A sparrow flying in air sits on a stretched telegraph wire If weight of the sparrow is `W` which of the following is true about the tension `T` produced in the wire ?

Options

A. T=W
B. `TgtW`
C. `TltW`
D. T=0

Answer

To determine the relationship between the tension \( T \) in the telegraph wire and the weight \( W \) of the sparrow, we can analyze the forces acting on the sparrow when it sits on the wire.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Sparrow**:
- The weight of the sparrow, denoted as \( W \), acts downward due to gravity.
- The tension \( T \) in the wire acts upward at an angle \( \theta \) from the horizontal.

2. **Analyze the Tension Components**:
- The tension \( T \) can be resolved into two components:
- Horizontal component: \( T \cos \theta \)
- Vertical component: \( T \sin \theta \)

3. **Balance the Forces**:
- In the vertical direction, the upward force must balance the downward force (weight of the sparrow):
\[ 2T \sin \theta = W \]
- Here, \( 2T \sin \theta \) accounts for the vertical components of tension from both sides of the wire.

4. **Rearranging the Equation**:
- From the equation \( 2T \sin \theta = W \), we can express \( T \):
\[ T = \frac{W}{2 \sin \theta} \]

5. **Understanding the Relationship**:
- Since \( \sin \theta \) is always between 0 and 1 for small angles \( \theta \), \( \frac{1}{\sin \theta} \) will always be greater than or equal to 1.
- Therefore, \( T \) must be greater than \( \frac{W}{2} \).

6. **Conclusion**:
- Since \( T \) is proportional to \( \frac{W}{2 \sin \theta} \) and \( \sin \theta \) is always positive and less than or equal to 1, we conclude that:
\[ T > \frac{W}{2} \]
- Thus, the total tension \( T \) in the wire is greater than the weight \( W \) of the sparrow.

### Final Answer:
The tension \( T \) produced in the wire when the sparrow sits on it is greater than the weight \( W \) of the sparrow.