Physics · Gravitation

JEE Advanced 2019 — Paper 1 — Question 1

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 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 $\rho(\mathrm{r})$ is constant in time, the particle number density $\mathrm{n}(\mathrm{r})=\rho(\mathrm{r}) / \mathrm{m}$ is[0pt] [ $G$ is universal gravitational constant]

Option A:
$\frac{3 \mathrm{~K}}{\pi \mathrm{r}^{2} \mathrm{~m}^{2} \mathrm{G}}$
Option B:
$\frac{\mathrm{K}}{2 \pi \mathrm{r}^{2} \mathrm{~m}^{2} \mathrm{G}}$
Correct
Option C:
$\frac{\mathrm{K}}{6 \pi \mathrm{r}^{2} \mathrm{~m}^{2} \mathrm{G}}$
Option D:
$\frac{K}{\pi r^{2} m^{2} G}$

Answer: B

Step-by-step solution

\begin{gathered} \rho=\frac{1}{4 \pi \mathrm{Gr}^{2}} \frac{\mathrm{~d}\left(\mathrm{gr}^{2}\right)}{\mathrm{dr}}…(i) \end{gathered}

Because $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg}$

so $\quad \frac{1}{2} \mathrm{mv}^{2}=\mathrm{K}=\frac{\mathrm{mgr}}{2}$

so from equation (i)

$\rho=\frac{1}{4 \pi \mathrm{Gr}^{2}} \frac{\mathrm{~d}}{\mathrm{dr}}\left(\frac{2 \mathrm{k}}{\mathrm{m}} \mathrm{r}\right)=\frac{\mathrm{K}}{2 \pi \mathrm{Gmr}^{2}}$

so, $\frac{\rho}{\mathrm{m}}=\frac{\mathrm{K}}{2 \pi \mathrm{Gm}^{2} \mathrm{r}^{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Gravitation
Topic: Gravitational Field and Gravity