A heavy particle is tied to the end A of a string of the length`1.6` m. Its other end O is fixed. It revolves as a conical pendulum with the string making `60^(@)` with the horizontal. Then,

Options

A. its period of revolution is `(4pi)/(7)s`
B. the tension is the string is `(1)/(sqrt3)` times the weight of the particle
C. the speed of the particle is `2.8sqrt3ms^(-1)`
D. the centripetal acceleration of the particle is `(9.8)/(sqrt3)ms^(-2)`

Explanation

To solve the problem of a heavy particle revolving as a conical pendulum, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
The length of the string (L) is given as 1.6 m, and the angle (θ) with the horizontal is 60°. The angle with the vertical will therefore be 30° (since 90° - 60° = 30°).

### Step 2: Resolve Forces Acting on the Particle
The forces acting on the particle are:
1. The tension (T) in the string.
2. The weight (mg) of the particle acting downwards.

We can resolve the tension into two components:
- The vertical component: \( T \cos(30°) \)
- The horizontal component: \( T \sin(30°) \)

### Step 3: Set Up the Equations of Motion
1. **Vertical Forces**: The vertical component of tension balances the weight of the particle.
\[ T \cos(30°) = mg \quad \text{(Equation 1)} \]

2. **Horizontal Forces**: The horizontal component of tension provides the centripetal force necessary for circular motion.
\[ T \sin(30°) = \frac{mv^2}{r} \quad \text{(Equation 2)} \]
where \( r \) is the radius of the circular path.

### Step 4: Calculate the Radius of the Circular Path
The radius \( r \) can be calculated using the length of the string and the angle:
\[ r = L \sin(60°) = 1.6 \sin(60°) = 1.6 \cdot \frac{\sqrt{3}}{2} = 0.8\sqrt{3} \, \text{m} \]

### Step 5: Substitute and Simplify
From Equation 1:
\[ T = \frac{mg}{\cos(30°)} = \frac{mg}{\frac{\sqrt{3}}{2}} = \frac{2mg}{\sqrt{3}} \]

Substituting \( T \) into Equation 2:
\[ \frac{2mg}{\sqrt{3}} \sin(30°) = \frac{mv^2}{r} \]
Since \( \sin(30°) = \frac{1}{2} \):
\[ \frac{2mg}{\sqrt{3}} \cdot \frac{1}{2} = \frac{mv^2}{0.8\sqrt{3}} \]
This simplifies to:
\[ \frac{mg}{\sqrt{3}} = \frac{mv^2}{0.8\sqrt{3}} \]

### Step 6: Cancel Mass and Solve for Speed
Cancel \( m \) from both sides:
\[ \frac{g}{\sqrt{3}} = \frac{v^2}{0.8\sqrt{3}} \]
Cross-multiplying gives:
\[ v^2 = 0.8g \]
Thus,
\[ v = \sqrt{0.8g} \]
Substituting \( g \approx 9.8 \, \text{m/s}^2 \):
\[ v = \sqrt{0.8 \cdot 9.8} \approx \sqrt{7.84} \approx 2.8 \, \text{m/s} \]

### Step 7: Calculate the Time Period
The time period \( T \) of revolution is given by:
\[ T = \frac{2\pi}{\omega} \]
where \( \omega = \frac{v}{r} \):
\[ \omega = \frac{2.8}{0.8\sqrt{3}} = \frac{2.8}{1.3856} \approx 2.02 \, \text{rad/s} \]
Thus,
\[ T = \frac{2\pi}{2.02} \approx 3.11 \, \text{s} \]

### Summary of Results
- Speed of particle \( v \approx 2.8 \, \text{m/s} \)
- Time period \( T \approx 3.11 \, \text{s} \)