Which of the following is dimensionless ?

Options

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

Answer

To determine which of the given options is dimensionless, we need to analyze the dimensions of each option step by step.

### Step-by-Step Solution

1. **Identify the dimensions of the quantities involved:**
- Velocity (V) has dimensions: \( [V] = L^1 T^{-1} \)
- Acceleration due to gravity (G) has dimensions: \( [G] = L^1 T^{-2} \)
- Radius (R) has dimensions: \( [R] = L^1 \)

2. **Analyze the first option: \( \frac{V^2}{RG} \)**
- Calculate \( V^2 \):
\[ [V^2] = (L^1 T^{-1})^2 = L^2 T^{-2} \]
- Now substitute the dimensions into the expression:
\[ [RG] = [R][G] = (L^1)(L^1 T^{-2}) = L^2 T^{-2} \]
- Now calculate \( \frac{V^2}{RG} \):
\[ \frac{V^2}{RG} = \frac{L^2 T^{-2}}{L^2 T^{-2}} = L^0 T^0 \]
- Result: This is dimensionless.

3. **Analyze the second option: \( \frac{V^2 G}{R} \)**
- Using the dimensions calculated earlier:
\[ [V^2 G] = [V^2][G] = (L^2 T^{-2})(L^1 T^{-2}) = L^3 T^{-4} \]
- Now calculate \( \frac{V^2 G}{R} \):
\[ \frac{V^2 G}{R} = \frac{L^3 T^{-4}}{L^1} = L^2 T^{-4} \]
- Result: This has dimensions and is not dimensionless.

4. **Analyze the third option: \( \frac{VG}{R} \)**
- Calculate \( [VG] \):
\[ [VG] = [V][G] = (L^1 T^{-1})(L^1 T^{-2}) = L^2 T^{-3} \]
- Now calculate \( \frac{VG}{R} \):
\[ \frac{VG}{R} = \frac{L^2 T^{-3}}{L^1} = L^1 T^{-3} \]
- Result: This has dimensions and is not dimensionless.

5. **Analyze the fourth option: \( V^2 R G \)**
- Calculate \( [V^2 R G] \):
\[ [V^2 R G] = [V^2][R][G] = (L^2 T^{-2})(L^1)(L^1 T^{-2}) = L^4 T^{-4} \]
- Result: This has dimensions and is not dimensionless.

### Conclusion
The only dimensionless quantity among the options is the first option \( \frac{V^2}{RG} \).

### Final Answer
**The dimensionless quantity is the first option: \( \frac{V^2}{RG} \)**.

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