A hot-air balloon of total mass 480 kg floats at 100 m. Its buoyant volume V is constant. After N sandbags of 1 kg each are dropped it floats at 150 m. Air density varies as ρ(h)=ρ₀ e − h h ο e^{\frac{-h}{h_{\omicron}}} e h ο − h , with ρ₀=1.25 kg m⁻³ and h₀=6000 m. Find N.
Physics · Fluid Mechanics
JEE Advanced 2020

Question

A hot-air balloon of total mass 480 kg floats at 100 m. Its buoyant volume V is constant. After N sandbags of 1 kg each are dropped it floats at 150 m. Air density varies as ρ(h)=ρ₀ e − h h ο e^{\frac{-h}{h_{\omicron}}} e h ο − h , with ρ₀=1.25 kg m⁻³ and h₀=6000 m. Find N.

Physics · Fluid Mechanics

JEE Advanced 2020

G1 Physics · Accepted Answer

Step 1: At equilibrium buoyancy equals total weight, so ρair(h)Vg = Mg. The same V applies before and after sand is thrown. Step 2: Using ρ(h)=ρ₀ e ( − h / h 0 ) , e^(-h/h₀), e ( − h / h 0 ​ ) , the mass reduction needed from 100 m to 150 m is about 4 kg. Final answer: 4.