JEE Advanced 2019 · Paper 1 · Physics Q1

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Consider a spherical gaseous cloud of mass density $\rho(r)$ 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 $\rho(r)$ is constant in time, the particle number density $n(r) = \rho(r)/m$ is

[ $G$ is universal gravitational constant]

\dfrac{K}{2\pi r^2m^2G}

\dfrac{K}{\pi r^2m^2G}

\dfrac{3K}{\pi r^2m^2G}

\dfrac{K}{6\pi r^2m^2G}

Answer and worked solution

Official answer: A

Orbit condition. A particle at radius $r$ feels only the mass $M(r)$ inside its orbit (shell theorem):
$\frac{mv^2}r=\frac{GM(r)m}{r^2}\quad\text{and}\quad\tfrac12mv^2=K\implies M(r)=\frac{2Kr}{Gm}.$

Density.
$\frac{dM}{dr}=4\pi r^2\rho=\frac{2K}{Gm}\implies\rho=\frac{K}{2\pi r^2Gm}.$

Number density.
$n=\frac\rho m=\frac{K}{2\pi r^2m^2G}.$
(A)

Topics:Gravitational field and potential Circular motion dynamics

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JEE Advanced 2019 · Paper 1, Physics question 1