JEE Advance - Physics (2020 - Paper 2 Offline - No. 3)
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
$$\rho \left( h \right) = {\rho _0}{e^{ - {h \over {{h_0}}}}}$$, where $$\rho $$0 = 1.25 kg m−3 and h0 = 6000 m, the value of N is _________.
$$\rho \left( h \right) = {\rho _0}{e^{ - {h \over {{h_0}}}}}$$, where $$\rho $$0 = 1.25 kg m−3 and h0 = 6000 m, the value of N is _________.
Respondre
4
Explicació
Weight = Upthrust
$$mg = {F_u} \Rightarrow 480 \times 10 = \rho Vg$$
$$480 \times 10 = {\rho _0}{e^{ - {h \over {{h_0}}}}} \Rightarrow 480 \times 10 = {\rho _0}{e^{ - {{100} \over {6000}}}}Vg$$ .... (i)
$$(480 - N \times 1)10 = \rho 'Vg$$
$$(480 - N)10 = {\rho _0}{e^{ - {{150} \over {6000}}}}Vg$$ .... (ii)
Dividing Eq. (i) by Eq. (ii), we get
$${{480} \over {480 - N}} = {e^{\left( {{{150 - 100} \over {6000}}} \right)}}$$
$${{480} \over {480 - N}} = {e^{{{50} \over {6000}}}} \Rightarrow {{480} \over {480 - N}} = {e^{{1 \over {120}}}}$$
N = 4
$$mg = {F_u} \Rightarrow 480 \times 10 = \rho Vg$$
$$480 \times 10 = {\rho _0}{e^{ - {h \over {{h_0}}}}} \Rightarrow 480 \times 10 = {\rho _0}{e^{ - {{100} \over {6000}}}}Vg$$ .... (i)
$$(480 - N \times 1)10 = \rho 'Vg$$
$$(480 - N)10 = {\rho _0}{e^{ - {{150} \over {6000}}}}Vg$$ .... (ii)
Dividing Eq. (i) by Eq. (ii), we get
$${{480} \over {480 - N}} = {e^{\left( {{{150 - 100} \over {6000}}} \right)}}$$
$${{480} \over {480 - N}} = {e^{{{50} \over {6000}}}} \Rightarrow {{480} \over {480 - N}} = {e^{{1 \over {120}}}}$$
N = 4