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 2 of 13.
SHUBHAM V said:
7 years ago
Here;
Water Content W1=70%,No. Of Blows N1=10.
Water Content W2=20%, No. Of Blows N2=100.
So For Liquid Limit,
W(L)=W(Y)(N/25)^1.21.
So For 1st 2nd Case Respectively,
W(L)1=70*(10/25)^1.21=23.09.
W(L)2=20*(100/25)^1.21=107.03.
So,
Liquid Limit=23.09+107.03/2=65.
Water Content W1=70%,No. Of Blows N1=10.
Water Content W2=20%, No. Of Blows N2=100.
So For Liquid Limit,
W(L)=W(Y)(N/25)^1.21.
So For 1st 2nd Case Respectively,
W(L)1=70*(10/25)^1.21=23.09.
W(L)2=20*(100/25)^1.21=107.03.
So,
Liquid Limit=23.09+107.03/2=65.
NARENDRA KATARA said:
8 years ago
Answer is 50.
For a particular soil, slope does not change in flow index curve to determine the liquid limit for a soil. In this question flow index is also 50 and put this 50 value equal to formula and find the liquid limit for given data.
For a particular soil, slope does not change in flow index curve to determine the liquid limit for a soil. In this question flow index is also 50 and put this 50 value equal to formula and find the liquid limit for given data.
Ashirwad said:
1 decade ago
Liquid limit is defined as the moisture content at which a groove made in a pat of soil will flow through a distance of 13mm under an impact of 25 blows. Using linear interpolation with the help of flow curve we can get the above.
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)
Suman Ghosh said:
8 years ago
Answer 50% is correct.
W = - I log N + C.
W = water content , I = flow index, N= no of blow.
Put value of w & N for both 70% & 20%, then results will be, I=50 & C=120. Now w= -50*log 25 + 120 = 50.1.
W = - I log N + C.
W = water content , I = flow index, N= no of blow.
Put value of w & N for both 70% & 20%, then results will be, I=50 & C=120. Now w= -50*log 25 + 120 = 50.1.
Asutosh said:
6 years ago
Liquid Limit= W * (N/25)^0.121.
Where,
W = Water content at N blows.
N = No.of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
Required liquid limit = 88-23 = 65%.
Where,
W = Water content at N blows.
N = No.of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
Required liquid limit = 88-23 = 65%.
V RAMESH said:
1 decade ago
Liquid Limit= W * (N/25)^0.121.
where,
W= water content at N blows.
N= no.of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
required liquid limit = 88-23 = 65%.
where,
W= water content at N blows.
N= no.of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
required liquid limit = 88-23 = 65%.
Akshay M Raut said:
7 years ago
Liquid limit WL is the one which is the water content at 25th blow. So simply interpolation is applied as;
Log 10 -> 0.7
Log 25 -> ?
Log 100 -> 0.2.
So I got answer 0.5....i.e. 50%.
Is that right?
Log 10 -> 0.7
Log 25 -> ?
Log 100 -> 0.2.
So I got answer 0.5....i.e. 50%.
Is that right?
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.
(26)
Swarnanka Das said:
1 decade ago
We can get it from flow index value, I.
n1 = 10; w1 = 70%.
n2 = 100; w2 = 20%.
Flow index (I) = (w1-w2)/log(n2/n1) = 50;
wL = Ilog (n2/25)+w2;
log(2) = 0.301.
wL = 50.10%.
OPTION B.
n1 = 10; w1 = 70%.
n2 = 100; w2 = 20%.
Flow index (I) = (w1-w2)/log(n2/n1) = 50;
wL = Ilog (n2/25)+w2;
log(2) = 0.301.
wL = 50.10%.
OPTION B.
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