Force `xx` velocity `=x`. The dimensional formula of `x` is

Options

A. `MLT^(-2)`
B. `ML^(2)T^(-2)`
C. `ML^(2)T`
D. `ML^(2)T^(-3)`

Answer

To find the dimensional formula of the quantity \( x \) given that \( \text{Force} \times \text{Velocity} = x \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dimensional Formula of Force**:
- Force is defined as mass times acceleration. The dimensional formula for mass \( m \) is \( [M] \) and for acceleration \( a \) (which is \( \text{velocity} / \text{time} \)) is \( [L][T^{-2}] \).
- Therefore, the dimensional formula for force \( F \) is:
\[ [F] = [M][L][T^{-2}] = [M L T^{-2}] \]

2. **Identify the Dimensional Formula of Velocity**:
- Velocity is defined as displacement divided by time. The dimensional formula for displacement \( [L] \) and for time \( [T] \) gives us:
\[ [V] = \frac{[L]}{[T]} = [L T^{-1}] \]

3. **Combine the Dimensional Formulas**:
- Now, we need to find the dimensional formula for \( x \) which is the product of force and velocity:
\[ x = F \times V \]
- Substituting the dimensional formulas we found:
\[ [x] = [F] \times [V] = [M L T^{-2}] \times [L T^{-1}] \]

4. **Multiply the Dimensional Formulas**:
- When multiplying the dimensions, we add the powers of the same base:
\[ [x] = [M^{1} L^{1+1} T^{-2-1}] = [M^{1} L^{2} T^{-3}] \]

5. **Final Result**:
- Therefore, the dimensional formula of \( x \) is:
\[ [x] = [M L^{2} T^{-3}] \]

### Conclusion:
The dimensional formula of \( x \) is \( [M L^{2} T^{-3}] \).