Question

A hot air balloon is carrying some passengers, and a few sandbags of mass $1\text{kg}$ each so that its total mass is $480\text{kg}$. Its effective volume giving the balloon its buoyancy is $V$. The balloon is floating at an equilibrium height of $100\text{m}$. When $N$ number of sandbags are thrown out, the balloon rises to a new equilibrium height close to $150\text{m}$ with its volume $V$ remaining unchanged. If the variation of the density of air with height $h$ from the ground is $\rho \left(h\right)={\rho }_0{e}^{\frac{h}{h_o}}$, where ${\rho }_0=1.25{\text{kg m}}^{-3}$ and $h_0=6000\text{m}$, the value of $N$ is

Answer 1

Applying equilibrium condition for the balloon at both height levels,
$480\times g=V{\rho }_{1}g$
$(480-N)g=V{\rho }_{2}g$
$\frac{480-N}{480}=\frac{{\rho }_2}{{\rho }_1}$
$\Rightarrow (1-\frac{N}{480})=\frac{e^{-h_2/h_0}}{e^{-h_1/h_0}}=e^{\frac{h_1-h_2}{h_0}}$
$\Rightarrow (1-\frac{N}{480})=e^{-\frac{50}{6000}}$
Using expansion for $e^{-x}=1-x$ and neglecting higher order powers
$\Rightarrow 1-\frac{N}{480}=1-\frac{50}{6000}\Rightarrow N=\frac{50\times 480}{6000}=4$