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 ρ( h )=ρ0 e -( h / h0), where ρ0=1.25 kg m -3 and h 0=6000 m, the value of N is .

Question

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 ρ( h )=ρ0 e -( h / h0), where ρ0=1.25 kg m -3 and h 0=6000 m, the value of N is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(480-N/480)=(ρ2/ρ1) 1-(N/480)=e-((150-100/6000)) ⇒ 1-(N/480)=1-(5/600) N=4

480480−N​=ρ1​ρ2​​ 1−480N​=e−(6000150−100​) ⇒1−480N​=1−6005​ N=4

$\frac{480 - N}{480} = \frac{ρ _{2}}{ρ _{1}}$
$1 - \frac{N}{480} = e^{- (\frac{150 - 100}{6000})}$
$\Rightarrow 1 - \frac{N}{480} = 1 - \frac{5}{600}$
$N = 4$