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A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be W. If the bird flies about inside the cage, the reading of the spring balance is ${{W}_{2}}$. Which of the following is true?
(A) ${{W}_{1}}={{W}_{2}}$
(B) ${{W}_{1}}>{{W}_{2}}$
(C) ${{W}_{1}}<{{W}_{2}}$
(D) Nothing definite can be predicted

Last updated date: 20th Jun 2024
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Hint: As we know that the spring balance actually measures the force. i.e. $W=mg$.Where $W$is the weight of the bird in the two respective cases when the bird is flying and it is not found the statement which is true.

Complete step by step solution Reading shown on the spring balance is weight of bird plus weight of cage. When a bird starts flying, the normal force exerted by the bird on the car becomes zero.
When a bird flies his weight is carried by air but the case is made of wire so an additional force of weight of bird acts on the cage so reading on the cage is less when the bird is flying.

Additional information It is verified that for the bird to stay in the air, the wings must push down on the air under the cage. When the cage is air tight, the air exerts an equal force on the floor of the cage. So, the net force down on the scale will remain constant.

Note: Be careful about the concept of balance of spring force with weight of body. Using these concepts, we can find the weight of the body(in our case bird) while flying and sitting in a wire cage.