Consider the equation `d/(dt)(int vec(F). dvec(S))=A(vec(F).vec(p))` where `vec(F) equiv` force, `vec(s) equiv` displacement, t `equiv` time and `vec(p)`= momentum. The dimensional formula of A will be

Options

A. `M^(0)L^(0)T^(0)`
B. `ML^(0)T^(0)`
C. `M^(-1)L^(0)T^(0)`
D. `M^(0)L^(0)T^(-1)`

Explanation

To find the dimensional formula of \( A \) in the equation

\[ \frac{d}{dt} \left( \int \vec{F} \cdot d\vec{S} \right) = A (\vec{F} \cdot \vec{p}), \]

we will analyze both sides of the equation step by step.

### Step 1: Analyze the left side of the equation

The left side of the equation is

\[ \frac{d}{dt} \left( \int \vec{F} \cdot d\vec{S} \right). \]

1. **Understanding \( \vec{F} \cdot d\vec{S} \)**:
- The force \( \vec{F} \) has the unit of Newton (N), which can be expressed dimensionally as \( \text{MLT}^{-2} \).
- The displacement \( d\vec{S} \) has the unit of meters (m), which can be expressed dimensionally as \( \text{L} \).
- Therefore, \( \vec{F} \cdot d\vec{S} \) has the unit of \( \text{N} \cdot \text{m} = \text{J} \) (Joules), which can be expressed dimensionally as \( \text{ML}^2\text{T}^{-2} \).

2. **Integrating over displacement**:
- The integral \( \int \vec{F} \cdot d\vec{S} \) will also have the same units as \( \vec{F} \cdot d\vec{S} \), which is \( \text{ML}^2\text{T}^{-2} \).

3. **Differentiating with respect to time**:
- When we differentiate with respect to time \( t \), we divide by time (s), so we get:
\[ \frac{d}{dt} \left( \int \vec{F} \cdot d\vec{S} \right) \text{ has the unit } \frac{\text{ML}^2\text{T}^{-2}}{\text{T}} = \text{ML}^2\text{T}^{-3}. \]

### Step 2: Analyze the right side of the equation

The right side of the equation is

\[ A (\vec{F} \cdot \vec{p}). \]

1. **Understanding \( \vec{F} \cdot \vec{p} \)**:
- The force \( \vec{F} \) has the unit of Newton (N), which is \( \text{MLT}^{-2} \).
- The momentum \( \vec{p} \) has the unit of \( \text{kg} \cdot \text{m/s} \) or \( \text{MLT}^{-1} \).
- Therefore, \( \vec{F} \cdot \vec{p} \) has the unit of:
\[ \text{N} \cdot \text{kg} \cdot \text{m/s} = \text{MLT}^{-2} \cdot \text{MLT}^{-1} = \text{M}^2\text{L}^2\text{T}^{-3}. \]

### Step 3: Equate both sides

Now we can equate the units from both sides of the equation:

\[ \text{ML}^2\text{T}^{-3} = A \cdot \text{M}^2\text{L}^2\text{T}^{-3}. \]

### Step 4: Solve for \( A \)

To isolate \( A \), we can rearrange the equation:

\[ A = \frac{\text{ML}^2\text{T}^{-3}}{\text{M}^2\text{L}^2\text{T}^{-3}} = \frac{1}{\text{M}\text{L}}. \]

### Step 5: Write the dimensional formula for \( A \)

Thus, the dimensional formula for \( A \) is:

\[ A = \text{M}^{-1} \text{L}^{-1} \text{T}^{0}. \]

### Final Answer

The dimensional formula of \( A \) is:

\[ \text{M}^{-1} \text{L}^{-1} \text{T}^{0}. \]