Question

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

Answer 1

Hint: The question is trying to confuse you by giving information about hydrostatic pressure and asking about hydrodynamic pressure, both will be calculated the exact same way as it’s only pressure. To solve, first put pressure proportional to power of the given factors and then remove the sign of proportionality using a dimensionless constant and then try to equate the powers of the dimensions of quantities to find the answer.

Complete step by step answer:
We will proceed with the solution exactly as described in the hint section of the solution to the question.
The question has told us that the hydrostatic pressure $P$ of a liquid column depends on density $d$ of the liquid column $h$ and acceleration due to gravity $g$ . We can write it mathematically as:
$P \propto {d^a}{h^b}{g^c}$
Now, first let us list the units and dimensions of each of the quantities in the above equation:
Unit of pressure is: $N/m$
Dimension of pressure is: $\left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]$
Unit of density is: $kg/{m^3}$
Dimensions of density is: $\left[ {M{L^{ - 3}}} \right]$
Unit of height is: $m$
Dimensions of height is: $\left[ L \right]$
Unit of acceleration due to gravity is: $m/{s^2}$
Dimensions of acceleration due to gravity is: $\left[ {L{T^{ - 2}}} \right]$
Now, let us remove the proportionality sign using a dimensionless constant as:
$P = k \times {d^a}{h^b}{g^c}$
Now, let us apply the dimension analysis method on the equation by only considering the dimensions of the quantities:

$\left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] = {\left[ {M{L^{ - 3}}} \right]^a}{\left[ L \right]^b}{\left[ {L{T^{ - 2}}} \right]^c}$
Upon solving, we get:
$\implies$ $\left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] = \left[ {{M^a}{L^{ - 3a + b + c}}{T^{ - 2c}}} \right]$
Upon equating the powers of the dimensions $\left[ M \right]$ , we get the following equation:
$\implies$ $1 = a$
Hence, we get the value of the power of density as:
$\implies$ $a = 1$
Now, equating the powers of the dimension $\left[ T \right]$ , we get the following equation:
$\implies$ $ - 2c = - 2$
Upon solving, we get the value of the power of acceleration due to gravity in the equation as:
$\implies$ $c = 1$
Finally, equating the powers of the dimension $\left[ L \right]$ , we get the following equation:
$\implies$ $ - 3a + b + c = - 1$
But we already know the value of $a,\,c$ , substituting them in the last equation, we get:
$\implies$ $b = 1$
So, we got the values of the powers as:
$\implies$ $a = 1$
$\implies$ $b = 1$
$\implies$ $c = 1$
Substituting them in the main equation, we get:
$\implies$ $P = kdhg$
Now, to find the value of the dimensionless proportionality constant, $k$ , we need to put the values of each of the quantities as its SI unit. After doing that, we get:
$\implies$ $k = 1$
Hence, the formula for the hydrodynamic pressure is given as:
$\implies$ $P = dhg$

Note: A major mistake in such questions is to not consider a proportionality constant while removing the proportionality sign, or directly developing the equation between the quantities without using the proportionality concept. The value of the proportionality constant is found experimentally.