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Question: If pressure 'p' depends upon velocity 'v' and density 'd', the relationship between p, v and d is: ...

If pressure 'p' depends upon velocity 'v' and density 'd', the relationship between p, v and d is:
A. pvdp \propto vd
B. pv2dp \propto {v^2}d
C. pv3dp \propto \dfrac{{{v^3}}}{d}
D. pv2d2p \propto \dfrac{{{v^2}}}{{{d^2}}}

Explanation

Solution

The above problem can be resolved using the dimensional analysis concepts and the dimensional formula of pressure, velocity, and density. The dimensional formulas are enlisted in fundamental units, and then these values are substituted in the significant relationship of the pressure, velocity, and density. This significant relationship involves the constant, whose values are identified using the mathematical technique of comparing the power terms.

Complete step by step answer:
Given
The pressure is, p
The velocity is, v
The density is, d
The dimensional formula for the pressure is, p=[M1L1T2]p = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right].
The dimensional formula for the velocity is, v=[LT1]v = \left[ {L{T^{ - 1}}} \right].
The dimensional formula for the density is, d=[ML3]d = \left[ {M{L^{ - 3}}} \right]
The significant relationship between the pressure, velocity and density is,
P=vα×dβP = {v^\alpha } \times {d^\beta }
Substituting the values of respective dimensional formulas in the above equation as,

P = {v^\alpha } \times {d^\beta }\\\ \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] = {\left[ {L{T^{ - 1}}} \right]^\alpha } \times {\left[ {M{L^{ - 3}}} \right]^\beta }\\\ {M^1}{L^{ - 1}}{T^{ - 2}} = {M^\beta }{L^{ - 3\beta + \alpha }}{T^{ - \alpha }}.........................\left( 1 \right) \end{array}$$ From equation 1, the value of $$\alpha $$ and $$\beta $$ are, $$\begin{array}{l} \beta = 1.......................\left( 2 \right)\\\ \- 3\beta + \alpha = - 1\\\ \- 3\left( 1 \right) + \alpha = - 1\\\ \alpha = 2.........................\left( 3 \right) \end{array}$$ Substituting the values of equation 2 and 3 in the significant relationship between the pressure, velocity and density as, $$\begin{array}{l} P = {v^\alpha } \times {d^\beta }\\\ P = {v^2} \times {d^1}\\\ P \propto {v^2} \times d \end{array}$$ Therefore, the relation between the pressure, velocity and density is $$p \propto {v^2}d$$ **So, the correct answer is “Option B”.** **Note:** To resolve the given problem, the dimensional analysis and its concept need to be remembered. Moreover, the basic dimensional formulas need to be remembered; this includes the dimensional formula for the distance, speed, force, and many. The significant applications of dimensional analysis include the prediction and formulation of various mathematical formulas and equations. Moreover, dimensional analysis is also useful in checking, whether the applied formula is accurate or not.