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JEE Advanced 2019
Paper-1 2019
Question
Consider a spherical gaseous cloud of mass density r(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 r(r) is constant in time, the particle number density n(r) = r(r)/m is :
[G is universal gravitational constant]
Select the correct option:
A
$\frac{\mathrm{K}}{\pi \mathrm{r}^2 \mathrm{~m}^2 \mathrm{G}}$
B
$\frac{\mathrm{K}}{6 \pi \mathrm{r}^2 \mathrm{~m}^2 \mathrm{G}}$
C
$\frac{3 \mathrm{~K}}{\pi \mathrm{r}^2 \mathrm{~m}^2 \mathrm{G}}$
D
$\frac{K}{2 \pi r^2 m^2 G}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Advance
Physics
Easy
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