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)

Options

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)`

Answer

B, C

Explanation

Consider a spherical element of linner radius r and outer radius (r + dr) as shown in

Let g be the ac celeration due to gravity at the location of the element which is at distance r from the center O. In equilibrium, the difference in pressure (dp) on the two sides of the element is pressure by gravity per unit area.
So `- dp = rho g dr`
If m is the mass of spherical body of radius r, then
`m=(4)/(3)pi r^(2) rho` and `g= (Gm)/(r^(2)) =(G)/(r^(2))xx(4)/(3)pi r^(3)rho = (4)/(3)pi G rho r`
`:. dp = - rho ((4)/(3)pi G rho r)dr = - (4)/(3)pi rho^(2) G r dr`
`int_(P(=R))^(P(=r)) dp = -(4)/(3)pi G rho^(2) int_(r=R)^(r=r) r dr`
or `P_((r)) - P_((R)) =(4)/(3)pi G rho^(2) ((R^(2)-r^(2))/(2))`
Put `P_((R)) =0` , then `P_((r)) =(4)/(3)pi G rho^(2) ((R^(2) - r^(2)))/(2)`
(a) When `r=0, P_((r)) =(4)/(3)pi G rho^(2) (R^(2))/(2)`.
Thus, option (a) is wrong.
(b) `(P_((r=3R//4)))/(P_((r=2R//3))) = (R^(2) -(3R//4)^(2))/(R^(2) -(2R//3)^(2)) = (7//16)/(5//9) =(63)/(80)`.
Thus, option (b) is true.
(c) `(P_((r=3R//5)))/(P_((r=2R//5))) =(R^(2) -(3R//5)^(2))/(R^(2) -(2R//5)^(2)) = (16//25)/(21//25) =(16)/(21)`
Thus, option (c) is true.
(d) `(P_((r=R//2)))/(P_((r=R//3))) =(R^(2) -(R//2)^(2))/(R^(2) - (R//3)^(2)) = (3//4)/(8//9) =(27)/(32)`.
Thus, option (d) is wrong.