# 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.
DaKiht said:
1 month 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%.

Yashasvi Vishwakarma said:
6 months 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%.

(3)

Vivek said:
2 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.

(20)

Neeraj Bhandari said:
2 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.

(10)

Haseeb said:
3 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%.

(15)

Dikesh don said:
3 years ago

The correct answer is B) 50%.

(10)

Noblen said:
3 years ago

50.10 is the correct answer.

It is not a triangle so don't use similar triangles. it is a logarithmic function. interpolate it using ln.

It is not a triangle so don't use similar triangles. it is a logarithmic function. interpolate it using ln.

(3)

Manjusha Shinde said:
3 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%.

(34)

Hareram sah said:
3 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.

(13)

Sneha Mathai said:
3 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.

(5)

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