The velocity of water wave `v` may depend on their wavelength `lambda`, the density of water `rho` and the acceleration due to gravity `g`. The method of dimensions gives the relation between these quantities as

Options

A. `v^(2) prop lambda g rho`
B. `v^(2) prop lambda^(-1) g^(-1) rho^(-1)`
C. `v^(2) prop g lambda`
D. `v^(2) prop rho lambda`

Explanation

To find the relationship between the velocity of water waves \( v \), their wavelength \( \lambda \), the density of water \( \rho \), and the acceleration due to gravity \( g \) using the method of dimensions, we can follow these steps:

### Step 1: Establish the relationship
Assume that the velocity \( v \) depends on the wavelength \( \lambda \), the density \( \rho \), and the acceleration due to gravity \( g \). We can express this relationship as:
\[ v \propto \lambda^A \rho^B g^C \]
where \( A \), \( B \), and \( C \) are the powers we need to determine.

### Step 2: Write down the dimensions
Next, we need to write down the dimensions of each quantity:
- The dimension of velocity \( v \) is \( [v] = L T^{-1} \).
- The dimension of wavelength \( \lambda \) is \( [\lambda] = L \).
- The dimension of density \( \rho \) is \( [\rho] = M L^{-3} \).
- The dimension of acceleration due to gravity \( g \) is \( [g] = L T^{-2} \).

### Step 3: Substitute dimensions into the equation
Substituting the dimensions into the proportionality equation gives:
\[ [L T^{-1}] = [L^A][(M L^{-3})^B][(L T^{-2})^C] \]
This expands to:
\[ L T^{-1} = L^A \cdot M^B \cdot L^{-3B} \cdot L^C \cdot T^{-2C} \]
Combining the dimensions on the right-hand side:
\[ L T^{-1} = M^B \cdot L^{A - 3B + C} \cdot T^{-2C} \]

### Step 4: Equate the dimensions
Now, we can equate the dimensions on both sides:
1. For mass \( M \): \( B = 0 \)
2. For length \( L \): \( A - 3B + C = 1 \)
3. For time \( T \): \( -2C = -1 \)

### Step 5: Solve for \( B \)
From the first equation, we have:
\[ B = 0 \]

### Step 6: Substitute \( B \) into the length equation
Substituting \( B = 0 \) into the length equation:
\[ A + C = 1 \]

### Step 7: Solve for \( C \)
From the time equation:
\[ -2C = -1 \implies C = \frac{1}{2} \]

### Step 8: Substitute \( C \) back to find \( A \)
Substituting \( C = \frac{1}{2} \) into \( A + C = 1 \):
\[ A + \frac{1}{2} = 1 \implies A = \frac{1}{2} \]

### Step 9: Write the final relationship
Now we have:
- \( A = \frac{1}{2} \)
- \( B = 0 \)
- \( C = \frac{1}{2} \)

Thus, we can express the relationship as:
\[ v \propto \lambda^{\frac{1}{2}} g^{\frac{1}{2}} \]
or equivalently,
\[ v^2 \propto \lambda g \]

### Final Answer
The final relationship is:
\[ v^2 \propto \lambda g \]