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 1 of 13.
Manjusha Shinde said:
4 years ago
LL = w*(N/25)^0.121.
LL = 70*(100/25)^0.121 = 82.78%.
LL = 20*(10/25)^0.121 = 17.90%.
LL = 83 - 18 = 65%.
LL = 70*(100/25)^0.121 = 82.78%.
LL = 20*(10/25)^0.121 = 17.90%.
LL = 83 - 18 = 65%.
(47)
Vivek said:
3 years ago
Using the formula of one point method.
WL= Wn(N/25)^0.1.
Where N should be between 15 to 35.
Using nearest value from data i.e. N=10 & Wn = 70.
WL = 0.70 * (10/25)^0.1 = 0.6387~0.65 ans.
WL= Wn(N/25)^0.1.
Where N should be between 15 to 35.
Using nearest value from data i.e. N=10 & Wn = 70.
WL = 0.70 * (10/25)^0.1 = 0.6387~0.65 ans.
(27)
Haseeb said:
4 years ago
The Correct Answer is 50%. Flow Index is 50. Flow Index = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
Now, using that Flow Index of 50 and the same formula to find moisture content at 25 No. of blows is 50%.
Flow Index = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50.1%.
Now, using that Flow Index of 50 and the same formula to find moisture content at 25 No. of blows is 50%.
Flow Index = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50.1%.
(19)
Neeraj Bhandari said:
4 years ago
According to me, the solution is;
10 -> 70%.
100 -> 20%.
25 -> ?.
Ans = [(70-20)/log (100/10)] * log25.
= 50*log25.
= 69.9.
10 -> 70%.
100 -> 20%.
25 -> ?.
Ans = [(70-20)/log (100/10)] * log25.
= 50*log25.
= 69.9.
(16)
Hareram sah said:
5 years ago
The formula of liquid limit =w* (N/25) ^0.121.
Find, LL for each condition and then subtract them. You will get 65% as answer.
Find, LL for each condition and then subtract them. You will get 65% as answer.
(14)
Tanmoy das said:
5 years ago
Liquid Limit = W*(N/25)^0.121.
Where,
W = water content at N blow.
N = number of the blow.
Liquid Limit = 70*(10/25)^0.121 =89.55%.
Liquid Limit = 20*(100/25)^ 0.121 = 23.65%.
Required Liquid Limit = ( 88-23) = 65%.
Where,
W = water content at N blow.
N = number of the blow.
Liquid Limit = 70*(10/25)^0.121 =89.55%.
Liquid Limit = 20*(100/25)^ 0.121 = 23.65%.
Required Liquid Limit = ( 88-23) = 65%.
(12)
Yashasvi Vishwakarma said:
2 years ago
Flow index = water content at 10 no of blows - water content at 100 no of blows.
So, flow index = W10-W100 = 70-20 = 50%.
So, flow index = W10-W100 = 70-20 = 50%.
(12)
Dikesh don said:
4 years ago
The correct answer is B) 50%.
(11)
DaKiht said:
1 year ago
@All.
Here, we don't use the Empirical formula i.e- WL=WP*(N/25)^x [where x = 0.1-0.3] the value of x is not given in the question.
The Main formula is preferred over the empirical.
Flow Index (If) = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
The same formula was used to find moisture content at 25 No. of blows is 50%.
If = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50%.
Here, we don't use the Empirical formula i.e- WL=WP*(N/25)^x [where x = 0.1-0.3] the value of x is not given in the question.
The Main formula is preferred over the empirical.
Flow Index (If) = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
The same formula was used to find moisture content at 25 No. of blows is 50%.
If = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50%.
(8)
Sneha Mathai said:
5 years ago
Take log(no of blows) only for finding flow index or for finding liquid limit using a flow index value.
If the graph is given and we want to find LL, then just do interpolate without taking the log of no of blows. Because we are plotting as such in semilog to get a straight line.
ie, (70-20)/(100-10) = (LL-20)/(100-25).
LL = 61.66.
If the graph is given and we want to find LL, then just do interpolate without taking the log of no of blows. Because we are plotting as such in semilog to get a straight line.
ie, (70-20)/(100-10) = (LL-20)/(100-25).
LL = 61.66.
(7)
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