Q166
General SciencePhysics
A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be $W_1$. If the bird flies about inside the cage, the reading of the spring balance becomes $W_2$. Which of the following is true?
- a.$W_1 > W_2$
- b.$W_1 < W_2$
- c.$W_1 = W_2$
- d.$W_1 = \dfrac{1}{W_2}$
Answer: (A) $W_1 > W_2$
The cage is made of wire, so it is open to the air. While the bird sits on its perch, its weight is carried by the cage and the balance reads cage plus bird ($W_1$). When it flies, its weight is supported by the air; the downward push it gives the air passes out through the open wires instead of pressing on the floor of the cage, so the balance reads less: $W_1>W_2$. If the bird were in a sealed, airtight box, the air inside would pass the bird's weight on to the box, and the average reading would not change ($W_1=W_2$). This open-cage versus closed-box contrast is a favourite exam trap.