Statement - 1 : A particle moves with a constant velocity parallel to the x-axis. Its angular momentum w.r.t. the origin will remains constant.
Because
Statement - 2 : Angular momentum is defined as `vec(L) = vec(r) xx vec(P)`

Options

A. Statement - 1 is true, statement - 2 is true, statement - 2 is correct explanation for statement - 1
B. Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for statement - 1
C. statement - 1 is true, statement - 2 is false
D. statement - 1 is false, statement - 2 is true.

Answer

To solve the question regarding the statements about angular momentum, we will analyze each statement step by step.

### Step 1: Understanding Statement 1
Statement 1 claims that a particle moving with a constant velocity parallel to the x-axis has a constant angular momentum with respect to the origin.

- **Analysis**:
- A particle moving with a constant velocity means that its linear momentum is constant.
- The angular momentum \( \vec{L} \) of a particle with respect to a point (in this case, the origin) is given by the formula:
\[ \vec{L} = \vec{r} \times \vec{P} \]
where \( \vec{r} \) is the position vector from the origin to the particle, and \( \vec{P} \) is the linear momentum of the particle.

### Step 2: Understanding Statement 2
Statement 2 states that angular momentum is defined as \( \vec{L} = \vec{r} \times \vec{P} \).

- **Analysis**:
- This is indeed the correct definition of angular momentum.
- The position vector \( \vec{r} \) is the distance from the origin to the particle, and \( \vec{P} \) is the momentum of the particle, which is given by \( \vec{P} = m\vec{v} \) (where \( m \) is mass and \( \vec{v} \) is velocity).

### Step 3: Evaluating the Statements
- Since the particle is moving parallel to the x-axis with constant velocity, the position vector \( \vec{r} \) changes in time but the direction of \( \vec{P} \) remains constant (as both are parallel).
- The angular momentum \( \vec{L} \) will remain constant because the cross product \( \vec{r} \times \vec{P} \) will not change in magnitude or direction as long as the particle moves along the x-axis and the origin remains fixed.

### Conclusion
- **Statement 1** is correct: The angular momentum remains constant.
- **Statement 2** is also correct: It correctly defines angular momentum.

Thus, both statements are true, and Statement 2 is a correct explanation of Statement 1.

### Final Answer
Both statements are correct, and Statement 2 is the correct explanation for Statement 1.