Question

The weight of an empty balloon on a spring balance is W₁. The weight becomes W₂ when the balloon is filled with air. Let the weight of the air itself be w. 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) W₂ = W₁. 

(b) W₂ = W₁ + w. 

(c) W₂ < W₁ + w. 

(d) W₂ > W₁.

Answer 1

(a) W₂ = W₁.

(c) W₂ < W₁+ W.

EXPLANATION:  

The weight of the empty balloon is W₁. The weight of the balloon filled with air should be W₁+w but the balloon is immersed fully in the air, so a force of buoyancy must act on it so the apparent weight W₂ < W₁+w, hence (c) is correct. Since the densities of air inside and outside are same so the weight of the displaced air by the balloon will also be w. Thus the apparent weight of air-filled balloon W₂ = (W₁+w)-w = W₁. Hence (a) is correct. Other options are not true.