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A spherical body of radius R consists of...

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(rltR), then the correct option(s) is (are)

A

P(r=0)=0

B

`(P(r=3R//4))/(P(r=2R//3))=63/80`

C

`(P(r=3R//5))/(P(r=2R//5))=16/21`

D

`(P(r=R//2))/(P(r=R//3))=20/27`

Text Solution

Verified by Experts

The correct Answer is:
B, C
Gravitational field at a distance r from the centre of a uniform sphere is
`E_r=(GM)/R^3 r =(G(4/3pR^3r))/R^3 r`
`rArr E_r =(4pGr)/3 r`
Balancing the forces on a thin spherical shell at a distance r. `4pr^2dP=-E_r dm`
`4pr^2dP=(-4pGr)/3 r(4pr^2dr r) rArr dP=(-4dGr^2rdr)/3`
Integrating `intdP=-(4pGr^2)/3 intrdr`
Hence pressure at a radial distance r.
`P(r)=(-4pGr^2)/3. r^2/2+C`
At surface P=O
So, `(2pGr^2R^2)/3=C therefore P(r)=(2pGr^2)/3 (R^2-r^2)`
`P("3R"/4)=(2pGr^2)/3 (1-9/16)R^2 =(2pGr^2)/3 . (7/16)R^2 P("2R"/3)=(2pGr^2)/3 (1-4/9)R^2 =(2pGr^2)/3 .(5/9)R^2`
Ratio `(P(r=3R//4))/(P(r=2R//3))=63/80` Hence (b)
`P("3R"/5)=(2pGr^2)/3 (1-9/25)=16/25` constant , `P("2R"/5)=(2pGr^2)/3 (1-4/25) =21/25` constant
Ratio `(P(r=3R//5))/(P(r=2R//3))=16/21` Hence (c )
`(P(R/2))/(P(R/3))=(1-1/4)/(1-1/9)=3/4xx9/8=(27/32)` Hence (d ) is not correct
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