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JEE Advanced Physics Mechanical Properties of Fluids 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

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 remain unchanged. If the variation of the density of air with height  h from the ground is ρ h = ρ 0 e - h h 0 , where  ρ 0 = 1 . 25   kg   m - 3 and  h 0 = 6000   m , the value of  N is _________.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Initially m g = F b 480 × g = V ρ 0 e - 100 6000 g       ....(1) Finally ( 480 - m ) g = V ρ 0 e - 150 6000 g      .....(2) dividing 480 - m 480 = e - 1 40 + 1 60 m = 480 1 - e - 1 / 120 Hence e - 1 / 120 ≈ 1 - 1 120 1 - e - 1 / 110 = 1 120 m = 4   kg Hence, four sandbags are thrown.

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