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

Options

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

Answer

To solve the problem of finding the dimensions of constants \( a \) and \( b \) in the equation \( F = \frac{a}{t} + bt^2 \), we will follow these steps:

### Step 1: Identify the dimensions of force \( F \)
Force \( F \) can be expressed in terms of mass \( m \) and acceleration \( a \):
\[ F = m \cdot a \]
Acceleration \( a \) has dimensions of length per time squared, which can be expressed as:
\[ a = \frac{L}{T^2} \]
Thus, the dimensions of force \( F \) are:
\[ [F] = [m] \cdot [a] = [M] \cdot \left[\frac{L}{T^2}\right] = [M L T^{-2}] \]

### Step 2: Find the dimensions of \( a \)
From the equation \( F = \frac{a}{t} + bt^2 \), we can isolate \( a \):
\[ \frac{a}{t} = F - bt^2 \]
For the term \( \frac{a}{t} \) to have the same dimensions as \( F \), we can express the dimensions of \( a \):
\[ \left[\frac{a}{t}\right] = [F] \]
This means:
\[ \frac{[a]}{[T]} = [M L T^{-2}] \]
Rearranging gives:
\[ [a] = [M L T^{-2}] \cdot [T] = [M L T^{-1}] \]

### Step 3: Find the dimensions of \( b \)
Now, we analyze the term \( bt^2 \):
\[ bt^2 = F \]
Thus, we can express the dimensions of \( b \):
\[ [b] \cdot [T^2] = [F] \]
This implies:
\[ [b] = \frac{[F]}{[T^2]} = \frac{[M L T^{-2}]}{[T^2]} = [M L T^{-4}] \]

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

### Final Answer
The dimensions of \( a \) and \( b \) are respectively:
- \( a: [M L T^{-1}] \)
- \( b: [M L T^{-4}] \)