JEE Challenger
More from Mechanical Properties of Fluids
PhyHardAdv 2015 • Paper 2

Pressure Distribution inside a Fluid Sphere under Self Gravity

A spherical body of radius RR consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r)P(r) is the pressure at distance rr (r<R)(r < R), then choose the correct option(s).

Options

A

P(r=0)=0P(r = 0) = 0

B

P(r=3R/4)P(r=2R/3)=6380\frac{P(r = 3R/4)}{P(r = 2R/3)} = \frac{63}{80}

Correct
C

P(r=3R/5)P(r=2R/5)=1621\frac{P(r = 3R/5)}{P(r = 2R/5)} = \frac{16}{21}

Correct
D

P(r=R/2)P(r=R/3)=2027\frac{P(r = R/2)}{P(r = R/3)} = \frac{20}{27}

Topics & Concepts

Mechanical Properties of FluidsFluid DynamicsGravitation

Step-by-Step Solution

To determine the pressure distribution P(r)P(r) inside a fluid sphere of radius RR and uniform density ρ\rho in equilibrium under its own gravity, we consider a concentric spherical fluid element of radius rr and thickness drdr.

1. Gravitational Field inside the Sphere

The mass of the fluid enclosed within a sphere of radius rr is: M(r)=43πr3ρM(r) = \frac{4}{3}\pi r^3 \rho

The gravitational field strength g(r)g(r) at a radial distance rr (r≤Rr \le R) is given by: g(r)=GM(r)r2=G(43πr3ρ)r2=43πGρrg(r) = \frac{G M(r)}{r^2} = \frac{G \left(\frac{4}{3}\pi r^3 \rho\right)}{r^2} = \frac{4}{3}\pi G \rho r

2. Condition for Hydrostatic Equilibrium

For the fluid element of thickness drdr to be in equilibrium, the hydrostatic equation gives: dPdr=−ρg(r)\frac{dP}{dr} = -\rho g(r)

Substituting g(r)g(r) into the equation: dPdr=−43πGρ2r\frac{dP}{dr} = -\frac{4}{3}\pi G \rho^2 r

3. Integrating to find P(r)P(r)

Assuming the pressure at the outer surface is zero, i.e., P(R)=0P(R) = 0, we integrate from rr to RR: ∫P(r)0dP=−43πGρ2∫rRr′ dr′\int_{P(r)}^{0} dP = -\frac{4}{3}\pi G \rho^2 \int_{r}^{R} r' \, dr'

−P(r)=−43πGρ2[R2−r22]-P(r) = -\frac{4}{3}\pi G \rho^2 \left[ \frac{R^2 - r^2}{2} \right]

P(r)=23πGρ2(R2−r2)P(r) = \frac{2}{3}\pi G \rho^2 \left(R^2 - r^2\right)

Let C=23πGρ2C = \frac{2}{3}\pi G \rho^2. Thus, the pressure distribution is: P(r)=C(R2−r2)P(r) = C \left(R^2 - r^2\right)

4. Evaluating the Options

  • Option (A): At the center (r=0r = 0): P(r=0)=CR2≠0P(r = 0) = C R^2 \neq 0 Therefore, Option (A) is incorrect.

  • Option (B): Calculating the pressures at r=3R4r = \frac{3R}{4} and r=2R3r = \frac{2R}{3}: P(r=3R4)=C[R2−(3R4)2]=CR2(1−916)=716CR2P\left(r = \frac{3R}{4}\right) = C \left[ R^2 - \left(\frac{3R}{4}\right)^2 \right] = C R^2 \left(1 - \frac{9}{16}\right) = \frac{7}{16} C R^2

    P(r=2R3)=C[R2−(2R3)2]=CR2(1−49)=59CR2P\left(r = \frac{2R}{3}\right) = C \left[ R^2 - \left(\frac{2R}{3}\right)^2 \right] = C R^2 \left(1 - \frac{4}{9}\right) = \frac{5}{9} C R^2

    Taking their ratio: P(r=3R4)P(r=2R3)=71659=7×916×5=6380\frac{P\left(r = \frac{3R}{4}\right)}{P\left(r = \frac{2R}{3}\right)} = \frac{\frac{7}{16}}{\frac{5}{9}} = \frac{7 \times 9}{16 \times 5} = \frac{63}{80} Therefore, Option (B) is correct.

  • Option (C): Calculating the pressures at r=3R5r = \frac{3R}{5} and r=2R5r = \frac{2R}{5}: P(r=3R5)=C[R2−(3R5)2]=CR2(1−925)=1625CR2P\left(r = \frac{3R}{5}\right) = C \left[ R^2 - \left(\frac{3R}{5}\right)^2 \right] = C R^2 \left(1 - \frac{9}{25}\right) = \frac{16}{25} C R^2

    P(r=2R5)=C[R2−(2R5)2]=CR2(1−425)=2125CR2P\left(r = \frac{2R}{5}\right) = C \left[ R^2 - \left(\frac{2R}{5}\right)^2 \right] = C R^2 \left(1 - \frac{4}{25}\right) = \frac{21}{25} C R^2

    Taking their ratio: P(r=3R5)P(r=2R5)=16252125=1621\frac{P\left(r = \frac{3R}{5}\right)}{P\left(r = \frac{2R}{5}\right)} = \frac{\frac{16}{25}}{\frac{21}{25}} = \frac{16}{21} Therefore, Option (C) is correct.

  • Option (D): Calculating the pressures at r=R2r = \frac{R}{2} and r=R3r = \frac{R}{3}: P(r=R2)=C[R2−(R2)2]=34CR2P\left(r = \frac{R}{2}\right) = C \left[ R^2 - \left(\frac{R}{2}\right)^2 \right] = \frac{3}{4} C R^2

    P(r=R3)=C[R2−(R3)2]=89CR2P\left(r = \frac{R}{3}\right) = C \left[ R^2 - \left(\frac{R}{3}\right)^2 \right] = \frac{8}{9} C R^2

    Taking their ratio: P(r=R2)P(r=R3)=3489=2732≠2027\frac{P\left(r = \frac{R}{2}\right)}{P\left(r = \frac{R}{3}\right)} = \frac{\frac{3}{4}}{\frac{8}{9}} = \frac{27}{32} \neq \frac{20}{27} Therefore, Option (D) is incorrect.

Correct Options:

(B) and (C)

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