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
35%
50%
65%
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
123 comments Page 2 of 13.

Niva said:   5 years ago
Liquid limit-water content by soil with small shear strength.
(3)

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%.
(12)

Anandhu Satheesan said:   5 years ago
(25-10)÷(100-25) = (X-70) ÷ (20-X),
After interpolate X = 61.66%.

Majid said:   5 years ago
=70-50/(log100-log10)*(log25-log10).

~~ 50%.

Raysz said:   6 years ago
(70/20)/(10-100) = (70-W)/(10-25).
W = 61.67.

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%.

Biniam said:   6 years ago
Using interpolation the liquid limit at 25 no of blow is 61.67%.

Hardik said:   6 years ago
How to find 4^0.12 without a calculator?
(1)

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%.
(2)

Prasantambc said:   6 years ago
70% correspond.

Linear rules:- 70-(((70-20)/(10-100))*(10-25)),
=70-8.33,
= 62.67.


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