The expression `S = a/b (1 - e^(-bt))` is dimensionally true. The dimensional formula of a and b are respectively, (if S is the displacement at any instant t,)

Options

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

Answer

To solve the problem, we need to analyze the given expression \( S = \frac{a}{b} (1 - e^{-bt}) \) and determine the dimensional formulas for \( a \) and \( b \) based on the condition that the expression is dimensionally true.

### Step-by-Step Solution:

1. **Identify the dimensions of \( S \)**:
- The problem states that \( S \) is the displacement at any instant \( t \). The dimensional formula for displacement is:
\[ [S] = L \]
where \( L \) represents length.

2. **Understand the term \( 1 - e^{-bt} \)**:
- The term \( e^{-bt} \) is an exponential function. For the exponent to be dimensionless, the quantity \( bt \) must also be dimensionless. Therefore:
\[ [b][t] = 1 \]
This implies that the dimensions of \( b \) must be the reciprocal of the dimensions of time \( t \):
\[ [b] = T^{-1} \]
where \( T \) represents time.

3. **Substituting \( b \) into the expression for \( S \)**:
- Now we can substitute the dimensions of \( b \) back into the expression for \( S \):
\[ [S] = \frac{[a]}{[b]} = \frac{[a]}{T^{-1}} = [a] \cdot T \]
Since we know that \( [S] = L \), we can equate the two:
\[ L = [a] \cdot T \]

4. **Solving for the dimensions of \( a \)**:
- Rearranging the equation gives us:
\[ [a] = \frac{L}{T} \]
Thus, the dimensional formula for \( a \) is:
\[ [a] = LT^{-1} \]

5. **Final Result**:
- We have determined the dimensional formulas:
- For \( a \): \( LT^{-1} \)
- For \( b \): \( T^{-1} \)

### Summary of Dimensional Formulas:
- \( [a] = LT^{-1} \)
- \( [b] = T^{-1} \)