The linear momentum p of a particle is given as a function of time t as `p=At^2+Bt+C`.The dimensions of constant B are

Options

A. `[ML^(-1)T^(-1)]`
B. `[ML^(-1)T^(-2)]`
C. `[MLT^(-2)]`
D. `[MLT^(-1)]`

Answer

To find the dimensions of the constant B in the equation of linear momentum \( p = At^2 + Bt + C \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation**: The equation given is \( p = At^2 + Bt + C \). Here, \( p \) represents linear momentum, and \( A \), \( B \), and \( C \) are constants.

2. **Identify the Dimensions of Linear Momentum**:
- Linear momentum \( p \) is defined as the product of mass and velocity.
- The dimension of mass is \( [M] \) and the dimension of velocity is \( [L][T]^{-1} \).
- Therefore, the dimensions of linear momentum \( p \) are:
\[ [p] = [M][L][T]^{-1} = M^1 L^1 T^{-1} \]

3. **Analyze Each Term in the Equation**:
- The term \( At^2 \) has dimensions \( [A][T]^2 \).
- The term \( Bt \) has dimensions \( [B][T] \).
- The term \( C \) is a constant and must also have dimensions of momentum, which is \( M^1 L^1 T^{-1} \).

4. **Equate the Dimensions**:
- Since all terms on the right-hand side must have the same dimensions as \( p \), we can set up the following equations:
\[ [A][T]^2 = [B][T] = [C] = M^1 L^1 T^{-1} \]

5. **Find the Dimensions of B**:
- From the equation \( [B][T] = M^1 L^1 T^{-1} \), we can express the dimensions of \( B \):
\[ [B] = \frac{[p]}{[T]} = \frac{M^1 L^1 T^{-1}}{T^1} = M^1 L^1 T^{-2} \]

6. **Conclusion**:
- The dimensions of the constant \( B \) are:
\[ [B] = M^1 L^1 T^{-2} \]

### Final Answer:
The dimensions of constant \( B \) are \( M^1 L^1 T^{-2} \).

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