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 for linear momentum \( p = At^2 + Bt + C \), we will follow these steps:

### Step 1: Understand the equation
The equation given is \( p = At^2 + Bt + C \). Here, \( p \) represents momentum, and \( A \), \( B \), and \( C \) are constants. Since the terms on the right-hand side are being added together, they must all have the same dimensions as \( p \).

### Step 2: Identify the dimensions of momentum
The dimension of momentum \( p \) is given by the product of mass and velocity. Therefore, we can express the dimensions of momentum as:
\[ [p] = [M][L][T^{-1}] \]
where:
- \( [M] \) is the dimension of mass,
- \( [L] \) is the dimension of length,
- \( [T] \) is the dimension of time.

### Step 3: Analyze the terms in the equation
Now, we will analyze the terms in the equation:
1. The term \( At^2 \) has dimensions of momentum, so \( A \) must have dimensions that, when multiplied by \( t^2 \), yield the dimensions of momentum.
2. The term \( Bt \) must also have dimensions of momentum, which means \( B \) must have dimensions that, when multiplied by \( t \), yield the dimensions of momentum.
3. The term \( C \) is a constant and must also have dimensions of momentum.

### Step 4: Determine the dimensions of B
Since \( Bt \) must have the same dimensions as \( p \), we can write:
\[ [B][T] = [p] \]
Rearranging gives us:
\[ [B] = \frac{[p]}{[T]} \]
Substituting the dimension of momentum:
\[ [B] = \frac{[M][L][T^{-1}]}{[T]} = [M][L][T^{-2}] \]

### Step 5: Conclusion
Thus, the dimensions of the constant \( B \) are:
\[ [B] = [M][L][T^{-2}] \]