Solveeit Logo
Login

Question

Question: Establish the relation \(P = h\rho g\) (derive dimensionally)....

Establish the relation P=hρgP = h\rho g (derive dimensionally).

Explanation

Solution

Dimensional analysis implies that we denote each physical quantity with respect to fundamental quantities. When the right-hand side of the equation dimensionally equals the left-hand side of the equation, we say the equation is consistent.

Complete step by step solution:
In the dimensional analysis, we represent each quantity with respect to other fundamental quantities. In this case, we consider fundamental quantities such as mass, length, and time.
As we can see, on the right-hand side, we have pressure. We know, the formula for pressure corresponds to:
P=FAP = \dfrac{F}{A}
where,
P=P = Pressure
F=F = Force exerted
A=A = Area
We know, the force can be written as:
F=maF = ma
F=F = Force
m=m = Mass
a=a = Acceleration
In dimensional analysis e write:
Mass, m=[M]m = [M]
Length, l=[L]l = [L]
Time, t=[T]t = [T]
Thus, dimensionally we write force is:
Force, F=[ML1T−2]F = \left[ {M{L^1}{T^{ - 2}}} \right]
Dimensionally, the area can be written as:
Area, A=[L2]A = \left[ {{L^2}} \right]
On combining, force and area, we get:
Pressure, P⁡=[ML−1T−2]\operatorname{P} = [M{L^{ - 1}}{T^{ - 2}}]
On the right-hand side we know:
h=h = Height
ρ=\rho = Density of the liquid
g=g = Acceleration due to gravity.
Dimensionally, we can write:
h=[L]h = [L]
We know, density is written as D=mVD = \dfrac{m}{V}
D=D = Density
m=m = Mass
V=V = Volume
Dimensionally, thus, density can be written as:
⇒ρ=[ML−3]\Rightarrow \rho = [M{L^{ - 3}}]
Again, acceleration due to gravity is written as:
⇒g=[ML−2]\Rightarrow g = [M{L^{ - 2}}]
In order to find the consistency of an equation, we write it as:
⇒\Rightarrow PP = [LaL^a ρb{\rho}^b gcg^c]
where, aa, bb vary as required.
If RHS equals LHS, we consider the equation to be consistent.
We consider aa, bb and cc as powers of hh, ρ\rho , gg respectively.
Thus we can write:
⇒[L]a[ML−3]b[ML−2]c\Rightarrow {[L]^a}{[M{L^{ - 3}}]^b}{[M{L^{ - 2}}]^c}
Now, this is our LHS.
Equating the LHS with RHS, we get:
a=1a = 1
b=1b = 1
c=1c = 1
Thus, on putting the values in LHS, we can derive:
P=hρgP = h\rho g
This is our required equation.
Therefore, we can also say, the equation is consistent.

Note: Dimensional analysis has a lot of applications, it is used to derive equations, check the correctness of the equations, and also to obtain relationships between various physical quantities. . However, dimensional analysis is not applicable for formulas having trigonometric, logarithmic, exponential notations.