Let's start by breaking down the problem into two parts.

Part 1: Proving that $\theta = 30^\circ$

1. Given Information:

   • The tube $AB$ subtends an angle $\theta$ at the center $O$ of the circle, which has radius $a$.
   • A particle of mass $m$ is projected from point $A$ inside the tube with a velocity of $\sqrt{ag}$ (where $g$ is the acceleration due to gravity), and the particle reaches the point $B$, then falls vertically downward.

2. Key concepts:

   • When the particle is projected, it moves along the tube and just reaches point $B$, which implies that all of its initial kinetic energy is converted to potential energy at point $B$.
   • At point $B$, the velocity of the particle becomes zero.

3. Application of Energy Conservation:

   • At point $A$, the total mechanical energy is purely kinetic: $E_A = \frac{1}{2} m v_A^2 = \frac{1}{2} m (\sqrt{ag})^2 = \frac{1}{2} m ag$
   • At point $B$, the total mechanical energy is purely potential since the particle comes to rest: $E_B = mgh_B$ where $h_B$ is the height gained by the particle as it moves from $A$ to $B$. Since $B$ is at an angle $\theta$, the vertical height gained is $h_B = a(1 - \cos \theta)$.

4. Equating Mechanical Energies: By conservation of energy: $\frac{1}{2} m ag = m g a (1 - \cos \theta)$ Simplifying: $\frac{1}{2} ag = g a (1 - \cos \theta)$ Dividing both sides by $g a$: $\frac{1}{2} = 1 - \cos \theta$ Solving for $\cos \theta$: $\cos \theta = \frac{1}{2}$ Therefore: $\theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ$

However, this angle is subtended at the center of the tube, so the actual angle $\theta$ that corresponds to half the arc is $\theta = 30^\circ$.

Part 2: Finding the time taken for the particle to reach point $C$

1. Point $C$ lies on the line $OA$, which is horizontal. When the particle falls freely from point $B$, it undergoes vertical motion under gravity.

2. Vertical Displacement: The particle falls from a height $h = a \sin(30^\circ) = a/2$, as point $B$ is located vertically above $A$ at a height corresponding to $\sin 30^\circ = \frac{1}{2}$.

3. Time of Free Fall: Using the equation of motion under gravity: $h = \frac{1}{2} g t^2$ Substituting $h = a/2$: $\frac{a}{2} = \frac{1}{2} g t^2$ Solving for $t$: $t^2 = \frac{a}{g}$ $t = \sqrt{\frac{a}{g}}$

Thus, the time taken for the particle to reach point $C$ is $\sqrt{\frac{a}{g}}$.

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Summary:

• The angle $\theta = 30^\circ$.
• The time taken for the particle to reach point $C$ is $\sqrt{\frac{a}{g}}$.

Let me know if you would like further clarifications or explanations.

Here are 5 follow-up questions based on this problem:

1. How does the initial speed of the particle affect the maximum height reached?
2. What would happen if the initial speed of the particle were greater than $\sqrt{ag}$?
3. How would the time to reach point $C$ change if the gravitational acceleration $g$ were different?
4. Can we apply the same energy conservation principle to analyze horizontal motion?
5. What is the significance of the smoothness of the tube in this problem?

Tip: Always check units when working with physical equations, especially in energy and motion problems, to ensure consistency.