### Discussion :: Soil Mechanics and Foundation Engineering - Section 1 (Q.No.1)

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Ashirwad said: (Apr 5, 2013) | |

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. |

Pratima Yadav said: (May 21, 2013) | |

Liquid Limit= W * (N/25)^0.121. where, W= water content at N blows. N= no.of blows. |

Apoorv said: (Jul 4, 2013) | |

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. |

Ravi Verma said: (Jul 24, 2013) | |

Just simply apply interpolation with 25 blows. |

Ashish Waghmare said: (Sep 13, 2013) | |

STEP1: just draw the rough graph on which, no. of blows on X-axis and moisture content on Y-axis. 80%. M C 70%. .A(10,70%) O O 60%. I N 50%. S T 40%. T E 30%. U N 20%. .B(100,20%) R T 10%. E(%)0% . . . . . . . . . . . . 0 10 20 30 40 50 60 70 80 90 100 110 NO. OF BLOWS STEP2: Join the coordinates A and B. STEP3: draw the vertical line from 25 no. of blows to the line AB and mark it. STEP4: Draw horizontal line from the marked point on line AB To the pt. on Y-axis, that same will give us the liquid limit of soil sample(ie. 65%). (NOTE: ALL STEPS ARE TAKEN FROM EXPERIMENT OF DETERMINATION OF LIQUID LIMIT BY USING LIQUID LIMIT DEVICE) |

Sagar Shah said: (Oct 11, 2013) | |

If we use flow index than the answer coming is 50. And plot shall be along log scale wouldn't rough idea from that be wrong? |

Sharda Charan said: (Jan 18, 2014) | |

Given; n1=10; w1=70%. n2=100; w2=20%. Flow index (I)=(w1-w2)/log(n2/n1) = 50;. wL = Ilog(n2/25)+w2;. wL = 50.10%. |

V Ramesh said: (Feb 4, 2014) | |

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%. |

Pappu said: (Feb 22, 2014) | |

Liquid limit is define as water content at which the soil possesses an arbitrarily fixed small amount of shear strength. |

Mudassir Abbas said: (Jun 1, 2014) | |

Just interpolate it on semi log graph paper. i.e No.of blows on logarithmic x-axis and m.c on y-axis and then find LL along 25 blows. |

Ashok said: (Jul 23, 2014) | |

Please anyone explain the proper and simple definition of the liquid limit? |

Priya said: (Sep 1, 2014) | |

The arbitrary limit of water content that represents the boundary between the liquid & plastic state is called as liquid limit. |

Amar Deep Tiwari said: (Sep 1, 2014) | |

Liquid Limit = (20-70)/(100-10) = (LL-70)/(25-10). So, LL = 61.67%. |

Adsam Gideon said: (Sep 23, 2014) | |

I tried it again and again, I got 50% as answer some one please clarify. |

Butera said: (Oct 28, 2014) | |

How do you get 89.55%? |

Pankaj said: (Dec 13, 2014) | |

Try it with coordinate geometry. My answer is 63.33%. |

Vishal Singh said: (Mar 4, 2015) | |

As per graph it comes near 65% and as per flow index its value come 50.10%. Please anyone explain right one method. |

Ankit Joshi said: (Apr 8, 2015) | |

I have a question to @Mr. V. Ramesh that how 89.55 becomes 88 and 23.65 becomes 23? |

K S Chetan said: (Apr 30, 2015) | |

Hello @Mr.V Ramesh. How come it possible Liquid limit = 70*(10/25)^0121 = 89.55%? |

Vasundhara said: (May 19, 2015) | |

I got 61.65% any one explain how it is 65%? |

Dipak said: (Jul 14, 2015) | |

By interpolation wL = 61.67. By flow index method wL= 50. |

Rahul Singh said: (Jul 20, 2015) | |

Please clarify the answer. |

Swarnanka Das said: (Jul 27, 2015) | |

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. |

Mantu Kumar Patel said: (Aug 7, 2015) | |

We can solve it by one point method formula of liquid limit determination. W1=Wn (N/25) ^n where, value of n is 0.12. 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%. |

Aakash Karoo said: (Aug 14, 2015) | |

STEP 1: Just draw the rough graph on which, no. of blows on X-axis and moisture content on Y-axis. STEP 2: Join the coordinates A (20, 70) and B (100, 20). STEP 3 : Draw the vertical line from 25 no. of blows to the line AB and mark it. STEP 4 : Draw horizontal line from the marked point on line AB to the point on Y-axis, that same will give us the liquid limit of soil sample (i.e. 65%). OR, Liquid Limit = Wl1 = W1*(N2/25)^0.121. Wl2 = W2*(N1/25)^0.121. Where, W = Water content at N blows. N = No. of blows. Liquid limit = 70*(100/25)^0.121 = 82.78%. Liquid Limit = 20*(10/25)^0.121 = 17.90%. Required liquid limit = 82.78-17.90 = 64.87%. = 65.00%. |

Moha Cool said: (Aug 19, 2015) | |

@Aakash karoo. Your simple steps helped me get the right answer which is 65%, the graph part is actually much easier! Thanks. |

Masud said: (Oct 2, 2015) | |

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%. |

Masud said: (Oct 2, 2015) | |

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%. |

Ishita Bhatnagar said: (Oct 8, 2015) | |

First the graph between water content and the corresponding number of blows is NOT a straight line! when the points are plotted on a SEMI LOG graph, that is, water content versus log (N), where N = no. of blows required to close the 2 mm groove at that water content, this water content versus log (N) plot comes out to be a STRAIGHT LINE! Hence the answer according to flow index will be the same as got from the semi log graph using interpolation. Because remember interpolation is valid only as long as the two quanties have a linear relationship!. Hence answer should be 50.1%. As far as the one point method is concerned, this method is used only when only one measurement is given to you. |

Mayank said: (Oct 25, 2015) | |

Correct answer is 50.1. Simply use equation ll = -if(logx)+c. Two equation, two unknown, solve it you will get ll = 50.1. |

Masoom said: (Dec 16, 2015) | |

50.1 is the correct answer, if solved by semi-log graph method. And this method is proper method to solve. |

Sukriti Dan said: (Mar 22, 2016) | |

As per flow index method I get 50.10. By the interpolation I get 61.6. But LIQUID LIMIT = Wn (N / 25)^n, I get 65. |

Sravya Iiit Basara said: (Mar 27, 2016) | |

No, the answer is wrong. |

Ammar said: (Jul 4, 2016) | |

Wrong answer. By linear interpolation, for blows 25 find L L. L. L = 61%. |

Rahul R K said: (Jul 9, 2016) | |

Why all of them are taking the difference between liquid limit for 10 and 100 Nos of blows? |

Ganesh Jadhav said: (Jul 26, 2016) | |

I think the one point method is not valid rather not necessary because two points are given for example. |

Shubham said: (Oct 17, 2016) | |

Why the graph is a straight line only? |

Maharnur Ajit Prakash said: (Oct 25, 2016) | |

Please tell me, why we have to give 25 blows for to find the liquid limit of soil? |

Pushkar Goyal said: (Oct 28, 2016) | |

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? |

Akshat Man said: (Dec 4, 2016) | |

We can also use w(n/25)^0,21. |

Kumar said: (Dec 7, 2016) | |

How it will be 70 * (10/25)^.121 = 89.58? |

Varsha said: (Dec 21, 2016) | |

Interpolation technique is very simple & easily understand to solve this.. |

Pradeep said: (Jan 11, 2017) | |

Moisture content 10 blows 20%. Moisture content 100 blows 70%. Answer is 65%. |

Harendra said: (Jan 22, 2017) | |

According to me, it is 61.70%. |

M Badra said: (Jan 27, 2017) | |

By interpolating the value for 25 blows the moisture content will linearly decrease up to 61.7%. Hence my answer is 61.7. |

Ritesh said: (Feb 5, 2017) | |

According to me, By using the method of interpolation, my answer is 50.10. |

Lokesh said: (Feb 27, 2017) | |

61.7 is the correct answer, So it is option D. |

Siva said: (Mar 12, 2017) | |

Why should we take the liquid limit value at 25 blows only? If there any relation, please explain it. |

Suman Ghosh said: (Mar 14, 2017) | |

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. |

Harshdeep said: (Mar 30, 2017) | |

Simply take x= 0.068. and WL = Wn (N÷25)^x . WL = 70 (10÷25 )^0.068 = 68.77 approx 65. |

Alekhya said: (Apr 5, 2017) | |

Water content at 25 blows gives you the liquid limit. Just interpolate the given values for 25 blows. You get the answer as 65. |

Brijesh Singh Satyal said: (Apr 8, 2017) | |

Flow index = -(W2-W1) / ln(N1/N2). From the given data, it comes out to be 50. |

Amit Kumar said: (May 1, 2017) | |

Plot it on semilog graph and find water contentbfor 25 blow. |

Madhur said: (May 7, 2017) | |

Answer is B, why are you taking the difference between 100 and 10 values? |

Bikashbiru said: (May 15, 2017) | |

Yes, I agree @Ramesh. |

Sasi said: (Jun 1, 2017) | |

Why we say liquid limit at 25 blows? What is the reason? |

Naresh said: (Jul 26, 2017) | |

I think aprox 65 is correct. |

Hatem Alhamaidi said: (Aug 21, 2017) | |

Answer is: LL = 61.7 % by interpolation. |

Hemant Maher said: (Aug 25, 2017) | |

W1=10, W2=20, N1=70, N2=100. So, I = (w2-w1)/log(n2/n1). I = 64.51= 65%. |

Tomin Nyodu said: (Sep 3, 2017) | |

@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. |

Amit said: (Sep 6, 2017) | |

Answer is 35. |

Deepak said: (Sep 13, 2017) | |

Thanks for your answers. |

Dhanraj Khatik said: (Sep 16, 2017) | |

Use interpolation. y-0.70=[(0.20-0.70)/(100-10)]*(25-10). y=62% ans. |

Surya Dev said: (Sep 20, 2017) | |

By interpolation, we can find the liquid limit corresponding 25 no of blows do not for the gate to take no of blows on lag scale. |

Sumeet Kulkarni said: (Sep 26, 2017) | |

@Pratima Yadav. The formula you put is right but followed that formula we supposed to get 39 (62.56-23.56) this option was not given. |

Shubham Suroliya said: (Oct 23, 2017) | |

By taking interpolation, Find value at 25 blows. And answer is 61.66%. |

Prasenjit said: (Nov 1, 2017) | |

(W1-w2)log (n2/n1). (70-20) log 100/10. 50 per Answer. |

Garry said: (Nov 28, 2017) | |

Answer is 50%. |

Narayan said: (Dec 12, 2017) | |

How do you get 89.55 by this formula? = W*(N/25)^0.121. |

Narendra Katara said: (Dec 16, 2017) | |

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. |

Ranjan said: (Dec 16, 2017) | |

50 % is the correct answer, interpolation is a wrong method. |

Mudassir said: (Dec 18, 2017) | |

You are right @Sharda Charan. |

Mohan Dev said: (Jan 12, 2018) | |

Which one correct (50%) or (65%)? |

Ashu said: (Jan 12, 2018) | |

It is 61%. |

Kranthi said: (Jan 28, 2018) | |

50% is correct. |

Mea said: (Feb 8, 2018) | |

65% is the right answer. Just draw the flow curve graph roughly. You will get it easily. |

Loki said: (Mar 22, 2018) | |

50.10 is correct and one point method is not applicable because it is valid for 20 to30 blows not for 100 blows. |

Akshay M Raut said: (Apr 4, 2018) | |

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? |

Abhay said: (Apr 5, 2018) | |

You cannot use interpolation because it is semilog graph. |

Ohoud said: (Apr 25, 2018) | |

LL= water content at 20 blow. 100-10/20-70= 100-20/20-LL, LL= 64.44. |

Yadav said: (Jun 11, 2018) | |

Yes, the Correct answer is 50%. |

Maharshi Saxena said: (Jun 18, 2018) | |

The correct Answer is 50%. |

Kasthuri said: (Jul 20, 2018) | |

By interpolation with 25 we get the correct answer as 61.67. |

Aita said: (Aug 4, 2018) | |

Liquid limit(1)=70*(100÷25)^.121=82.78. Liquid limit (2)=20*(10÷25)^.121=17.90. Total =82.78-17.90 =64.88%. Say 65%. |

Vicki said: (Sep 3, 2018) | |

L.L = 70%-50%(0.4) = 50%. |

Maha said: (Oct 10, 2018) | |

Thanks for the answer @Aakash Karoo. |

Nagendra said: (Oct 29, 2018) | |

@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. |

Shubham V said: (Nov 3, 2018) | |

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. |

Ganesh Kawal said: (Jan 2, 2019) | |

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. |

Ashwani Bhandari said: (Jan 22, 2019) | |

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. |

Navya said: (Mar 8, 2019) | |

Interpolation method is correct method for liquid limit at 25 blows. So the answer is 50%. |

Anoni said: (Mar 10, 2019) | |

Thank you @Akash. |

Rahi said: (May 11, 2019) | |

70x(10/25)^0.121 = 62.65378306. Am I right? Thanks. |

Khushi said: (Jun 9, 2019) | |

How come 0 .121? Please explain. |

Abdul Quader said: (Jun 11, 2019) | |

N1=10, W1=70% & N2=100, W2=20%. WL1 = W1*(N1/25)^0.121 = 70*(10/25)^0.121 = 62.65%, WL2 = W2*(N2/25)^0.121 = 20*(100/25)^0.121 = 23.65, Required Liquid Limit= WL1 - WL2. = 62.65 - 23.65, = 39%. So, Answer (D) None of These. |

Kishan said: (Jun 27, 2019) | |

The correct answer is 50%, because we can't apply interpolation to a semilog graph, so need to solve using flow index. |

Shubham said: (Jul 8, 2019) | |

By interpolation, the correct answer is 50%. |

Faruque Abdullah said: (Jul 23, 2019) | |

tanβ = 0.121 is for the range of 20 to 30 blows. As the blows number is out of this range this value should be adjusted for the individual case. For 20 to 30 blows, tanβ = 20/(30^1.5) = 0.121. Case-I: tanβ = 10/(25^1.5)=0.08. w(LL) = 70 x (10/25)^0.08 = 65.05 which is tends to 65%. Case-II: tanβ = 100/(25^1.5) = 0.8. w(LL) = 20 x (100/25)^0.08 = 60.62 which is close to 65%. |

Kicktony said: (Aug 2, 2019) | |

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. |

Rahul said: (Aug 15, 2019) | |

Go flow index or similarities of traingle and the answer is 50.1. |

Chandu said: (Aug 20, 2019) | |

The correct Answer is 50.1. |

Avnit said: (Aug 28, 2019) | |

Flow index = (70-20)/log(100/10) = 50. 50 = (W-20)/log(100/25). Therefore W = 50. |

Md. Suruzzaman said: (Sep 24, 2019) | |

The water content at which soil changes from liquid limit to plastic limit is called liquid limit. |

Prasantambc said: (Sep 25, 2019) | |

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

Pintu Kumar(Govt.Poly Saharsa,Bihar said: (Oct 2, 2019) | |

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%. |

Hardik said: (Oct 12, 2019) | |

How to find 4^0.12 without a calculator? |

Biniam said: (Oct 23, 2019) | |

Using interpolation the liquid limit at 25 no of blow is 61.67%. |

Asutosh said: (Feb 23, 2020) | |

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%. |

Raysz said: (Feb 23, 2020) | |

(70/20)/(10-100) = (70-W)/(10-25). W = 61.67. |

Majid said: (Jun 7, 2020) | |

=70-50/(log100-log10)*(log25-log10). ~~ 50%. |

Anandhu Satheesan said: (Aug 2, 2020) | |

(25-10)÷(100-25) = (X-70) ÷ (20-X), After interpolate X = 61.66%. |

Tanmoy Das said: (Sep 26, 2020) | |

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%. |

Niva said: (Nov 3, 2020) | |

Liquid limit-water content by soil with small shear strength. |

Sneha Mathai said: (Jan 2, 2021) | |

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. |

Hareram Sah said: (Feb 13, 2021) | |

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. |

Manjusha Shinde said: (Mar 22, 2021) | |

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%. |

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