If force (F) is given by `F = Pt^(-1) + alpha t,` where t is time. The unit of P is same as that of

Options

A. velocity
B. displacement
C. acceleration
D. momentum

Answer

To solve the problem, we need to analyze the given equation for force and determine the unit of \( P \) in the expression \( F = Pt^{-1} + \alpha t \).

### Step-by-step Solution:

1. **Identify the given equation**:
The equation provided is:
\[ F = Pt^{-1} + \alpha t \]
Here, \( F \) is the force, \( P \) is a variable we need to analyze, \( t \) is time, and \( \alpha \) is another variable.

2. **Write the unit of force**:
The unit of force \( F \) in the SI system is given by:
\[ [F] = \text{kg} \cdot \text{m/s}^2 = \text{MLT}^{-2} \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

3. **Analyze the terms in the equation**:
Since \( F \) is expressed as the sum of \( Pt^{-1} \) and \( \alpha t \), both terms must have the same dimensions as \( F \).

4. **Determine the dimension of \( Pt^{-1} \)**:
The term \( Pt^{-1} \) can be expressed as:
\[ [P] \cdot [t^{-1}] = [P] \cdot T^{-1} \]
Since this term must have the same dimension as force, we can set up the equation:
\[ [P] \cdot T^{-1} = \text{MLT}^{-2} \]

5. **Solve for the dimension of \( P \)**:
Rearranging the equation gives:
\[ [P] = \text{MLT}^{-2} \cdot T = \text{MLT}^{-1} \]
Thus, the dimension of \( P \) is:
\[ [P] = \text{MLT}^{-1} \]

6. **Identify the physical quantity corresponding to the dimension of \( P \)**:
The dimension \( \text{MLT}^{-1} \) corresponds to momentum, as momentum is defined as mass times velocity:
\[ \text{Momentum} = \text{mass} \times \text{velocity} = \text{M} \cdot \text{L} \cdot \text{T}^{-1} \]

7. **Conclusion**:
Therefore, the unit of \( P \) is the same as that of momentum.

### Final Answer:
The unit of \( P \) is the same as that of momentum.