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

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 \( F = at + bt^2 \), we will use the principle of dimensional homogeneity. Let's break down the solution step by step.

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

### Step 2: Analyze the term \( at \)
In the equation \( F = at + bt^2 \), the term \( at \) must have the same dimensions as \( F \). Therefore, we can write:
\[ [at] = [F] \]
Substituting the dimensions, we have:
\[ [a][t] = M^1 L^1 T^{-2} \]
Since the dimension of time \( t \) is \( [t] = T^1 \), we can rewrite the equation as:
\[ [a][T^1] = M^1 L^1 T^{-2} \]
To find the dimensions of \( a \), we rearrange the equation:
\[ [a] = \frac{M^1 L^1 T^{-2}}{T^1} = M^1 L^1 T^{-3} \]

### Step 3: Analyze the term \( bt^2 \)
Similarly, for the term \( bt^2 \), we also have:
\[ [bt^2] = [F] \]
Substituting the dimensions, we get:
\[ [b][t^2] = M^1 L^1 T^{-2} \]
The dimension of \( t^2 \) is \( [t^2] = T^2 \), so we can rewrite the equation as:
\[ [b][T^2] = M^1 L^1 T^{-2} \]
Rearranging gives us:
\[ [b] = \frac{M^1 L^1 T^{-2}}{T^2} = M^1 L^1 T^{-4} \]

### Step 4: Summarize the dimensions of \( a \) and \( b \)
Thus, we find the dimensions of \( a \) and \( b \):
\[ [a] = M^1 L^1 T^{-3} \]
\[ [b] = M^1 L^1 T^{-4} \]

### Final Answer
The dimensions of \( a \) are \( M^1 L^1 T^{-3} \) and the dimensions of \( b \) are \( M^1 L^1 T^{-4} \).

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