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QJEEAdv20Phd7c1: JEE Advanced 2020 Physics - Class 11 JEEAdv20Phd7c1 Solution: A hot air balloon is carrying some passengers, and a few…

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 remaining unchanged. If the variation of the density of air with height h from the ground is
ρ(h)=ρ0e−hh0\rho \left( h \right) = {\rho _0}{e^{ - {h \over {{h_0}}}}} , where ρ\rho 0 = 1.25 kg m−3 and h0 = 6000 m, the value of N is _________.

Step-by-Step Verified Solution

Weight = Upthrust

mg=Fu⇒480×10=ρVgmg = {F_u} \Rightarrow 480 \times 10 = \rho Vg

480×10=ρ0e−hh0⇒480×10=ρ0e−1006000Vg480 \times 10 = {\rho _0}{e^{ - {h \over {{h_0}}}}} \Rightarrow 480 \times 10 = {\rho _0}{e^{ - {{100} \over {6000}}}}Vg .... (i)

(480−N×1)10=ρ′Vg(480 - N \times 1)10 = \rho 'Vg

(480−N)10=ρ0e−1506000Vg(480 - N)10 = {\rho _0}{e^{ - {{150} \over {6000}}}}Vg .... (ii)

Dividing Eq. (i) by Eq. (ii), we get

480480−N=e(150−1006000){{480} \over {480 - N}} = {e^{\left( {{{150 - 100} \over {6000}}} \right)}}

480480−N=e506000⇒480480−N=e1120{{480} \over {480 - N}} = {e^{{{50} \over {6000}}}} \Rightarrow {{480} \over {480 - N}} = {e^{{1 \over {120}}}}

N = 4

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