The hydro static pressure P of a liquid column depends on density d of the liquid column h and acceleration due to gravity g. Using dimensional analysis suggest a formula for hydrodynamic pressure.

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Solution

Let $P = k d^{a} h^{b} g^{c}$ where $k =$ dimensionless proportionality constant

$[P] = M L^{- 1} T^{- 2}$
$[d] = M L^{- 3}$
$[h] = L$
$[g] = L T^{- 2}$

Hence, On dimensional analysis-

$M L^{- 1} T^{- 2} = (M L^{- 3})^{a} L^{b} (L T^{- 2})^{c} = M^{a} L^{- 3 a + b + c} T^{- 2 c}$

Hence,

$a = 1$
$- 3 a + b + c = - 1$
$- 2 c = - 2$

Hence, on solving-

$a = 1, b = 1, c = 1$

Thus, $P = k d g h$

On calculation $k$ has found to be unity for $S I$ units.

$P = d g h$

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