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 6 of 13.
Alekhya said:
8 years ago
Water content at 25 blows gives you the liquid limit.
Just interpolate the given values for 25 blows. You get the answer as 65.
Just interpolate the given values for 25 blows. You get the answer as 65.
Brijesh Singh Satyal said:
8 years ago
Flow index = -(W2-W1) / ln(N1/N2).
From the given data, it comes out to be 50.
From the given data, it comes out to be 50.
Amit Kumar said:
8 years ago
Plot it on semilog graph and find water contentbfor 25 blow.
Madhur said:
8 years ago
Answer is B, why are you taking the difference between 100 and 10 values?
Bikashbiru said:
8 years ago
Yes, I agree @Ramesh.
Sasi said:
8 years ago
Why we say liquid limit at 25 blows? What is the reason?
Naresh said:
8 years ago
I think aprox 65 is correct.
Hatem Alhamaidi said:
8 years ago
Answer is:
LL = 61.7 % by interpolation.
LL = 61.7 % by interpolation.
Hemant maher said:
8 years ago
W1=10, W2=20, N1=70, N2=100.
So,
I = (w2-w1)/log(n2/n1).
I = 64.51= 65%.
So,
I = (w2-w1)/log(n2/n1).
I = 64.51= 65%.
Tomin Nyodu said:
8 years ago
@Hemant Mahar.
Method 1
I = (w1-w2) /log(n2/n1)
I= (70-20)/log(100/10)
I= 50
Hence, the ans is option (b)
Method 2
W1=70(10/25)^0.121=62.65
W2=20(100/25)^0.121=23.65
W1-W2=38.99 (Wrong)
Why? This is an empherical Formula and the value of the exponent is between 0.068 to 1.210.
Method 1
I = (w1-w2) /log(n2/n1)
I= (70-20)/log(100/10)
I= 50
Hence, the ans is option (b)
Method 2
W1=70(10/25)^0.121=62.65
W2=20(100/25)^0.121=23.65
W1-W2=38.99 (Wrong)
Why? This is an empherical Formula and the value of the exponent is between 0.068 to 1.210.
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