The dimensional formula for Reynold's number is

Options

A. `["L"^(0)"M"^(0)"T"^(0)]`
B. `["LMT"]`
C. `["L"^(-1)"MT"]`
D. `["LMT"^(-1)]`

Explanation

To find the dimensional formula for Reynolds number, we start by understanding its definition and the quantities involved.

### Step-by-Step Solution:

1. **Understanding Reynolds Number**:
Reynolds number (Re) is defined as the ratio of inertial forces to viscous forces in a fluid flow. Mathematically, it can be expressed as:
\[ Re = \frac{\text{Inertial Force}}{\text{Viscous Force}} \]

2. **Identifying Forces**:
- The **Inertial Force** can be represented as \( F_i = \rho v^2 L \), where:
- \( \rho \) is the density of the fluid (mass per unit volume, \( [M L^{-3}] \)),
- \( v \) is the velocity (length per time, \( [L T^{-1}] \)),
- \( L \) is a characteristic length (dimension of length, \( [L] \)).
- The **Viscous Force** is given by \( F_v = \mu \frac{v}{L} A \), where:
- \( \mu \) is the dynamic viscosity (force per unit area per unit velocity, \( [M L^{-1} T^{-1}] \)),
- \( A \) is the area (dimension of area, \( [L^2] \)).

3. **Calculating Dimensions**:
- The **Inertial Force** has dimensions:
\[ [F_i] = [\rho v^2 L] = [M L^{-3}] [L^2 T^{-2}] [L] = [M L T^{-2}] \]
- The **Viscous Force** has dimensions:
\[ [F_v] = [\mu \frac{v}{L} A] = [M L^{-1} T^{-1}] [L T^{-1}] [L^2] = [M L T^{-2}] \]

4. **Ratio of Forces**:
Since both inertial force and viscous force have the same dimensions, their ratio will be dimensionless:
\[ Re = \frac{F_i}{F_v} = \frac{[M L T^{-2}]}{[M L T^{-2}]} = 1 \]

5. **Conclusion**:
Therefore, the dimensional formula for Reynolds number is:
\[ [Re] = [M^0 L^0 T^0] = [1] \]
This indicates that Reynolds number is dimensionless.

### Final Answer:
The dimensional formula for Reynolds number is \( M^0 L^0 T^0 \).