An object of mass 10kg is whirled round a horizontal circle of radius 4m by a revolving string inclined `30^(@)` to vertical. If the uniform speed of the object is `5 m//sec,` the tension in the string (approximately) is

Options

A. 720 N
B. 960N
C. 114N
D. 125N

Answer

To solve the problem of finding the tension in the string when an object is whirled in a horizontal circle, we can follow these steps:

### Step 1: Identify the Forces Acting on the Object
The forces acting on the object are:
1. The gravitational force (weight) acting downwards, \( F_g = mg \).
2. The tension in the string, \( T \), which has two components:
- A vertical component \( T \cos(30^\circ) \) that balances the weight.
- A horizontal component \( T \sin(30^\circ) \) that provides the centripetal force necessary for circular motion.

### Step 2: Write the Equations for Forces
1. **Vertical Forces**: The vertical component of the tension must balance the weight of the object:
\[ T \cos(30^\circ) = mg \]
2. **Horizontal Forces**: The horizontal component of the tension provides the centripetal force:
\[ T \sin(30^\circ) = \frac{mv^2}{r} \]

### Step 3: Substitute Known Values
Given:
- Mass \( m = 10 \, \text{kg} \)
- Radius \( r = 4 \, \text{m} \)
- Speed \( v = 5 \, \text{m/s} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equations:

1. For the vertical forces:
\[ T \cos(30^\circ) = 10 \times 10 \]
\[ T \cos(30^\circ) = 100 \]

2. For the horizontal forces:
\[ T \sin(30^\circ) = \frac{10 \times 5^2}{4} \]
\[ T \sin(30^\circ) = \frac{10 \times 25}{4} = \frac{250}{4} = 62.5 \]

### Step 4: Solve for Tension \( T \)
Now we can express \( T \) from both equations:

1. From the vertical forces:
\[ T = \frac{100}{\cos(30^\circ)} \]
Knowing that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ T = \frac{100}{\frac{\sqrt{3}}{2}} = \frac{200}{\sqrt{3}} \approx 115.47 \, \text{N} \]

2. From the horizontal forces:
\[ T = \frac{62.5}{\sin(30^\circ)} \]
Knowing that \( \sin(30^\circ) = \frac{1}{2} \):
\[ T = \frac{62.5}{\frac{1}{2}} = 125 \, \text{N} \]

### Final Answer
Thus, the tension in the string is approximately \( T \approx 125 \, \text{N} \).