A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be W1 . If the bird flies about inside the cage, the reading of the spring balance is W2 . Which of the following is true

The correct option is W₁ > W₂.

Let us analyze the situation step-by-step to understand the physical principles involved:

1. Bird sitting inside the cage (Reading W₁):
When the bird is resting on the floor or perch of the cage, the spring balance supports both the weight of the cage and the weight of the bird. Therefore, the reading is the sum of both weights:

$W_1=W_{\text{cage}}+W_{\text{bird}}$
where $W_{\text{cage}}$ is the weight of the cage and $W_{\text{bird}}$ is the weight of the bird.

2. Bird flying inside the wire cage (Reading W₂):
When the bird starts flying, it is no longer in direct physical contact with the cage. To lift itself and fly, the bird's wings push the surrounding air downwards.
In a closed, airtight container, this downward-moving air would exert a force on the bottom of the container equal to the bird's weight, keeping the total weight reading unchanged. However, in this scenario, the bird is inside a wire cage, which has open spaces and is not sealed. The downward current of air produced by the bird's wings passes through the wire mesh and escapes into the outer atmosphere. As a result, the downward force of the air is not transferred to the bottom of the cage.

3. Comparison:
Since the weight of the bird is no longer transmitted to the spring balance, the scale will register a lower weight, which is essentially just the weight of the cage:

$W_2<W_1$
This is equivalent to:

$W_1>W_2$