Q. 2

# The weight of an empty balloon on a spring balance is W_{1}. The weight becomes W_{2} when the balloon is filled with air. Let the weight of the air itself be ω. Neglect the thickness of the balloon when it is filled with air. Also neglect the difference in the densities of air inside and outside the balloon.

A. W_{2} = W_{1}

B. W_{2} = W_{1} + ω

C. W_{2} < W_{1} + ω

D. W_{2} > W_{1}

Answer :

Here, we need to take into the fact that the air filled balloon is immersed in “air”. Thus, a force of buoyancy will act on the balloon which will make the weight of the air filled balloon always less than the actual weight, W_{2} < W_{1} + ω.

As it is completely immersed and the densities of air are equal, the buoyant force of air will be equal to the weight of the air. This means, W_{2} = W_{1} + ω - ω = W_{1}. Thus option (A) and (C) are correct.

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