The speed `(v)` of ripples on the surface of waterdepends on surface tension `(sigma)`, density `(rho)` and wavelength `(lambda)`. The square of speed `(v)` is proportional to

Options

A. `(sigma)/(rho lambda)`
B. `(rho)/(sigma lambda)`
C. `(lambda )/(sigma rho)`
D. `rho lambda sigma`

Answer

To solve the problem of determining how the square of the speed \( v \) of ripples on the surface of water depends on surface tension \( \sigma \), density \( \rho \), and wavelength \( \lambda \), we will use dimensional analysis. Here’s a step-by-step solution:

### Step 1: Identify the relationship
We know that the speed \( v \) depends on \( \sigma \), \( \rho \), and \( \lambda \). We can express this relationship as:
\[ v \propto \sigma^A \cdot \rho^B \cdot \lambda^C \]
where \( A \), \( B \), and \( C \) are the powers we need to determine.

### Step 2: Write the dimensions
We need to write down the dimensions of each quantity:
- The dimension of speed \( v \) is \( [v] = M^0 L^1 T^{-1} \).
- The dimension of surface tension \( \sigma \) is \( [\sigma] = M^1 T^{-2} \).
- The dimension of density \( \rho \) is \( [\rho] = M^1 L^{-3} \).
- The dimension of wavelength \( \lambda \) is \( [\lambda] = L^1 \).

### Step 3: Write the dimensional equation
Now we can substitute the dimensions into our proportionality:
\[ [M^0 L^1 T^{-1}] = [M^1 T^{-2}]^A \cdot [M^1 L^{-3}]^B \cdot [L^1]^C \]
This expands to:
\[ M^0 L^1 T^{-1} = M^{A+B} L^{-3B+C} T^{-2A} \]

### Step 4: Set up equations for each dimension
Now we equate the powers of \( M \), \( L \), and \( T \) on both sides:
1. For mass \( M \):
\[ 0 = A + B \quad \text{(Equation 1)} \]
2. For length \( L \):
\[ 1 = -3B + C \quad \text{(Equation 2)} \]
3. For time \( T \):
\[ -1 = -2A \quad \text{(Equation 3)} \]

### Step 5: Solve for \( A \)
From Equation 3:
\[ -1 = -2A \implies A = \frac{1}{2} \]

### Step 6: Solve for \( B \)
Substituting \( A \) into Equation 1:
\[ 0 = \frac{1}{2} + B \implies B = -\frac{1}{2} \]

### Step 7: Solve for \( C \)
Now substitute \( B \) into Equation 2:
\[ 1 = -3(-\frac{1}{2}) + C \implies 1 = \frac{3}{2} + C \implies C = 1 - \frac{3}{2} = -\frac{1}{2} \]

### Step 8: Write the final expression for \( v \)
Now we have:
\[ v \propto \sigma^{\frac{1}{2}} \cdot \rho^{-\frac{1}{2}} \cdot \lambda^{-\frac{1}{2}} \]
This can be rewritten as:
\[ v \propto \frac{\sqrt{\sigma}}{\sqrt{\rho} \cdot \sqrt{\lambda}} \]

### Step 9: Find \( v^2 \)
To find \( v^2 \), we square both sides:
\[ v^2 \propto \frac{\sigma}{\rho \cdot \lambda} \]

### Final Answer
Thus, the square of the speed \( v^2 \) is proportional to:
\[ v^2 \propto \frac{\sigma}{\rho \lambda} \]