Which of the following is dimensionless?

Options

A. `(v^(2))/(rg)`
B. `(v^(2)g)/(r)`
C. `(vg)/(r)`
D. `v^(2)rg`

Explanation

To determine which of the given options is dimensionless, we need to analyze each option by calculating their dimensions. A dimensionless quantity has no physical dimensions and is often represented as a pure number.

### Step-by-Step Solution:

1. **Understand the Dimensions of Each Quantity:**
- Velocity (v) has dimensions: \( [v] = L T^{-1} \)
- Acceleration due to gravity (g) has dimensions: \( [g] = L T^{-2} \)
- Distance (r) has dimensions: \( [r] = L \)

2. **Option 1: \( \frac{v^2}{r \cdot g} \)**
- Calculate \( v^2 \):
\[ v^2 = (L T^{-1})^2 = L^2 T^{-2} \]
- Now calculate \( r \cdot g \):
\[ r \cdot g = L \cdot (L T^{-2}) = L^2 T^{-2} \]
- Therefore, the dimensions of \( \frac{v^2}{r \cdot g} \) are:
\[ \frac{L^2 T^{-2}}{L^2 T^{-2}} = 1 \quad (\text{dimensionless}) \]

3. **Option 2: \( \frac{v^2 g}{r} \)**
- Calculate \( v^2 g \):
\[ v^2 g = L^2 T^{-2} \cdot (L T^{-2}) = L^3 T^{-4} \]
- Now calculate \( \frac{v^2 g}{r} \):
\[ \frac{L^3 T^{-4}}{L} = L^2 T^{-4} \quad (\text{not dimensionless}) \]

4. **Option 3: \( \frac{vg}{r} \)**
- Calculate \( vg \):
\[ vg = (L T^{-1}) \cdot (L T^{-2}) = L^2 T^{-3} \]
- Now calculate \( \frac{vg}{r} \):
\[ \frac{L^2 T^{-3}}{L} = L T^{-3} \quad (\text{not dimensionless}) \]

5. **Option 4: \( v^2 r g \)**
- Calculate \( v^2 r g \):
\[ v^2 r g = (L^2 T^{-2}) \cdot L \cdot (L T^{-2}) = L^4 T^{-4} \]
- This is clearly not dimensionless.

### Conclusion:
The only option that is dimensionless is **Option 1: \( \frac{v^2}{r \cdot g} \)**.