Given `F = (a//t) + bt^2` where F denotes force and t time. The diamensions of a and b are respectively :

Options

A. `[MLT^(-1)] and [MLT^(-4)]`
B. `[LT^(-1)] and [T^(-2)]`
C. `[T] and [T^(-2)]`
D. `[LT^(-2)] and [T^(-2)]`

Explanation

To solve the problem, we need to determine the dimensions of the constants \( a \) and \( b \) in the equation \( F = \frac{a}{t} + bt^2 \), where \( F \) represents force and \( t \) represents time.

### Step-by-Step Solution:

1. **Identify the dimensions of force (F)**:
The dimension of force \( F \) is given by:
\[ [F] = [M][L][T^{-2}] \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

2. **Analyze the term \( \frac{a}{t} \)**:
The term \( \frac{a}{t} \) must have the same dimensions as \( F \). Therefore:
\[ \left[\frac{a}{t}\right] = [F] \]
This implies:
\[ [a] = [F] \cdot [t] = [M][L][T^{-2}] \cdot [T] = [M][L][T^{-1}] \]
Thus, the dimensions of \( a \) are:
\[ [a] = [M][L][T^{-1}] \]

3. **Analyze the term \( bt^2 \)**:
Similarly, the term \( bt^2 \) must also have the same dimensions as \( F \):
\[ [bt^2] = [F] \]
This implies:
\[ [b] = \frac{[F]}{[t^2]} = \frac{[M][L][T^{-2}]}{[T^2]} = [M][L][T^{-4}] \]
Thus, the dimensions of \( b \) are:
\[ [b] = [M][L][T^{-4}] \]

4. **Final Result**:
The dimensions of \( a \) and \( b \) are:
- \( [a] = [M][L][T^{-1}] \)
- \( [b] = [M][L][T^{-4}] \)

### Summary of Dimensions:
- Dimensions of \( a \): \( [M][L][T^{-1}] \)
- Dimensions of \( b \): \( [M][L][T^{-4}] \)