A vertical rectangular plane surface is submerged in water such that its top and bottom surfaces are 1.5 m and 6.0 m res-pectively below the free surface. The position of center of pressure below the free surface will be at a distance of

h=(d^2/12*x)+x, where x is center of gravity from the surface of fluid
=4.5^2/(12*3.75)+3.75
=(20.25)/45 + 3.75
=4.2

h=1.5+6-1.5/2
h*=3.75+b×(4.5)^3/12×b×4.5×3.75=4.2m

h=1.5+6-1.5/2=3.75
h*=3.75+b×4.5^3/12×b×4.5×3.75=4.2m

Center of pressure for rectangle = 2* h/3 = 2*6/3 = 4

center of pressure= (b x 4.5² )/(12x 3.75x4.5) + 3.75
= 4.2

H°=(lg/A*h) +h'
Ig=1*4.5^3/12=7.59
A=1*4.5=4.5
h'=1.5+4.5/2=3.75
h=4.5
Put all values
H°=(7.59/4.5*4.5) +3.75=4.12

jafar & mahfuzar both are wrong....
soln...
centre of pressure= c.g+(moi/(area×c.g) )
considering per metre breadth....
moi=(1/12)×1×4.5^3=7.59
cg=1.5+(6-1.5)/2 =3.75
so ....
cp = 3.75+(7.59/(4.5×3.75) )
=4.2m

h = 6-1.5 =3.5 m
* hc = 3.5 + 1.5/2 + 1.5/4 = 4.12 m

1.5+ (6-1.5)/2= 3.75

How ?

Solution for the sum