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QJEEAdv19Ph6230: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph6230 Solution: Consider a spherical gaseous cloud of mass density (r) in free…

Consider a spherical gaseous cloud of mass density ρ\rho (r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ\rho (r) is constant in time, the particle number density n(r) = ρ\rho (r)/m is [G is universal gravitational constant]

Multiple Choice Options

  • A K6πr2m2G{K \over {6\pi {r^2}{m^2}G}}
  • B Kπr2m2G{K \over {\pi {r^2}{m^2}G}}
  • C 3Kπr2m2G{3K \over {\pi {r^2}{m^2}G}}
  • D K2πr2m2G{K \over {2\pi {r^2}{m^2}G}}

Step-by-Step Verified Solution

JEE Advanced 2019 Paper 1 Offline Physics - Gravitation Question 10 English Explanation

Gravitational force = Centripetal force of the earth

GMmr2=mv2r{{GMm} \over {{r^2}}} = {{m{v^2}} \over r}

( ∵\because M = total mass from 0 to r)

= 2r(12mv2){2 \over r}\left( {{1 \over 2}m{v^2}} \right)

⇒GMmr2=2Kr⇒M=2KrGm\Rightarrow {{GMm} \over {{r^2}}} = {{2K} \over r} \Rightarrow M = {{2Kr} \over {Gm}}

∵\because (12mv2=K)\left( {{1 \over 2}m{v^2} = K} \right)

Differentiate on both sides, we get

⇒dM=2KGmdr\Rightarrow dM = {{2K} \over {Gm}}dr

⇒4πr2drρ=2KGmdr\Rightarrow 4\pi {r^2}dr\rho = {{2K} \over {Gm}}dr

( ∵\because volume = mass ×\times density)

∴\therefore ρ=K2πGmr2\rho = {K \over {2\pi Gm{r^2}}}

∴\therefore ρm=K2πr2m2G{\rho \over m} = {K \over {2\pi {r^2}{m^2}G}} ∵\because (ρm=volume)\left( {{\rho \over m} = volume} \right)

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