JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Gravitation

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)

Answer: (B), (C)

Solution

For a self-gravitating fluid sphere, consider a shell at radius

r

with thickness

d r

.

Gravitational field at radius

r

g (r) = \frac{GM ( r )}{r ^{2}} = \frac{G \cdot \frac{4}{3} π r ^{3} ρ}{r ^{2}} = \frac{4 G π ρ r}{3}

Pressure balance:

- d P \cdot A = ρ g (r) \cdot A \cdot d r

- d P = \frac{4 G π ρ ^{2} r}{3} d r

Integrating from surface (

r = R

P = 0

) to radius

r

P (r) = \frac{2 G π ρ ^{2}}{3} (R^{2} - r^{2})

Let

C = \frac{2 G π ρ ^{2}}{3}

, so

P (r) = C (R^{2} - r^{2})

.

(A)

P (0) = C R^{2} \neq = 0

✗

(B)

\frac{P ( 3 R /4 )}{P ( 2 R /3 )} = \frac{C ( R ^{2} - 9 R ^{2} /16 )}{C ( R ^{2} - 4 R ^{2} /9 )} = \frac{7 R ^{2} /16}{5 R ^{2} /9} = \frac{7 \times 9}{16 \times 5} = \frac{63}{80}

✓

(C)

\frac{P ( 3 R /5 )}{P ( 2 R /5 )} = \frac{C ( R ^{2} - 9 R ^{2} /25 )}{C ( R ^{2} - 4 R ^{2} /25 )} = \frac{16 R ^{2} /25}{21 R ^{2} /25} = \frac{16}{21}

✓

(D)

\frac{P ( R /2 )}{P ( R /3 )} = \frac{C ( R ^{2} - R ^{2} /4 )}{C ( R ^{2} - R ^{2} /9 )} = \frac{3 R ^{2} /4}{8 R ^{2} /9} = \frac{3 \times 9}{4 \times 8} = \frac{27}{32} \neq = \frac{20}{27}

✗

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options