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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2020 · Paper 2 · Physics · Mechanical Properties of Fluids

A hot air balloon is carrying some passengers, and a few sandbags...

A hot air balloon is carrying some passengers, and a few sandbags of mass 1kg1kg each so that its total mass is 480kg480kg. Its effective volume giving the balloon its buoyancy is VV. The balloon is floating at an equilibrium height of 100m100m. When NN number of sandbags are thrown out, the balloon rises to a new equilibrium height close to 150m150m with its volume VV remain unchanged. If the variation of the density of air with height hh from the ground is ρ(h)=ρ0e−hh0,ρ\left(h\right) = {ρ}_{0}{e}^{- \dfrac{h}{{h}_{0}}}, where ρ0=1.25kgm−3{ρ}_{0} = 1 . 25kg{m}^{- 3} and h0=6000m,{h}_{0} = 6000m, the value of NN is _________.

Correct answer

0

Step-by-step solution

Initially
mg=Fbmg = {F}_{b}

480×g=V(ρ0e−1006000)g480 \times g = V\left({ρ}_{0}{e}^{- \dfrac{100}{6000}}\right)g ....(1)

Finally (480−m)g=V(ρ0e−1506000)g( 480 - m ) g = V\left({ρ}_{0}{e}^{- \dfrac{150}{6000}}\right)g .....(2)

dividing 480−m480=e−140+160\dfrac{480 - m}{480} = {e}^{- \dfrac{1}{40} + \dfrac{1}{60}}

m=480(1−e−1/120)m = 480\left(1 - {e}^{- 1 / 120}\right)

Hence e−1/120≈1−1120{e}^{- 1 / 120} \approx 1 - \dfrac{1}{120}

1−e−1/110=11201 - {e}^{- 1 / 110} = \dfrac{1}{120}

m=4kgm = 4kg

Hence, four sandbags are thrown.

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