A particle of mass 200 g is attached to an ideal string of length 1.30 m whose upper end is fixed to ceiling. The particle is made to revolve in a horizontal circle of radius 50 cm. The tension in the string is (g = 10 m/`s^2`)

Options

A. `frac{13}{6}N`
B. `frac{11}{6}N`
C. `frac{26}{5]N`
D. Zero

Answer

To find the tension in the string when the particle is revolving in a horizontal circle, we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (since 1 g = 0.001 kg)
- Length of the string, \( L = 1.30 \, \text{m} \)
- Radius of the horizontal circle, \( r = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the vertical and horizontal components of forces
When the particle is revolving in a horizontal circle, the tension in the string has two components:
1. A vertical component that balances the weight of the particle.
2. A horizontal component that provides the centripetal force required for circular motion.

### Step 3: Calculate the weight of the particle
The weight \( W \) of the particle can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 0.2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 2 \, \text{N} \]

### Step 4: Use the geometry of the system
In this scenario, we can use the geometry of the string. The vertical component of the tension \( T \) must balance the weight of the particle, and the horizontal component provides the centripetal force.

The vertical component of the tension is given by:
\[ T_v = T \cdot \cos(\theta) \]
And the horizontal component is given by:
\[ T_h = T \cdot \sin(\theta) \]

Where \( \theta \) is the angle the string makes with the vertical.

### Step 5: Find \( \theta \)
Using the right triangle formed by the string, the vertical line, and the radius:
\[ \cos(\theta) = \frac{L - r}{L} = \frac{1.30 - 0.5}{1.30} = \frac{0.8}{1.30} \approx 0.6154 \]
\[ \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - (0.6154)^2} \approx \sqrt{1 - 0.3787} \approx \sqrt{0.6213} \approx 0.789 \]

### Step 6: Set up the equations
From the vertical forces:
\[ T \cdot \cos(\theta) = W \]
Substituting the known values:
\[ T \cdot 0.6154 = 2 \]
Thus,
\[ T = \frac{2}{0.6154} \approx 3.24 \, \text{N} \]

### Step 7: Conclusion
The tension in the string is approximately \( 3.24 \, \text{N} \).

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