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)=klamda^(-1)g^(-1)rho^(-1)`
B. `v^(2)=kglamda`
C. `v^(2)=kglamda rho`
D. `v^(2)=klamda^(3)g^(-1)rho^(-1)`

Answer

To solve the problem, we will use the method of dimensional analysis 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 \( g \).

### Step-by-Step Solution:

1. **Identify the quantities and their dimensions**:
- Velocity \( v \): The dimension of velocity is given by:
\[ [v] = L T^{-1} \]
- Wavelength \( \lambda \): The dimension of wavelength is:
\[ [\lambda] = L \]
- Density \( \rho \): The dimension of density is:
\[ [\rho] = M L^{-3} \]
- Acceleration due to gravity \( g \): The dimension of acceleration is:
\[ [g] = L T^{-2} \]

2. **Assume a relationship**:
We assume that the velocity \( v \) can be expressed in terms of \( \lambda \), \( \rho \), and \( g \):
\[ v = k \lambda^a \rho^b g^c \]
where \( k \) is a dimensionless constant, and \( a \), \( b \), and \( c \) are the powers we need to determine.

3. **Write the dimensions of both sides**:
The dimensions of the left-hand side (velocity) are:
\[ [v] = L T^{-1} \]
The dimensions of the right-hand side can be expressed as:
\[ [\lambda^a] = L^a, \quad [\rho^b] = (M L^{-3})^b = M^b L^{-3b}, \quad [g^c] = (L T^{-2})^c = L^c T^{-2c} \]
Therefore, the dimensions of the right-hand side become:
\[ [v] = k L^a M^b L^{-3b} L^c T^{-2c} = k M^b L^{a - 3b + c} T^{-2c} \]

4. **Set up the equation for dimensions**:
Equating the dimensions from both sides gives:
\[ L^{a - 3b + c} M^b T^{-2c} = L^1 M^0 T^{-1} \]
This leads to the following system of equations:
- For \( L \): \( a - 3b + c = 1 \)
- For \( M \): \( b = 0 \)
- For \( T \): \( -2c = -1 \) or \( c = \frac{1}{2} \)

5. **Solve the equations**:
From \( b = 0 \), substitute into the first equation:
\[ a + c = 1 \quad \text{(since \( b = 0 \))} \]
Substituting \( c = \frac{1}{2} \):
\[ a + \frac{1}{2} = 1 \implies a = \frac{1}{2} \]

6. **Final relationship**:
Now we have:
- \( a = \frac{1}{2} \)
- \( b = 0 \)
- \( c = \frac{1}{2} \)

Therefore, the relationship can be expressed as:
\[ v = k \lambda^{\frac{1}{2}} g^{\frac{1}{2}} = k \sqrt{g \lambda} \]
Squaring both sides gives:
\[ v^2 = k g \lambda \]

### Conclusion:
The final relationship derived from dimensional analysis is:
\[ v^2 = k g \lambda \]
where \( k \) is a constant.