Q1: A soil sample has passing 0.075 mm sieve = 60% liquid limit = 65% and plastic limit = 40%. The group index of the soil, is

ANS:D - none of these.

GI= { 0. 2 a + 0. 005 ac + 0. 01 bd } ------> eq. (1).

A = % of soil passing through 0. 075 sieve - 35. {limit 1 < a < 40}.
A = 60 - 35 = 25 (ok within limit).

B = % of soil passing through 0. 075 sieve - 15. {limit 1 < b < 40}.
B = 60 - 15 = 45 (It's over limit) so we will take = 40.

C = Liquid limit - 40. {limit 1 < c < 20}.
C = 65 - 40 = 25 (It's over limit) so we will take = 20.

D = plasticity index - 10. {limit 1 < d < 20}.
{Plasticity index = Liquid limit - Plastic limit}.
D = (65 - 40) - 10 = 15 (ok within limit).

Putting the value of a, b, c, d in eq. (1).

GI = 0. 2 a + 0. 005 ac + 0. 01 bd.
GI = (0. 2 x 25) + (0. 005 x 25 x 20) + (0. 01 x 40 x 15).
GI = 13. 5.
So, The answer will be 13. 5.
And None of these is the correct one. GI = 0.2a + 0.005ac + 0.01bd.
a = F-35 = 60-35 = 25.

Which is less than 40 so a=25
b = F-15=60-15 = 45 which is greater than 40 so take b=40
c = LL-40=65-40 = 25 which must not be greater than 20 so take c=20
d = Ip-10=(LL-PL)-10 = (65-40)-10=15
GI = 0.2*25 + 0.005*25*20 + 0.01*40*15,
= 5 + 2.5 + 6 = 13.5. GI = .2 * a +.005* a*c +.01*b*d.

a = P (% passing from 75-micron sieve ) - 40 (value not more than 40.. if more than 40 then take 40).

a= 60-35= 25 ok not more than 40.
b= P (% passing from 75 micron sieve ) - 15 (value not more than 40.. if more than 40 then take 40),
b= 60 - 15 =45 ( not ok ) (take 40).

c= wL (liquid limit ) -40 (value not more than 20.. if more than 20 then take 20),
c= 65- 40 = 25 (not ok ) ( take 20).

d= IP ( plasiticity index ) - 10,
d=( WL liquid limit( 65) - (40)WP plastic limit )- 10,
= 25-10 = 15 ok value less than 20.

So , G.I = .2*25 +.005*25 * 25 + .01*40 * 15,
and 14.125 is the correct answer.