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.
Pushkar Goyal said:
9 years ago
Since liquid limit is limit corresponding to 25 nos of blow. Hence by solving it through interpolation I am getting 61.66%. By which the answer should be D Option.
Am I right?
Am I right?
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.
Nagendra said:
7 years ago
@All.
The exact Answer was obtained by interpolation of no of blows in logarithmic format right, % of water content. Hence exact Answer is 50.107 So, Option B is Right.
The exact Answer was obtained by interpolation of no of blows in logarithmic format right, % of water content. Hence exact Answer is 50.107 So, Option B is Right.
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)
Pintu kumar(Govt.poly saharsa,bihar said:
6 years ago
Given,
n1=10, w1=70%.
n2=100, w2=20%.
liquid limit(1) = 70*(100÷25)^.121 = 82.78.
liquid limit(2) = 20*(10÷25)^.121 = 17.90.
Total = (82.78-17.90) = 65%.
Ans 65%.
n1=10, w1=70%.
n2=100, w2=20%.
liquid limit(1) = 70*(100÷25)^.121 = 82.78.
liquid limit(2) = 20*(10÷25)^.121 = 17.90.
Total = (82.78-17.90) = 65%.
Ans 65%.
(2)
Masud said:
10 years ago
W1 = 70, N1 = 10, W2 = 20, N2 = 100, W3 = ?, N3 = 25.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
Masud said:
10 years ago
W1 = 70, N1 = 10, W2 = 20, N2 = 100, W3 = ?, N3 = 25.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
Apoorv said:
1 decade ago
We can find the flow index 1st using.
I = (w1 - w2)/log(n2/n1) and then find the water content corresponding to the 25 blows using this flow index value.
I = (w1 - w2)/log(n2/n1) and then find the water content corresponding to the 25 blows using this flow index value.
Ashwani bhandari said:
7 years ago
Liquid limit (1) = 70 * (10/25)^0.121 = 62.65.
Liquid limit (2) =20 * (100/25)^0.121 = 23.65.
Required limit =62.65-23.65=39.
Option D is right I think.
Liquid limit (2) =20 * (100/25)^0.121 = 23.65.
Required limit =62.65-23.65=39.
Option D is right I think.
Ganesh kawal said:
7 years ago
Plot graph liquid limit vs no of the blow.
From graph.
(70-20) ÷ (100-10)=(LL-20)÷ (100-20),
Answer is 61.
Still we assume graph is linear.
From graph.
(70-20) ÷ (100-10)=(LL-20)÷ (100-20),
Answer is 61.
Still we assume graph is linear.
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