The hydraulic head that would produce a quick condition in a sand stratum of thickness 1.5 m, specific gravity 2.67 and voids ratio 0.67 is equal to

i=(G-1)/(1+e)
i=(2.67-1)/(1+0.67)
i=1.67/1.67
i=1
we know i = H/L
H/L=1
H=L
H= 1.5m

ie=G-1/1+e
Specific gravity G=2.67
Voids ratio e=0.67
ie=2.67-1/1+067=1.67/1.67=1
Hydraulic gradient=i=h/L
1=h/1.5
1*1.5=h, h=1.5

h/L =G-1/1+e

B

Yes c option is correct

Critical hydraulic gradient,
(ic)=(G-1)/(1+e)
=(2.67-1)/(1+0.67)=1
We know,
Hydraulic gradient,
i=h/L
=>1=h/1.5
=>h=1.5m

soil has bulky densityof 22kN/m and water content 10% the dry density of the soil is
15KN/m
20KN/m
30KN/m
35KN/m

h/l=g-1/1+e
h/1.5=2.67-1/1+0.67
h/1.5=1
h=1.5

h/l=g-1/1+e
h/1.5=2.67-1/1+0.67
h/1.5=1
h=1.5

Critical hydraulic gradient(i)=(G-1)/(1+e)=1 again i=h/L so here i=1,L=1.5 so h=1.5m

i/h=(G-1)/(1+e)

wrong answer its 1m.
i=(G-1)/(1+e)

What was the formula.?