Question

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

Answer 1

Step1: Given data.

Weight of each sand bag$=1kg$

Total mass$=480kg$

Effective volume of balloon$=V$

Equilibrium height of floating balloon $h_1=100m$

New equilibrium height of floating balloon after rises, $h_2=150m$

Variation of the density of air with height $\rho ={\rho }_0{e}^{-\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}$

Where ${\rho }_0=1.25kg{m}^{-3}$and $h_0=6000m$

Step2: Finding number of sandbags $N$ thrown out after balloon rises.

Applying equilibrium in above figure:

${\rho }_{1}Vg=480g$

${\rho }_0{e}^{-\raisebox{1ex}{$h_1$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}Vg=480g$ $\left[\because \rho ={\rho }_0{e}^{-\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}\right]$

${\rho }_0{e}^{\raisebox{1ex}{$h_1$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}V=480$ ….$\left(i\right)$

Now,

${\rho }_0{e}^{-\raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}V=480-N$ $\left[\because \rho ={\rho }_0{e}^{\raisebox{1ex}{$-h$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}\right]$ ….$\left(ii\right)$

Dividing equation $\left(ii\right)$ by $\left(i\right)$ we get.

$\frac{e^{-\raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}}{e^{-\raisebox{1ex}{$h_1$}\!\left/ \!\raisebox{-1ex}{$h_0$}\right.}}=\frac{480-N}{480}$

$e^{\frac{h_1-h_2}{h_0}}=\left(1-\frac{N}{480}\right)$

$\frac{N}{480}=1-e^{-\frac{1}{120}}$

$N=480\left(1-0.9917\right)$

$N=4\left(approx\right)$

Hence, the value of $N$ is $4$.