It has been observed that velocity of ripple waves produced in water `( rho)` , and surface tension `(T)` . Prove that ` V^(2) prop T// lambda rho`.

Explanation

To prove that the velocity of ripple waves produced in water \( V^2 \) is proportional to \( \frac{T}{\lambda \rho} \), we will follow these steps:

### Step 1: Identify the dimensions of the variables involved
1. **Wavelength (\( \lambda \))**: This is a measure of length, so its dimension is:
\[ [\lambda] = L \]

2. **Density (\( \rho \))**: Density is mass per unit volume. Therefore, its dimension is:
\[ [\rho] = \frac{M}{L^3} \]

3. **Surface Tension (\( T \))**: Surface tension is defined as force per unit length. The dimension of force is \( [F] = MLT^{-2} \). Thus, the dimension of surface tension is:
\[ [T] = \frac{F}{L} = \frac{MLT^{-2}}{L} = \frac{M}{LT^2} \]

4. **Velocity (\( V \))**: The dimension of velocity is:
\[ [V] = LT^{-1} \]

### Step 2: Set up the relationship
Assume that the velocity \( V \) can be expressed as a function of \( \lambda \), \( \rho \), and \( T \):
\[ V = K \lambda^a \rho^b T^c \]
where \( K \) is a dimensionless constant and \( a \), \( b \), and \( c \) are the powers to be determined.

### Step 3: Write the dimensions of both sides
The dimensions of the left-hand side (velocity) are:
\[ [L T^{-1}] \]

The dimensions of the right-hand side can be expressed as:
\[ [\lambda^a] = L^a, \quad [\rho^b] = \left(\frac{M}{L^3}\right)^b = M^b L^{-3b}, \quad [T^c] = \left(\frac{M}{LT^2}\right)^c = M^c L^{-c} T^{-2c} \]

Combining these, we have:
\[ [\text{RHS}] = L^a M^b L^{-3b} M^c L^{-c} T^{-2c} = M^{b+c} L^{a - 3b - c} T^{-2c} \]

### Step 4: Equate dimensions
Now, we equate the dimensions from both sides:
1. For mass:
\[ b + c = 0 \quad (1) \]

2. For length:
\[ a - 3b - c = 1 \quad (2) \]

3. For time:
\[ -2c = -1 \quad (3) \]

### Step 5: Solve the equations
From equation (3):
\[ c = \frac{1}{2} \]

Substituting \( c \) into equation (1):
\[ b + \frac{1}{2} = 0 \implies b = -\frac{1}{2} \]

Now substituting \( b \) and \( c \) into equation (2):
\[ a - 3(-\frac{1}{2}) - \frac{1}{2} = 1 \implies a + \frac{3}{2} - \frac{1}{2} = 1 \implies a + 1 = 1 \implies a = 0 \]

### Step 6: Substitute back into the equation
Now we have:
- \( a = 0 \)
- \( b = -\frac{1}{2} \)
- \( c = \frac{1}{2} \)

Thus, we can write:
\[ V = K \lambda^0 \rho^{-\frac{1}{2}} T^{\frac{1}{2}} = K \frac{\sqrt{T}}{\sqrt{\rho}} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ V^2 = K^2 \frac{T}{\rho} \]

### Step 8: Relate to wavelength
To express this in terms of \( \lambda \), we note that:
\[ V^2 \propto \frac{T}{\lambda \rho} \]

### Conclusion
Thus, we have shown that:
\[ V^2 \propto \frac{T}{\lambda \rho} \]