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(B) At equilibrium,the buoyant force equals the weight of the balloon: $Mg = V \rho(h) g$,which simplifies to $M = V \rho(h)$.For the initial state at

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$4$

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(B) At equilibrium,the buoyant force equals the weight of the balloon: $Mg = V \rho(h) g$,which simplifies to $M = V \rho(h)$.For the initial state at $h_1 = 100 m$: $480 = V \rho_0 e^{-\frac{100}{6000}}$.For the final state at $h_2 = 150 m$ after removing $N$ sandbags: $(480 - N) = V \rho_0 e^{-\frac{150}{6000}}$.Dividing the two equations: $\frac{480 - N}{480} = \frac{e^{-\frac{150}{6000}}}{e^{-\frac{100}{6000}}} = e^{-\frac{50}{6000}}$.Using the approximation $e^{-x} \approx 1 - x$ for small $x$: $1 - \frac{N}{480} \approx 1 - \frac{50}{6000}$.Therefore,$\frac{N}{480} = \frac{50}{6000} = \frac{1}{120}$.$N = \frac{480}{120} = 4$.

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