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Consider a spherical gaseous cloud of mass density $ρ (r)$ 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 $ρ (r)$ is constant with time. The particle number density $n (r)$ = $ρ (r) / m$ is:
[ G is universal gravitational constant]

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$\frac{K}{π r^{2} m^{2} G}$

$\frac{3 K}{π r^{2} m^{2} G}$

$\frac{K}{6 π r^{2} m^{2} G}$

$\frac{K}{2 π r^{2} m^{2} G}$

answer is D.

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Detailed Solution

M(r) = mass of gas in a sphere of radius r.

For any particle of mass m located at radius r, $F_{g} = F_{c p}$

$\Rightarrow \frac{G M (r) m}{r^{2}} = \frac{m v^{2}}{r}$

$\Rightarrow G M (r) m = 2 (K) r [h e r e K = \frac{1}{2} m v^{2}]$

$\Rightarrow M (r) = \frac{2 K}{G m} r$
On differentiating w.r.t r,

$\frac{d M}{d r} = \frac{2 K}{G m}$ ................ (1)
Here dM = mass of gas in a spherical shell of radius r and thickness dr.

$∴ d M = [ρ (r)] (d V) = [ρ (r)] 4 π r^{2} d r$ ............... (2)

Substitute (2) in (1)

$\frac{[ρ (r)] 4 π r^{2} d r}{d r} = \frac{2 K}{G m}$

$∴ ρ (r) = \frac{K}{2 π r^{2} G m}$

$\Rightarrow n (r) = \frac{ρ (r)}{m} = \frac{K}{2 π r^{2} G m^{2}}$

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