JEE Class main Answered

Let us derive a relation for absolute pressure P at a point A in liquid

P = P_o + p 

where P_o 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 ρ^a g^b h^c ..........................(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 = M^a 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 = P_o + ( ρ × g × h )