If `v-(A)/(t)+Bt^(2)+Ct^(3)` where v is velocity, t is time and A, B and C are constants, then the dimensional formula of B is

Options

A. `[M^(0)LT^(0)]`
B. `[ML^(0)T^(0)]`
C. `[M^(0)L^(0)T]`
D. `[M^(0)LT^(-3)]`

Answer

To find the dimensional formula of \( B \) in the equation \( v - \frac{A}{t} + Bt^2 + Ct^3 = 0 \), we will use the principle of homogeneity. This principle states that we can only add or subtract quantities that have the same dimensions.

### Step-by-step Solution:

1. **Identify the dimensions of each term**:
- The term \( v \) represents velocity. The dimensional formula for velocity is:
\[ [v] = L^1 T^{-1} \]
- The term \( \frac{A}{t} \) involves \( A \) and time \( t \). The dimensional formula for time is:
\[ [t] = T^1 \]
- Therefore, the dimensional formula for \( \frac{A}{t} \) is:
\[ \left[\frac{A}{t}\right] = \frac{[A]}{[t]} = \frac{[A]}{T^1} \]

2. **Equate the dimensions**:
Since \( v \) and \( \frac{A}{t} \) must have the same dimensions, we can write:
\[ L^1 T^{-1} = \frac{[A]}{T^1} \]
Rearranging gives:
\[ [A] = L^1 T^0 = L^1 \]
(This shows that \( A \) has the dimension of length.)

3. **Analyze the term \( Bt^2 \)**:
The dimensional formula for \( t^2 \) is:
\[ [t^2] = T^2 \]
Thus, the dimensional formula for \( Bt^2 \) is:
\[ [Bt^2] = [B][t^2] = [B]T^2 \]

4. **Set the dimensions equal**:
Since \( v \) and \( Bt^2 \) must also have the same dimensions, we can equate them:
\[ L^1 T^{-1} = [B] T^2 \]

5. **Solve for the dimensional formula of \( B \)**:
Rearranging gives:
\[ [B] = \frac{L^1 T^{-1}}{T^2} = L^1 T^{-3} \]

6. **Express the final dimensional formula**:
The dimensional formula of \( B \) is:
\[ [B] = M^0 L^1 T^{-3} \]
This means that \( B \) has dimensions of length per time cubed, with no mass dimension.

### Final Answer:
The dimensional formula of \( B \) is \( M^0 L^1 T^{-3} \).