We use Archimides' principle, which states that the buoyant force on an object is always directed upward and its magnitude is equal to the weight of the fluid displaced. Recall that a fluid can be either a gas or a liquid. So, let $V$ denote that volume of the balloon. We neglect the volume of the suspended mass. The buoyant force is then $1.29\;\mathrm{kg}\mathrm{/}\mathrm{m}^3 \cdot V \cdot g$ . This must equal the weight of the helium-filled balloon plus the weight of the $300 \mbox{ kg}$ mass. That is
$\begin{align*}1.29\;\mathrm{kg}\mathrm{/}\mathrm{m}^3 \cdot V \cdot g = 300 \mbox{ kg} \cdot g + 0.18 \;\mathrm{kg}\mathrm{/...$
This can be easily solved for $V$ :
$\begin{align*}V = \frac{300 \mbox{ kg}}{1.29 \;\mathrm{kg}\mathrm{/}\mathrm{m}^3- 0.18 \;\mathrm{kg}\mathrm{/}\mathrm{m}^3} =...$
Therefore, answer (D) is correct.

Solution to 2001 Problem 56