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. . . . .
IIT 2020, Medium
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$480 \times g=v \rho_1 g$
$(480-N) g=v \rho_2 g$
$\frac{480-N}{480}=\frac{\rho_2}{\rho_1}$
$\left(1-\frac{N}{480}\right)=\frac{e^{-h_2 / h_0}}{e^{-h_1 / h_0}}=e^{\frac{h_1-h_2}{h_0}}=e^{-\frac{50}{6000}}$
$1-\frac{N}{480}=1-\frac{50}{6000} \Rightarrow N=\frac{50 \times 480}{6000}=4$
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