Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 5 (Q.No. 35)
35.
If the natural moisture content, the liquid limit and plastic limit of a soil sample are stated as 30.5%, 42.5% and 22.5% respectively, the ratio of liquidity index and plastic index, is
2
Discussion:
10 comments Page 1 of 1.
Shahid khan said:
4 years ago
In IL, the percentage cancels out automatically as both nominator and denominator both are in percentage but in PI we must have to decide by 100.
(2)
Jyoutirmoi Sarma said:
4 years ago
You are correct. @Rakesh.
The answer should be 0.02.
The answer should be 0.02.
Nil said:
4 years ago
Thanks @Abhay.
Rakesh Ojha said:
7 years ago
@Abhay.
Why did you take 20% as 0.2 and left 0.4 as it is?
The Right answer would be 0.02.
Why did you take 20% as 0.2 and left 0.4 as it is?
The Right answer would be 0.02.
Priya said:
7 years ago
.02 should be the answer. I also agree.
Aamir said:
8 years ago
.02 should be the answer.
Kalp said:
9 years ago
Nice explanation @Abhay.
Abhay said:
10 years ago
Plasticity Index (Ip) = liquid limit(Wl) - plastic limit(Wp).
Ip = 42.5 - 22.5 = 22% => 0.2.
Liquidity Index(Il) = {Natural water content(W)- Plastic limit}/Plasticity Index(Ip).
Il = {30.5-22.5}/20 = 0.4.
Ratio Il/Ip = 0.4/0.2 = 2.
Ip = 42.5 - 22.5 = 22% => 0.2.
Liquidity Index(Il) = {Natural water content(W)- Plastic limit}/Plasticity Index(Ip).
Il = {30.5-22.5}/20 = 0.4.
Ratio Il/Ip = 0.4/0.2 = 2.
(8)
Mayuri said:
10 years ago
IP = WL-WP = 42.5-22.5 = 20.
LI = wc-WP/IP = (30.5-22.5)/20 = 0.40*100 = 40.
IL/IP = 40/20 = 2.
LI = wc-WP/IP = (30.5-22.5)/20 = 0.40*100 = 40.
IL/IP = 40/20 = 2.
(3)
CH SWAMY said:
1 decade ago
Ip = Wl - Wp = 42.5 - 22.5 = 0.20.
Il = w - Wp / Ip = (30.5 - 22.5)/20 = 0.40.
Il/Ip = 0.4/0.2 = 2.0 answer - C.
Il = w - Wp / Ip = (30.5 - 22.5)/20 = 0.40.
Il/Ip = 0.4/0.2 = 2.0 answer - C.
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