The dimensions of `(a)/(b)` in the relatin F = ax + bt are

Options

A. `LT^(-1)`
B. `L^(-1)T`
C. LT
D. `L^(-1)T^(-1)`

Answer

To find the dimensions of \( \frac{a}{b} \) in the relation \( F = ax + bt \), we will apply the principle of homogeneity. This principle states that the dimensions on both sides of an equation must be the same, especially when terms are added or subtracted.

### Step-by-Step Solution:

1. **Identify the dimensions of \( F \)**:
- 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. **Identify the dimensions of \( x \)**:
- The dimension of displacement \( x \) is \( [x] = [L] \).

3. **Write the dimensions of the term \( ax \)**:
- Since \( ax \) is a term in the equation, its dimensions must equal the dimensions of \( F \).
- Therefore, \( [ax] = [a][x] = [a][L] \).
- Setting the dimensions equal gives us:
\[ [F] = [a][L] \implies [a] = \frac{[F]}{[L]} = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}] \]

4. **Identify the dimensions of \( t \)**:
- The dimension of time \( t \) is \( [t] = [T] \).

5. **Write the dimensions of the term \( bt \)**:
- Similarly, since \( bt \) is also a term in the equation, its dimensions must also equal the dimensions of \( F \).
- Therefore, \( [bt] = [b][t] = [b][T] \).
- Setting the dimensions equal gives us:
\[ [F] = [b][T] \implies [b] = \frac{[F]}{[T]} = \frac{[M L T^{-2}]}{[T]} = [M L T^{-3}] \]

6. **Calculate the dimensions of \( \frac{a}{b} \)**:
- Now we can find the dimensions of \( \frac{a}{b} \):
\[ \frac{[a]}{[b]} = \frac{[M T^{-2}]}{[M L T^{-3}]} = \frac{[M T^{-2}]}{[M]} \cdot \frac{1}{[L T^{-3}]} = \frac{T^{-2}}{L T^{-3}} = \frac{T^{-2} \cdot T^{3}}{L} = \frac{T^{1}}{L} \]
- Thus, the dimensions of \( \frac{a}{b} \) are \( [L^{-1} T] \).

### Final Answer:
The dimensions of \( \frac{a}{b} \) are \( [L^{-1} T] \).