An object of mass 10kg is attached to roof by string whirled round a horizontal circle of radius 4 and `30^@` to the vertical. The tension in the string (approximately) is

Options

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

Answer

To solve the problem of finding the tension in the string when an object of mass 10 kg is attached to the roof and whirled in a horizontal circle at an angle of 30 degrees to the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Object**:
- The object has a weight (W) acting downwards, given by \( W = mg \), where \( m = 10 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
- The tension (T) in the string acts at an angle of 30 degrees to the vertical.

2. **Calculate the Weight of the Object**:
\[ W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

3. **Resolve the Tension into Components**:
- The vertical component of the tension (T) can be expressed as:
\[ T_y = T \cos(30^\circ) \]
- The horizontal component of the tension (T) is:
\[ T_x = T \sin(30^\circ) \]

4. **Apply the Condition for Vertical Forces**:
- Since there is no vertical acceleration, the sum of the vertical forces must equal zero:
\[ T \cos(30^\circ) - mg = 0 \]
- Rearranging gives:
\[ T \cos(30^\circ) = mg \]

5. **Substituting Known Values**:
- Substitute \( mg \) and \( \cos(30^\circ) \):
\[ T \cos(30^\circ) = 98 \, \text{N} \]
- We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ T \left(\frac{\sqrt{3}}{2}\right) = 98 \]

6. **Solve for Tension (T)**:
\[ T = \frac{98}{\frac{\sqrt{3}}{2}} = \frac{98 \times 2}{\sqrt{3}} = \frac{196}{\sqrt{3}} \]
- To approximate \( \sqrt{3} \approx 1.732 \):
\[ T \approx \frac{196}{1.732} \approx 113.5 \, \text{N} \]

7. **Final Result**:
- The tension in the string is approximately \( 114 \, \text{N} \).