JEE Class main Answered

Let us derive a relation for absolute pressure P at a point A in liquid
P = Po + p
where Po is atmospheric pressure and p is hydrostatic pressure at the point A
Hydrostatic pressure p depends on three parameters namely
(i) height h of liquid column above point A
(ii) density ρ of liquid
(iii) acceleration due to gravity g
Hence we write the relation for hydrostatic pressure as
p = k ρa gb hc ..........................(1)
where k, a, b and c are constant numbers to be determined .
LHS of eqn.(1) is pressure :- Force/ area = ( mass × acceleration ) / area
Dimension of LHS = ( M × L T-2 ) / L2 = M L-1 T-2
Dimension of density ρ = mass / volume = M L-3
Dimension of acceleration due to gravity g = L T-2
Dimension of height h = L
Hence if we equate the dimension on both sides of eqn.(1), then we get
M L-1 T-2 = [ M L-3 ]a [ L T-2 ]b [ L ]c
M L-1 T-2 = Ma L-3a+b+c T-2b
By equating powers , we get
a = 1
-3a+b+c = -1
-2b = -2
By solving above expressions , we get , a= b = c = 1
Hence from eqn.(1), we get hydrostatic pressure p as
p = k × ρ × g × h
From experiments it can be determined that k = 1
Hence , we get , p = ρ × g × h
Absolute pressure P at a point in liquid is
P = Po + ( ρ × g × h )