If the velocity is represented by V, force by F and work by W . These quantities are taken as fundamental quantities then the correct dimensions of mass will be

Options

A. `W//V^2`
B. `W//V`
C. `F//V^2`
D. `V//W`

Answer

To find the correct dimensions of mass when velocity (V), force (F), and work (W) are taken as fundamental quantities, we will perform dimensional analysis.

### Step-by-Step Solution:

1. **Identify the Dimensions of Each Quantity**:
- **Velocity (V)**: The dimension of velocity is given by \( [V] = [L][T]^{-1} \) (length per time).
- **Force (F)**: The dimension of force is given by \( [F] = [M][L][T]^{-2} \) (mass times acceleration).
- **Work (W)**: The dimension of work is given by \( [W] = [M][L^2][T]^{-2} \) (force times distance).

2. **Set Up the Dimensional Equation**:
- We can express mass (M) in terms of the other quantities:
\[ [M] = [F]^\alpha [W]^\beta [V]^\gamma \]
Substituting the dimensions of F, W, and V:
\[ [M] = ([M][L][T]^{-2})^\alpha \cdot ([M][L^2][T]^{-2})^\beta \cdot ([L][T]^{-1})^\gamma \]

3. **Expand the Equation**:
- Expanding this gives:
\[ [M] = [M]^{\alpha + \beta} \cdot [L]^{\alpha + 2\beta + \gamma} \cdot [T]^{-2\alpha - 2\beta - \gamma} \]

4. **Equate the Exponents**:
- Since the left side is just [M], we can set up the following equations based on the exponents:
- For mass:
\[ \alpha + \beta = 1 \quad \text{(1)} \]
- For length:
\[ \alpha + 2\beta + \gamma = 0 \quad \text{(2)} \]
- For time:
\[ -2\alpha - 2\beta - \gamma = 0 \quad \text{(3)} \]

5. **Solve the Equations**:
- From equation (1), we have:
\[ \beta = 1 - \alpha \]
- Substitute \(\beta\) in equation (2):
\[ \alpha + 2(1 - \alpha) + \gamma = 0 \]
Simplifying gives:
\[ \alpha + 2 - 2\alpha + \gamma = 0 \implies -\alpha + \gamma + 2 = 0 \implies \gamma = \alpha - 2 \quad \text{(4)} \]
- Substitute \(\beta\) and \(\gamma\) in equation (3):
\[ -2\alpha - 2(1 - \alpha) - (\alpha - 2) = 0 \]
Simplifying gives:
\[ -2\alpha - 2 + 2\alpha - \alpha + 2 = 0 \implies -\alpha = 0 \implies \alpha = 0 \]
- From \(\alpha = 0\), we find:
\[ \beta = 1 \quad \text{and} \quad \gamma = -2 \]

6. **Final Expression for Mass**:
- Substituting back, we find:
\[ [M] = [W]^1 [V]^{-2} = \frac{W}{V^2} \]
Thus, the correct dimension of mass is:
\[ M = \frac{W}{V^2} \]

### Conclusion:
The correct dimensions of mass when velocity, force, and work are taken as fundamental quantities is given by:
\[ M = \frac{W}{V^2} \]