JEE Challenger
More from Mechanical Properties of Fluids
PhyHardAdv 2020 • Paper 2

Number of Sandbags Thrown Out to Reach New Equilibrium Height

A hot air balloon containing passengers and several 1 kg1\text{ kg} sandbags has an initial combined mass of 480 kg480\text{ kg}. Its constant displacement volume providing buoyancy is VV. Initially, it floats in equilibrium at an altitude of 100 m100\text{ m}. After discarding NN sandbags, the balloon ascends to a new equilibrium altitude of approximately 150 m150\text{ m} while maintaining the same volume VV. Assuming the atmospheric air density varies with altitude hh as ρ(h)=ρ0e−hh0\rho(h) = \rho_0 e^{-\frac{h}{h_0}}, with ρ0=1.25 kg m−3\rho_0 = 1.25\text{ kg m}^{-3} and h0=6000 mh_0 = 6000\text{ m}, find the integer value of NN.

Official Numerical Answer4

Topics & Concepts

Mechanical Properties of FluidsFluid Dynamics

Step-by-Step Solution

To find the number of discarded sandbags NN, we apply the equilibrium condition where the buoyant force equals the total weight of the balloon at each altitude.

Since the volume VV remains constant, the mass of the balloon at any height hh is directly proportional to the atmospheric density ρ(h)=ρ0e−h/h0\rho(h) = \rho_0 e^{-h/h_0}.

Taking the ratio of the equilibrium masses at h1=100 mh_1 = 100\text{ m} and h2=150 mh_2 = 150\text{ m}: 480−N480=ρ(150)ρ(100)=e−150−1006000=e−1120\frac{480 - N}{480} = \frac{\rho(150)}{\rho(100)} = e^{-\frac{150 - 100}{6000}} = e^{-\frac{1}{120}}

Using the first-order approximation e−x≈1−xe^{-x} \approx 1 - x for small xx: 480−N480≈1−1120\frac{480 - N}{480} \approx 1 - \frac{1}{120}

Solving for NN: 480−N=480−4  ⟹  N=4480 - N = 480 - 4 \implies N = 4

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