A bird is sitting on stretched telephone wires. If its weight is W then the additional tension produced by it in the wires will be

The correct answer is T > W.

When a bird of weight W sits on a stretched telephone wire, the wire sags slightly at the point of contact, forming a small angle θ with the horizontal on each side. Let's analyze the forces involved using the equilibrium condition.

Step 1: Understand the geometry

Before the bird sits, the wire is nearly horizontal. When the bird lands, the wire bends downward at that point, creating a "V"-shape. Let θ be the small angle each side of the wire makes with the horizontal at the point where the bird sits.

Step 2: Draw the free-body diagram

At the point where the bird sits, two tension forces act — one along each side of the wire, both directed upward and outward at angle θ above the horizontal. The weight W acts vertically downward.

Step 3: Apply vertical equilibrium

For the bird (and the contact point) to be in equilibrium, the net vertical component of the two tension forces must balance the weight W:

Tsin\theta +Tsin\theta =W

2Tsin\theta =W

Solving for T:

T=\frac{W}{2sin\theta }

Step 4: Analyze the value of sinθ

The telephone wire is stretched (taut), which means the sag is very small. This means the angle θ is a very small angle, much less than 90°. Therefore:

sin\theta <1

which means:

2sin\theta <2

and therefore:

T=\frac{W}{2sin\theta }>\frac{W}{2}

Step 5: Compare T with W

For a stretched wire, the angle of sag is very small. As θ becomes small, sinθ becomes very small (approaching 0), which makes the denominator 2sin\theta much less than 1. This means:

T=\frac{W}{2sin\theta }>>W

Even without the extreme small-angle case, as long as sin\theta <\frac{1}{2} (i.e., θ < 30°), we have T > W.

Physical Insight: This is a classic result showing that a single concentrated vertical load on a nearly horizontal wire produces a tension far exceeding the load itself. The wire cannot remain perfectly horizontal under any vertical load — even a tiny weight forces a sag, and the shallow angle means the wire must carry enormous tension to provide even a small upward vertical component. This is why telephone wires are never perfectly horizontal and why tightrope walkers put enormous stress on the supporting structure.

Conclusion: Since 2sin\theta <1 for any realistic sag angle (θ < 30°), we get:

T=\frac{W}{2sin\theta }>W

Therefore, the additional tension produced in the wire is always greater than the weight W of the bird, confirming the answer T > W.