The velocity of a body is given by the equation
`v = (b)/(t) + ct^(2) + dt^(3)`.
The dimensional formula of `b` is

Options

A. `[M^(0)LT^(0)]`
B. `[ML^(0)T^(0)]`
C. `[M^(0)L^(0)T]`
D. `[MLT^(-1)]`

Explanation

To find the dimensional formula of \( b \) in the equation \( v = \frac{b}{t} + ct^2 + dt^3 \), we will follow these steps:

### Step 1: Understand the equation
The equation given is:
\[ v = \frac{b}{t} + ct^2 + dt^3 \]
where \( v \) is the velocity of the body, \( b \) is a constant, \( c \) and \( d \) are also constants, and \( t \) is time.

### Step 2: Identify the dimensions
We know that:
- The dimension of velocity \( v \) is given by:
\[ [v] = [L^1 T^{-1}] \]
where \( L \) is the dimension of length and \( T \) is the dimension of time.

### Step 3: Analyze each term in the equation
1. **Term \( \frac{b}{t} \)**:
- The dimension of time \( t \) is:
\[ [t] = [T^1] \]
- Therefore, the dimension of \( \frac{b}{t} \) can be expressed as:
\[ \left[\frac{b}{t}\right] = \frac{[b]}{[T^1]} = [b][T^{-1}] \]

2. **Term \( ct^2 \)**:
- The dimension of \( ct^2 \) is:
\[ [ct^2] = [c][T^2] \]
- Since \( ct^2 \) must also have the dimension of velocity, we can equate:
\[ [c][T^2] = [L^1 T^{-1}] \]

3. **Term \( dt^3 \)**:
- The dimension of \( dt^3 \) is:
\[ [dt^3] = [d][T^3] \]
- Similarly, we can equate:
\[ [d][T^3] = [L^1 T^{-1}] \]

### Step 4: Equate dimensions
Since all terms on the right side of the equation must have the same dimensions as \( v \), we can set the dimensions of \( \frac{b}{t} \) equal to the dimensions of \( v \):
\[ [b][T^{-1}] = [L^1 T^{-1}] \]

### Step 5: Solve for the dimension of \( b \)
From the equation above, we can isolate \( [b] \):
\[ [b] = [L^1 T^{-1}] \cdot [T^1] = [L^1] \]
Thus, the dimensional formula of \( b \) is:
\[ [b] = [L^1] \]

### Final Answer
The dimensional formula of \( b \) is:
\[ [L] \]