Question

The hydrostatic pressure ‘P’ of a liquid column depends upon the density, height ‘h’ of liquid column and also an acceleration ‘g’ due to gravity. Using dimensional analysis, derive a formula for pressure P.

Answer 1

Explanation :

It is given that the hydrostatic pressure depends on the height, density and acceleration due to gravity.

$P\propto \rho^x\ g^y\ h^z$

$P=k \rho^x\ g^y\ h^z$.......(1)

We have to derive the formula for pressure using dimensional analysis.

The dimensional formula of pressure is $[P]=[ML^{-1}T^{-2}]$

The dimensional formula of density is $[\rho]=[ML^{-3}]$

The dimensional formula of acceleration due to gravity is $[g]=[LT^{-2}]$

The dimensional formula of height is $[h]=[L]$

So, equation (1) becomes :

$[ML^{-1}T^{-2}]=k[ML^{-3}]^x[LT^{-2}]^y[L]^z$

$[ML^{-1}T^{-2}]=k[M]^xL^{-3x+y+z}T^{-2y}$

On equating both sides :

x = 1

y = 1

z = 1

Hydrostatic pressure becomes $P=\rho gh$.

Hence, this is the required solution.

Answer 2

"Given:

Hydrostatic Pressure: P

Density of fluid: D

Height of liquid column: H

Acceleration due to gravity: g

Solution:

Answer: P = DHg. (Hydrostatic pressure)

Initially, the dimensions of P, D, H and g in terms of basic units are:

$P = MLT^-2$

$D = ML^-3$

H = L

$g = LT^-2$

Relationship between hydrostatic pressure and the parameters can be established as follows:

$P = k Dx Hy gz$

Where, k is the constant.

Equating the dimensions:

$MLT^-2 = (ML^-3)x (L)y (LT^-2)z$

Comparing the powers of same parameters;

$x =1; z=1and y=1$

On substituting the values,

$P = kDHg$

If the proportionality constant is k=1;

Then,

Hydrostatic Pressure = DHg."