Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 19)
19.
A soil sample has passing 0.075 mm sieve = 60% liquid limit = 65% and plastic limit = 40%. The group index of the soil, is
Discussion:
48 comments Page 1 of 5.
Deep said:
1 decade ago
F = fineness constant.
L.l = liquid limit.
P.l = plastic limit.
Group index = (F-35)[0.2+(0.005)(Ll-40)]+0.01(F-15)(Pl-10).
G I = (60-35)[.2+(0.005)(65-40)]+0.01(60-15)(40-10).
Group index = 21.625.
L.l = liquid limit.
P.l = plastic limit.
Group index = (F-35)[0.2+(0.005)(Ll-40)]+0.01(F-15)(Pl-10).
G I = (60-35)[.2+(0.005)(65-40)]+0.01(60-15)(40-10).
Group index = 21.625.
Vivek said:
1 decade ago
Pl-10 = Plasticity index-10 = 25-10 = 15.
GI = 14.875.
GI = 14.875.
Harshitha said:
10 years ago
Which answer is correct. Can I know how to solve?
Gokul said:
9 years ago
GI = 0.2a + 0.005ac + 0.01bd.
a = P - 35 = 60 - 35 = 35.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PL - 10 = 40 - 10 = 30.
GI = (0.2 * 35) + (0.05 * 35 * 25) + (0.01 * 45 * 30) = 24.875.
Therefore the correct option is D.
a = P - 35 = 60 - 35 = 35.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PL - 10 = 40 - 10 = 30.
GI = (0.2 * 35) + (0.05 * 35 * 25) + (0.01 * 45 * 30) = 24.875.
Therefore the correct option is D.
RANJAN PATRA,BBSR said:
9 years ago
GI = 0.2a + 0.005ac + 0.01bd.
a = P - 35 = 60 - 35 = 35.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 25 - 10 = 15.
GI = (0.2 * 35) + (0.05 * 35 * 25) + (0.01 * 45 * 15) = 18.125
Therefore the correct option is D.
a = P - 35 = 60 - 35 = 35.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 25 - 10 = 15.
GI = (0.2 * 35) + (0.05 * 35 * 25) + (0.01 * 45 * 15) = 18.125
Therefore the correct option is D.
Vishu said:
9 years ago
The correct answer is 37.
Roop Narayan (nit surat) said:
9 years ago
GI = 0.2a + 0.005ac + 0.01bd.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125 ans.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125 ans.
Roop Narayan (NIT SURAT) said:
9 years ago
GI = 0.2 a + 0.005 ac + 0.01 bd.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125.
Therefore the correct answer is (D)
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125.
Therefore the correct answer is (D)
GiRi said:
9 years ago
GI = (F-35) (0.2 + 0.05(wL - 40)) + 0.01(F - 15)(Ip - 10).
Where,
Ip = wL - Wp.
We get answer is 12.
Where,
Ip = wL - Wp.
We get answer is 12.
SAMKIT said:
9 years ago
The correct method is;
A = p-35 shud not be greater than 40,
B = p-15 shud not be greater than 40,
C = LL-40 shud not be greater than 20,
D = Ip-10 shud not be greater than 20,
Put in GI = 0.2a + 0.01bd + 0.005ac.
We get GI = 0.2*25+0.01 * 40 * 15 + 0.005 * 25 * 25.
= 14.125.
A = p-35 shud not be greater than 40,
B = p-15 shud not be greater than 40,
C = LL-40 shud not be greater than 20,
D = Ip-10 shud not be greater than 20,
Put in GI = 0.2a + 0.01bd + 0.005ac.
We get GI = 0.2*25+0.01 * 40 * 15 + 0.005 * 25 * 25.
= 14.125.
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