A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If is the pressure at , then the correct option(s) is(are) [2015]

Question

A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If $P (r)$ is the pressure at $r (r < R)$ , then the correct option(s) is(are) [2015]

Smart Achievers · Accepted Answer

(2, 3)

$Let us consider an elemental mass d m shown in the shaded portion of the figure.$

$Here, P (4 π r^{2}) - (P + d P) (4 π r^{2}) = \frac{G M r}{R^{3}} ρ (4 π r^{2}) d r$

$∴$ $- \int_{0}^{P} d P = \frac{G M ρ}{R^{3}} \int_{r}^{R} r d r$

$∴$ $P = \frac{G M ρ}{2 R^{3}} [R^{2} - r^{2}]$

$∴$ $\frac{P (r = 3 R / 4)}{P (r = 2 R / 3)} = \frac{[R^{2} - \frac{9 R^{2}}{16}]}{[R^{2} - \frac{4 R^{2}}{9}]} = \frac{\frac{7 R^{2}}{16}}{\frac{5 R^{2}}{9}} = \frac{63}{80}$

$and \frac{P (r = 3 R / 5)}{P (r = 2 R / 5)} = \frac{[R^{2} - \frac{9 R^{2}}{25}]}{[R^{2} - \frac{4 R^{2}}{25}]} = \frac{16}{21}$

Solution

Q.
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If $P (r)$ is the pressure at $r (r < R)$ , then the correct option(s) is(are) [2015]

1 $P (r = 0) = 0$

2 $\frac{P (r = 3 R / 4)}{P (r = 2 R / 3)} = \frac{63}{80}$

3 $\frac{P (r = 3 R / 5)}{P (r = 2 R / 5)} = \frac{16}{21}$

4 $\frac{P (r = R / 2)}{P (r = R / 3)} = \frac{20}{27}$

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Solution

Q. A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r (r<R), then the correct option(s) is(are) [2015]

1 P(r=0)=0

2 P(r=3R/4)P(r=2R/3)=6380

3 P(r=3R/5)P(r=2R/5)=1621

4 P(r=R/2)P(r=R/3)=2027