The dynamic pressure P exerted by a moving fluid depends on its density \rho and flow velo… — JAMB UTME

The dynamic pressure $P$ exerted by a moving fluid depends on its density $\rho$ and flow velocity $v$ according to the relationship $P = k \rho^a v^b$ , where $k$ is a dimensionless constant. What is the value of the exponent $b$ ?

Answer: 2

Answer

The value of the exponent

b

is 2.

Equating the exponent of time (T) on both sides of the dimensional equation

\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}

yields

-2 = -b

, giving

b = 2

Step-by-Step Solution

Determine the base dimensions of all physical quantities in the equation.

Pressure

[P] = \text{M L}^{-1} \text{T}^{-2}

, Density

[\rho] = \text{M L}^{-3}

, and Velocity

[v] = \text{L T}^{-1}

Dimensional analysis requires converting derived physical quantities into fundamental dimensions of Mass (M), Length (L), and Time (T).

Apply dimensional homogeneity by substituting the dimensions into

P = k \rho^a v^b

\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}

The principle of dimensional homogeneity states that exponents of M, L, and T must match on both sides of a physically correct equation.

Equate the corresponding exponents for time (T) and solve for

b

-2 = -b \implies b = 2

Matching the powers of T directly yields the numerical value of exponent

b

Key Concept

Principle of Dimensional Homogeneity and Dimensional Analysis

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