
What is the dimensional formula of the pressure head?
Answer
220.5k+ views
Hint: In physical science, we can express every physical quantity in terms of the basic fundamental quantities: mass $\left[ M \right]$, length $\left[ L \right]$ , and time $\left[ T \right]$. It is called the dimensional formula of a quantity.
We will think fundamentally about how to describe a physical quantity with the fundamental quantities. This idea will then lead us to find out the dimensional formula of Pressure Head.
Complete solution:
Here we write, ${P_H}$to denote ‘Pressure Head.’
Now,
${P_H} = \dfrac{P}{{D \times g}}$.......……..$(1)$
Where
$P$= pressure
$D$= density
$g$= gravitational acceleration
We will first find out the dimensional formula of Pressure and Density by converting them into a combination of fundamental physical quantities, and then we will use them to find out the dimensional formula of Pressure Head.
So our fast task is to find the dimensional formula of pressure.
And we know,
$P = \dfrac{F}{S}$
Where $F$= force
$S$= surface area
Therefore, $P = \dfrac{{M \times a}}{S}$
Where, $M$= mass
$a$= acceleration
By putting the dimensional formula for all the quantities on the right-hand side-
$\Rightarrow \left[ P \right] = \dfrac{{\left[ M \right] \times \left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}$
By simplifying this, we have-
$\left[ P \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
Now we will find the dimensional formula of density.
And we know,
$D = \dfrac{M}{V}$
Where, $V$ = volume
Now we put the dimensional formula of mass and volume-
$\Rightarrow \left[ D \right] = \dfrac{{\left[ M \right]}}{{\left[ {{L^3}} \right]}}$
Hence we get:
$\Rightarrow \left[ D \right] = \left[ {M{L^{ - 3}}} \right]$
At last, we just put all the dimensional formula in $(1)$,
$\Rightarrow \left[ {{P_H}} \right] = \dfrac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]}}{{\left[ {M{L^{ - 3}}} \right] \times \left[ {L{T^{ - 2}}} \right]}}$
$\Rightarrow \left[ {{M^0}{L^1}{T^0}} \right]$
Hence we get our desired answer of the dimensional formula of Pressure Head, i.e. $\left[ {{M^0}{L^1}{T^0}} \right]$.
Note: We can find out the dimensional formula of any physical quantity by following this process. It is crucial to understand the dimensional formula of a physical quantity because that gives a broad idea about the quantity despite various measurement systems. There are some other fundamental physical quantities other than mass, length, and time such as Charge $\left[ Q \right]$ and Angle $\left[ \Theta \right]$.
We will think fundamentally about how to describe a physical quantity with the fundamental quantities. This idea will then lead us to find out the dimensional formula of Pressure Head.
Complete solution:
Here we write, ${P_H}$to denote ‘Pressure Head.’
Now,
${P_H} = \dfrac{P}{{D \times g}}$.......……..$(1)$
Where
$P$= pressure
$D$= density
$g$= gravitational acceleration
We will first find out the dimensional formula of Pressure and Density by converting them into a combination of fundamental physical quantities, and then we will use them to find out the dimensional formula of Pressure Head.
So our fast task is to find the dimensional formula of pressure.
And we know,
$P = \dfrac{F}{S}$
Where $F$= force
$S$= surface area
Therefore, $P = \dfrac{{M \times a}}{S}$
Where, $M$= mass
$a$= acceleration
By putting the dimensional formula for all the quantities on the right-hand side-
$\Rightarrow \left[ P \right] = \dfrac{{\left[ M \right] \times \left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}$
By simplifying this, we have-
$\left[ P \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
Now we will find the dimensional formula of density.
And we know,
$D = \dfrac{M}{V}$
Where, $V$ = volume
Now we put the dimensional formula of mass and volume-
$\Rightarrow \left[ D \right] = \dfrac{{\left[ M \right]}}{{\left[ {{L^3}} \right]}}$
Hence we get:
$\Rightarrow \left[ D \right] = \left[ {M{L^{ - 3}}} \right]$
At last, we just put all the dimensional formula in $(1)$,
$\Rightarrow \left[ {{P_H}} \right] = \dfrac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]}}{{\left[ {M{L^{ - 3}}} \right] \times \left[ {L{T^{ - 2}}} \right]}}$
$\Rightarrow \left[ {{M^0}{L^1}{T^0}} \right]$
Hence we get our desired answer of the dimensional formula of Pressure Head, i.e. $\left[ {{M^0}{L^1}{T^0}} \right]$.
Note: We can find out the dimensional formula of any physical quantity by following this process. It is crucial to understand the dimensional formula of a physical quantity because that gives a broad idea about the quantity despite various measurement systems. There are some other fundamental physical quantities other than mass, length, and time such as Charge $\left[ Q \right]$ and Angle $\left[ \Theta \right]$.
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