A

Step 1: Understand the setup
The particle is released from point A and moves down the groove AB, which has a length L and makes an angle θ with the radius OA.

Step 2: Forces acting on the particle
The only forces acting on the particle are gravitational force and the normal force from the groove. The particle will accelerate down the groove due to the component of gravitational force acting along it.

Step 3: Determine acceleration
The acceleration a of the particle down the groove can be found using:
$$a = g imes ext{sin}( heta)$$
where g is acceleration due to gravity.

Step 4: Apply equations of motion
The initial velocity (u) of the particle at point A is 0 (it is released).

Using the second equation of motion, which states:
$$s = ut + \frac{1}{2} a t^2$$
where:
- s is the distance (L),
- u = 0,
- a = g imes ext{sin}( heta),
we can rewrite the equation to find time t:
$$L = 0 + \frac{1}{2} (g \times \text{sin}(\theta)) t^2$$
Therefore, we have:
$$L = \frac{1}{2} (g \times \text{sin}(\theta)) t^2$$

Step 5: Solve for time t
Rearranging the equation, we find:
$$t^2 = \frac{2L}{g \times \text{sin}(\theta)}$$
$$t = \sqrt{\frac{2L}{g \times \text{sin}(\theta)}}$$

Conclusion
The time taken by the particle to reach point B is given by the equation derived above. Thus, the answer is the expression for t as derived.