
1.
In a liquid limit test, the moisture content at 10 blows was 70% and that at 100 blows was 20%. The liquid limit of the soil, is
Discussion:
123 comments Page 3 of 13.
Md. Suruzzaman said:
6 years ago
The water content at which soil changes from liquid limit to plastic limit is called liquid limit.
Avnit said:
6 years ago
Flow index = (70-20)/log(100/10) = 50.
50 = (W-20)/log(100/25).
Therefore W = 50.
50 = (W-20)/log(100/25).
Therefore W = 50.
Chandu said:
6 years ago
The correct Answer is 50.1.
Rahul said:
6 years ago
Go flow index or similarities of traingle and the answer is 50.1.
KickTony said:
6 years ago
Do iteration, with N=25 blows=x.
(x1,y1) = (10,70),
(x2,y2) = (100,20),
(x,y) = (25,LL).
Substitute the values in below equation.
(y2-y1)/(x2-x1)=(y-y1)/(x-x1).
LL=65.
(x1,y1) = (10,70),
(x2,y2) = (100,20),
(x,y) = (25,LL).
Substitute the values in below equation.
(y2-y1)/(x2-x1)=(y-y1)/(x-x1).
LL=65.
Faruque Abdullah said:
6 years ago
tanβ = 0.121 is for the range of 20 to 30 blows.
As the blows number is out of this range this value should be adjusted for the individual case.
For 20 to 30 blows, tanβ = 20/(30^1.5) = 0.121.
Case-I:
tanβ = 10/(25^1.5)=0.08.
w(LL) = 70 x (10/25)^0.08 = 65.05 which is tends to 65%.
Case-II:
tanβ = 100/(25^1.5) = 0.8.
w(LL) = 20 x (100/25)^0.08 = 60.62 which is close to 65%.
As the blows number is out of this range this value should be adjusted for the individual case.
For 20 to 30 blows, tanβ = 20/(30^1.5) = 0.121.
Case-I:
tanβ = 10/(25^1.5)=0.08.
w(LL) = 70 x (10/25)^0.08 = 65.05 which is tends to 65%.
Case-II:
tanβ = 100/(25^1.5) = 0.8.
w(LL) = 20 x (100/25)^0.08 = 60.62 which is close to 65%.
Shubham said:
6 years ago
By interpolation, the correct answer is 50%.
Kishan said:
6 years ago
The correct answer is 50%, because we can't apply interpolation to a semilog graph, so need to solve using flow index.
Abdul Quader said:
6 years ago
N1=10, W1=70% & N2=100, W2=20%.
WL1 = W1*(N1/25)^0.121 = 70*(10/25)^0.121 = 62.65%,
WL2 = W2*(N2/25)^0.121 = 20*(100/25)^0.121 = 23.65,
Required Liquid Limit= WL1 - WL2.
= 62.65 - 23.65,
= 39%.
So, Answer (D) None of These.
WL1 = W1*(N1/25)^0.121 = 70*(10/25)^0.121 = 62.65%,
WL2 = W2*(N2/25)^0.121 = 20*(100/25)^0.121 = 23.65,
Required Liquid Limit= WL1 - WL2.
= 62.65 - 23.65,
= 39%.
So, Answer (D) None of These.
(1)
Khushi said:
6 years ago
How come 0 .121? Please explain.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers