Question

Drag force $F$ depends on density $\rho$, velocity $V$, and diameter $D$: $F = k \rho^a V^b D^c$. Find exponents $a, b, c$.

• A. $(1, 2, 1)$
• B. $(2, 2, 1)$
• C. $(1, 1, 2)$
• D. $(1, 2, 2)$

Hint:

The standard drag force formula is:
$F_d = \frac{1}{2} C_d \rho A V^2$
Since area $A \propto D^2$, the force is proportional to $\rho^1 V^2 D^2$. This matches $(1, 2, 2)$ perfectly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to find the exponents $a$, $b$, and $c$ in the drag force equation using dimensional analysis (Rayleigh's method).

Step 2: Key Formula or Approach:
Write down the dimensional formulas for each variable in the SI system:
- Force ($F$): $[\text{M L T}^{-2}]$
- Density ($\rho$): $[\text{M L}^{-3}]$
- Velocity ($V$): $[\text{L T}^{-1}]$
- Diameter ($D$): $[\text{L}]$
- Constant ($k$): Dimensionless ($[1]$)

Step 3: Detailed Explanation:
Substitute the dimensional formulas into the equation:
\[ F = k \rho^a V^b D^c \]
\[ [\text{M L T}^{-2}] = [\text{M L}^{-3}]^a [\text{L T}^{-1}]^b [\text{L}]^c \]
\[ [\text{M L T}^{-2}] = \text{M}^a \text{L}^{-3a + b + c} \text{T}^{-b} \]
Now, equate the exponents on both sides of the equation:
1. For Mass ($\text{M}$):
\[ a = 1 \]
2. For Time ($\text{T}$):
\[ -b = -2 \implies b = 2 \]
3. For Length ($\text{L}$):
\[ -3a + b + c = 1 \]
Substitute the values of $a$ and $b$:
\[ -3(1) + 2 + c = 1 \]
\[ -1 + c = 1 \implies c = 2 \]
Thus, the values are:
$a = 1$, $b = 2$, $c = 2$.

Step 4: Final Answer:
The exponents $(a, b, c)$ are $(1, 2, 2)$.