JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Fluid Mechanics

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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) = ρ_{0} e^{- \frac{h}{h _{0}}}

, where

ρ_{0} = 1.25 kg m^{- 3}

and

h_{0} = 6000

m, the value of

N

is ____.

Answer: 4

Solution

At equilibrium, buoyant force equals weight:

At height

h_{1} = 100

480 g = V ρ_{1} g

At height

h_{2} = 150

(480 - N) g = V ρ_{2} g

Dividing:

\frac{480 - N}{480} = \frac{ρ _{2}}{ρ _{1}} = \frac{ρ _{0} e ^{- h_{2} / h_{0}}}{ρ _{0} e ^{- h_{1} / h_{0}}} = e^{(h_{1} - h_{2}) / h_{0}} = e^{- 50/6000}

Using

e^{x} \approx 1 + x

for small x:

1 - \frac{N}{480} = 1 - \frac{50}{6000}

\frac{N}{480} = \frac{50}{6000}

N = \frac{50 \times 480}{6000} = 4

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Integer answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Fluid Mechanics

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