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 $w$ .Neglect the thickness of the balloon when it is filled with air. Also neglect the difference in the densities of air inside $\&$ outside the balloon. Then :
Diffcult
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The weight of the empty balloon is $\mathrm{W}_{1}$. The weight of the balloon filled with air should be $W_{1}+w$ but the balloon is immersed fully in the air, so a force of buoyancy must act on it so the apparent weight $\mathrm{W}_{2}<\mathrm{W}_{1}+\mathrm{w},$ 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 $\mathrm{W}_{2}=$ $\left(\mathrm{W}_{1}+\mathrm{w}\right)-\mathrm{w}=\mathrm{W}_{1} \cdot$ Hence $(\mathrm{a})$ is correct. Other options are not true.
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