  Question

Consider a spherical gaseous cloud of mass density $$\rho (r)$$ in a free space where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre 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$$=$$ universal gravitational constant)

A
K6πr2m2G  B
Kπr2m2G  C
3Kπr2m2G  D
K2πr2m2G  Solution

The correct option is D $$\dfrac{K}{2\pi r^2m^2G}$$$$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$$$$=\dfrac{2}{r}\left(\dfrac{1}{2}mv^2\right)$$$$\Rightarrow \dfrac{GMm}{r^2}=\dfrac{2K}{r}$$$$\Rightarrow M=\dfrac{2Kr}{Gm}$$$$\Rightarrow dM=\dfrac{2K}{Gm}dr$$$$\Rightarrow 4\pi r^2dr\rho =\dfrac{2K}{Gm}dr$$$$\therefore \rho =\dfrac{K}{2\pi Gmr^2}$$$$\Rightarrow \dfrac{\rho}{m}=\dfrac{K}{2\pi Gm^2r^2}$$ Alternative:$$\dfrac{GM(r)}{r^2}=\dfrac{V^2}{r}$$Where $$M(r)=$$ total mass upto radius(r)$$\Rightarrow K=\dfrac{GMm}{2r}$$$$\Rightarrow M(r)=\dfrac{2Kr}{Gm}$$$$\Rightarrow dM(r)=\dfrac{2K}{Gm}dr=\rho dV=\rho 4\pi r^2dr$$$$\Rightarrow \rho =\dfrac{K}{G2\pi r^2m}$$$$\Rightarrow \dfrac{\rho}{m}=\dfrac{K}{2\pi Gm^2r^2}$$Correct option $$4$$. Physics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 