A hot air balloon is carrying some passengers, and a few sandbags of mass 1 kg each so that its total mass is 480 kg. Its effective volume giving the balloon its buoyancy is 𝑉. The balloon is floating at an equilibrium height of 100 m. When 𝑁 number of sandbags are thrown out, the balloon rises to a new equilibrium height close to 150 m with its volume 𝑉 remaining unchanged. If the variation of the density of air with height ℎ from the ground is $$\rho(h) = \rho_0 e^{-\frac{h}{h_0}}$$, where $$\rho_0 = 1.25$$ kg m$$^{-3}$$ and $$h_0 = 6000$$ m, the value of N is ________________.

The upward buoyant force acting on the balloon at any height $$h$$ is

$$B = \rho(h)\,V\,g$$

At equilibrium this buoyant force equals the total weight of the balloon-system.

Initial equilibrium at $$h_1 = 100\;{\rm m}$$
Total mass (balloon + passengers + sandbags) $$M_1 = 480\;{\rm kg}$$.

$$\rho(h_1)\,V\,g = M_1\,g \quad \Rightarrow \quad \rho(h_1)\,V = 480 \;-(1)$$

Final equilibrium at $$h_2 = 150\;{\rm m}$$ after dropping $$N$$ sandbags
Each sandbag has mass $$1\;{\rm kg}$$, so the new mass is $$M_2 = 480 - N\;{\rm kg}$$.

$$\rho(h_2)\,V\,g = M_2\,g \quad \Rightarrow \quad \rho(h_2)\,V = 480 - N \;-(2)$$

Divide $$(2)$$ by $$(1)$$ to eliminate $$V$$:

$$\frac{480 - N}{480} = \frac{\rho(h_2)}{\rho(h_1)} \;-(3)$$

The air-density profile is given as $$\rho(h) = \rho_0\,e^{-h/h_0}$$, so

$$\frac{\rho(h_2)}{\rho(h_1)} = e^{-(h_2-h_1)/h_0}$$

Here, $$h_2 - h_1 = 150 - 100 = 50\;{\rm m}$$ and $$h_0 = 6000\;{\rm m}$$, hence

$$e^{-(h_2-h_1)/h_0} = e^{-50/6000} = e^{-0.008333\ldots}$$

For a small exponent we may expand: $$e^{-x} \approx 1 - x + \frac{x^2}{2}$$.
With $$x = 0.008333$$,

$$e^{-0.008333} \approx 1 - 0.008333 + \frac{(0.008333)^2}{2} \approx 0.9917$$

Substitute this value into $$(3)$$:

$$\frac{480 - N}{480} \approx 0.9917$$

$$480 - N \approx 480 \times 0.9917 = 476.016$$

$$N \approx 480 - 476.016 = 3.984$$

Since $$N$$ must be an integer, one must drop $$N = 4$$ sandbags.

Answer: 4