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← JEE Advanced 2020 · Paper 2
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JEE Advanced 2020 · Paper 2, Physics question 4

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 VV. The balloon is floating at an equilibrium height of 100 m. When NN number of sandbags are thrown out, the balloon rises to a new equilibrium height close to 150 m with its volume VV remaining unchanged. If the variation of the density of air with height hh from the ground is ρ(h)=ρ0e−hh0\rho(h) = \rho_0 e^{-\frac{h}{h_0}}, where ρ0=1.25 kg m−3\rho_0 = 1.25\ \mathrm{kg\,m^{-3}} and h0=6000h_0 = 6000 m, the value of NN is ______.

Answer and worked solution

Official answer: 4

At each equilibrium the buoyant force equals the weight, with the same volume VV:
ρ(100)V=480,ρ(150)V=480−N.\rho(100)V=480,\qquad\rho(150)V=480-N.

Dividing.
480−N480=e−50/6000  ⟹  N=480(1−e−1/120)≈480×0.0083=3.98.\frac{480-N}{480}=e^{-50/6000}\implies N=480\left(1-e^{-1/120}\right)\approx480\times0.0083=3.98.
Answer: 4