The mechanical power P dissipated by a circular disc of radius R rotating at an angular ve… — JAMB UTME

The mechanical power $P$ dissipated by a circular disc of radius $R$ rotating at an angular velocity $\omega$ in a fluid of density $\rho$ is given by the relation $P = k \cdot \rho^a \cdot \omega^b \cdot R^c$ , where $k$ is a dimensionless constant. Using dimensional analysis, which of the following represents the correct values of the exponents $a$ , $b$ , and $c$ ?

$a = 1, b = 3, c = 5$ Answer
B
$a = 1, b = 2, c = 4$
C
$a = 1, b = 3, c = 2$
D
$a = 1, b = 1, c = 5$

Answer

a = 1, b = 3, c = 5

By writing the dimensions of power as

[M L^2 T^{-3}]

, density as

[M L^{-3}]

, angular velocity as

[T^{-1}]

, and radius as

[L]

, dimensional homogeneity requires that

M L^2 T^{-3} = M^a L^{-3a + c} T^{-b}

. Equating powers gives

a = 1

b = 3

, and

c = 5

Step-by-Step Solution

Express the base dimensions for each physical quantity

Power

[P] = M L^2 T^{-3}

, Density

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

, Angular velocity

[\omega] = T^{-1}

, and Radius

[R] = L

Dimensional analysis requires decomposing derived physical quantities into fundamental dimensions of mass (

M

), length (

L

), and time (

T

Substitute dimensions into the given equation

P = k \cdot \rho^a \cdot \omega^b \cdot R^c

M L^2 T^{-3} = (M L^{-3})^a \cdot (T^{-1})^b \cdot (L)^c = M^a \cdot L^{-3a + c} \cdot T^{-b}

The constant

k

is dimensionless, so its dimension is 1.

Equate the powers of

M

L

, and

T

on both sides of the equation

For mass

M

a = 1

. For time

T

-b = -3 \implies b = 3

. For length

L

-3a + c = 2 \implies -3(1) + c = 2 \implies c = 5

According to the principle of dimensional homogeneity, powers of fundamental dimensions must be equal on both sides of a physically correct equation.

Key Concept

Dimensional Homogeneity and Formula Derivation

Estimated Time:2m 0s

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