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 the constant \( B \) in the equation \( v - \frac{A}{t} + Bt^2 + Ct^3 = 0 \), we need to ensure that all terms in the equation have the same dimensions, as they are being added or subtracted.

### Step-by-Step Solution:

1. **Identify the dimensions of velocity \( v \)**:
- The dimensional formula for velocity is given by:
\[ [v] = \frac{[L]}{[T]} = L^1 T^{-1} \]

2. **Analyze the term \( \frac{A}{t} \)**:
- Since \( t \) is time, its dimensional formula is:
\[ [t] = T^1 \]
- Therefore, the dimensional formula of \( \frac{A}{t} \) is:
\[ \left[\frac{A}{t}\right] = \frac{[A]}{[T]} = [A] T^{-1} \]

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

4. **Analyze the term \( Ct^3 \)**:
- The dimensional formula for \( t^3 \) is:
\[ [t^3] = T^3 \]
- Therefore, the dimensional formula of \( Ct^3 \) is:
\[ [Ct^3] = [C] T^3 \]

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

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

7. **Final result**:
- The dimensional formula of \( B \) is:
\[ [B] = M^0 L^1 T^{-3} \]

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

---