JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020
Answers
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 . The balloon is
floating at an equilibrium height of 100 m. When 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) = ρ₀e^{-h/h₀}, where ρ₀ = 1.25 kg/m³ and h₀ = 6000m, the value of N is ...
solution : at equilibrium,
Buoyancy force = weight
⇒Vρ₁g = mg
⇒Vρ₁ = 480.......(1)
where ρ₁ is the initial density of air.
given number of N sandbags are thrown out. then density of air becomes ρ₂
and net mass of balloon = (480kg × g - N × 1 kg × g)
= (480 - N)g
now Vρ₂g = (480 - N)g
⇒Vρ₂ = (480 - N) .......(2)
from equations (1) and (2) we get,
ρ₂/ρ₁ = (480 - N)/480 = 1 - N/480.....(2)
but from equation, ρ(h) = ρ₀e^{-h/h₀},
ρ₂/ρ₁ = e^{-h₂/h₀}/e^{-h₁/h₀} = e^{(h₁ - h₂)/h₀} = e^{(100 - 150)/6000} = e^{-50/6000}.......(3)
from equations (2) and (3) we get,
1 - N/480 = e^{-1/120}
from binomial expansion, e^{-1/120} = 1 - 1/120
⇒1 - N/480 = 1 - 1/120
N = 4
Therefore the value of N is 4.