Check the correctness of the equation `v=(K eta)/(r rho)` using dimensional analysis, where v is the critical velocity `eta` is the coefficient of viscosity, r is the radius and `rho` is the density?

Explanation

To check the correctness of the equation \( v = \frac{K \eta}{r \rho} \) using dimensional analysis, we will analyze the dimensions of each variable involved in the equation.

### Step-by-Step Solution:

1. **Identify the Variables and Their Dimensions**:
- \( v \) (critical velocity): The dimension of velocity is given by:
\[ [v] = [L T^{-1}] \]
- \( K \) (Reynolds number): This is a dimensionless quantity, so:
\[ [K] = 1 \]
- \( \eta \) (coefficient of viscosity): The dimension of viscosity is:
\[ [\eta] = [M L^{-1} T^{-1}] \]
- \( r \) (radius): The dimension of radius is:
\[ [r] = [L] \]
- \( \rho \) (density): The dimension of density is:
\[ [\rho] = [M L^{-3}] \]

2. **Substitute the Dimensions into the Right-Hand Side of the Equation**:
The right-hand side of the equation is \( \frac{K \eta}{r \rho} \). We will substitute the dimensions we found:
\[ [K \eta] = [1] \cdot [M L^{-1} T^{-1}] = [M L^{-1} T^{-1}] \]
Now, substituting for \( r \) and \( \rho \):
\[ [r \rho] = [L] \cdot [M L^{-3}] = [M L^{-2}] \]

3. **Combine the Dimensions**:
Now, we can find the dimensions of \( \frac{K \eta}{r \rho} \):
\[ \left[\frac{K \eta}{r \rho}\right] = \frac{[M L^{-1} T^{-1}]}{[M L^{-2}]} = [L^{1} T^{-1}] \]
Here, the \( M \) in the numerator and denominator cancels out.

4. **Final Comparison**:
Now, we compare the dimensions of the left-hand side \( [v] \) and the right-hand side \( \left[\frac{K \eta}{r \rho}\right] \):
- Left-hand side: \( [v] = [L T^{-1}] \)
- Right-hand side: \( \left[\frac{K \eta}{r \rho}\right] = [L T^{-1}] \)

Since both sides have the same dimensions, we conclude that the equation is dimensionally correct.

### Conclusion:
The equation \( v = \frac{K \eta}{r \rho} \) is dimensionally correct.