The volume V of water passing and any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation `V prop A^(alpha)u^(beta)t^(gamma)`
Which one of the following will be true?

Options

A. `alpha=beta=gamma`
B. `alpha ne beta = gamma`
C. `alpha= beta ne gamma`
D. `alpha ne beta ne gamma`

Explanation

To solve the problem, we need to analyze the relationship given in the question:

\[ V \propto A^{\alpha} u^{\beta} t^{\gamma} \]

### Step 1: Identify the dimensions of each variable
- **Volume (V)**: The dimensions of volume are \( [V] = L^3 \).
- **Cross-sectional area (A)**: The dimensions of area are \( [A] = L^2 \).
- **Velocity (u)**: The dimensions of velocity are \( [u] = L T^{-1} \).
- **Time (t)**: The dimensions of time are \( [t] = T \).

### Step 2: Write the dimensional equation
We can express the relationship dimensionally as:

\[ [V] = [A^{\alpha}] \cdot [u^{\beta}] \cdot [t^{\gamma}] \]

Substituting the dimensions we found:

\[ L^3 = (L^2)^{\alpha} \cdot (L T^{-1})^{\beta} \cdot (T)^{\gamma} \]

### Step 3: Simplify the right-hand side
Expanding the right-hand side gives us:

\[ L^3 = L^{2\alpha} \cdot L^{\beta} \cdot T^{-\beta} \cdot T^{\gamma} \]

Combining the terms, we have:

\[ L^3 = L^{2\alpha + \beta} \cdot T^{\gamma - \beta} \]

### Step 4: Equate the dimensions
For the dimensions to be equal, the powers of \( L \) and \( T \) on both sides must match:

1. For \( L \):
\[ 3 = 2\alpha + \beta \] (Equation 1)

2. For \( T \):
\[ 0 = \gamma - \beta \] (Equation 2)

### Step 5: Solve the equations
From Equation 2, we can express \( \gamma \) in terms of \( \beta \):

\[ \gamma = \beta \]

Substituting \( \gamma \) into Equation 1 gives:

\[ 3 = 2\alpha + \beta \]

Now, we have two equations:
1. \( \gamma = \beta \)
2. \( 3 = 2\alpha + \beta \)

### Step 6: Analyze the relationships
From \( \gamma = \beta \), we can substitute \( \beta \) back into the equation:

\[ 3 = 2\alpha + \gamma \]

This indicates that \( \alpha \) and \( \beta \) cannot be equal unless they are both zero, which contradicts our dimensional analysis since \( \alpha \) must contribute to the volume. Thus, we conclude:

- \( \alpha \neq \beta \)

### Conclusion
The correct relationship that holds true is that \( \alpha \) is not equal to \( \beta \).