An object of mass 10kg is attached to roof by string whirled round a horizontal circle of radius 4 and and 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 find the tension in the string when an object of mass 10 kg is whirled in a horizontal circle at an angle of 30 degrees to the vertical, we can follow these steps:

### Step 1: Identify the forces acting on the object
The forces acting on the object are:
1. The weight of the object (mg) acting downwards.
2. The tension (T) in the string acting at an angle of 30 degrees to the vertical.

### Step 2: Resolve the tension into components
The tension can be resolved into two components:
- The vertical component \( T \cos(30^\circ) \)
- The horizontal component \( T \sin(30^\circ) \)

### Step 3: Apply the equilibrium condition in the vertical direction
Since the object is in equilibrium in the vertical direction (no vertical motion), we can set up the following equation:
\[ T \cos(30^\circ) - mg = 0 \]
This implies:
\[ T \cos(30^\circ) = mg \]

### Step 4: Substitute the known values
Given:
- Mass \( m = 10 \, \text{kg} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

Substituting these values into the equation:
\[ T \cdot \frac{\sqrt{3}}{2} = 10 \cdot 9.8 \]
\[ T \cdot \frac{\sqrt{3}}{2} = 98 \]

### Step 5: Solve for tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = \frac{98 \cdot 2}{\sqrt{3}} = \frac{196}{\sqrt{3}} \]

### Step 6: Calculate the numerical value of \( T \)
Now, we can calculate the value of \( T \):
\[ T \approx \frac{196}{1.732} \approx 113.5 \, \text{N} \]
Rounding this gives us approximately \( 114 \, \text{N} \).

### Final Answer
The tension in the string is approximately \( 114 \, \text{N} \).

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