A force F is given by F = `at+ bt^(2)` , where t is time. The dimensions of a and b are

Options

A. `["MLT"^(-3)]and["MLT"^(-4)]`
B. `["MLT"^(-4)]and["MLT"^(-3]`
C. `["MLT"^(-1)]and["MLT"^(-2)]`
D. `["MLT"^(-2)]and["MLT"^(0)]`

Answer

To find the dimensions of the constants \( a \) and \( b \) in the equation \( F = at + bt^2 \), where \( F \) is force and \( t \) is time, we will follow these steps:

### Step 1: Understand the dimensions of force
The dimensional formula for force \( F \) is given by:
\[ [F] = MLT^{-2} \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

### Step 2: Analyze the term \( at \)
In the equation \( F = at + bt^2 \), we can analyze the term \( at \). The dimensions of \( t \) (time) are:
\[ [t] = T^1 \]
Thus, the dimensions of \( at \) can be expressed as:
\[ [at] = [a][t] = [a]T^1 \]
Since \( at \) must have the same dimensions as \( F \), we can equate:
\[ [a][T] = MLT^{-2} \]
From this, we can solve for the dimensions of \( a \):
\[ [a] = \frac{MLT^{-2}}{T^1} = MLT^{-3} \]

### Step 3: Analyze the term \( bt^2 \)
Next, we analyze the term \( bt^2 \). The dimensions of \( t^2 \) are:
\[ [t^2] = T^2 \]
Thus, the dimensions of \( bt^2 \) can be expressed as:
\[ [bt^2] = [b][t^2] = [b]T^2 \]
Again, since \( bt^2 \) must have the same dimensions as \( F \), we can equate:
\[ [b][T^2] = MLT^{-2} \]
From this, we can solve for the dimensions of \( b \):
\[ [b] = \frac{MLT^{-2}}{T^2} = ML T^{-4} \]

### Summary of Results
- The dimensional formula of \( a \) is:
\[ [a] = MLT^{-3} \]
- The dimensional formula of \( b \) is:
\[ [b] = MLT^{-4} \]

### Final Answer
Thus, the dimensions of \( a \) and \( b \) are:
- \( a: MLT^{-3} \)
- \( b: MLT^{-4} \)

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