Step-by-step Derivation of Pressure P(r)

1. Consider Hydrostatic Equilibrium Let's consider a spherical shell of fluid of radius r and thickness dr. For the fluid to be in equilibrium, the forces acting on this shell must balance. The condition for hydrostatic equilibrium is given by the equation: ${dP \over dr} = -\rho(r) g(r)$ where P(r) is the pressure, $\rho(r)$ is the density, and g(r) is the gravitational acceleration at a distance r from the center.

2. Calculate Gravitational Acceleration g(r) The problem states that the fluid has a constant density $\rho$. The gravitational force at a distance r from the center is due only to the mass enclosed within the sphere of radius r. Let this enclosed mass be $M_{enc}$. $M_{enc}(r) = \text{density} \times \text{volume} = \rho \times \left({4 \over 3}\pi r^3\right)$ The gravitational acceleration g(r) at distance r is: $g(r) = {G M_{enc}(r) \over r^2} = {G \left(\rho {4 \over 3}\pi r^3\right) \over r^2} = \left({4 \over 3}\pi G \rho\right) r$

3. Formulate the Differential Equation for Pressure Substitute $\rho(r) = \rho$ and the expression for g(r) into the hydrostatic equilibrium equation: ${dP \over dr} = -\rho \left[\left({4 \over 3}\pi G \rho\right) r\right] = -{4 \over 3}\pi G \rho^2 r$

4. Integrate to find P(r) To find the pressure P(r) at a radius r, we integrate the above differential equation from r to the surface of the sphere R. The pressure at the surface of the sphere is zero, i.e., P(R) = 0. $\int_{P(r)}^{P(R)} dP = -\int_{r}^{R} {4 \over 3}\pi G \rho^2 r' dr'$ $P(R) - P(r) = -{4 \over 3}\pi G \rho^2 \left[ {r'^2 \over 2} \right]_r^R$ $0 - P(r) = -{2 \over 3}\pi G \rho^2 (R^2 - r^2)$ $P(r) = {2 \over 3}\pi G \rho^2 (R^2 - r^2)$

5. Analyze the Pressure Ratios From the derived expression, we can see that P(r) is proportional to $(R^2 - r^2)$. Let $C = {2 \over 3}\pi G \rho^2$ be the constant of proportionality. $P(r) = C(R^2 - r^2)$ Now we can evaluate the given options.

Evaluation of Options

A: $P(r = 0) = 0$ $P(0) = C(R^2 - 0^2) = C R^2 = {2 \over 3}\pi G \rho^2 R^2$ Since G, $\rho$, and R are non-zero, P(0) is non-zero. In fact, it's the maximum pressure. So, option A is incorrect.

**B: {{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{63} \over {80}}$`** {{P(3R/4)} \over {P(2R/3)}} = {{C(R^2 - (3R/4)^2)} \over {C(R^2 - (2R/3)^2)}} = {{R^2(1 - 9/16)} \over {R^2(1 - 4/9)}} = {{7/16} \over {5/9}} = {7 \over 16} \times {9 \over 5} = {63 \over 80} $$ So, option B is correct.

**C: {{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{16} \over {21}}$`** {{P(3R/5)} \over {P(2R/5)}} = {{C(R^2 - (3R/5)^2)} \over {C(R^2 - (2R/5)^2)}} = {{R^2(1 - 9/25)} \over {R^2(1 - 4/25)}} = {{16/25} \over {21/25}} = {16 \over 21} $$ So, option C is correct.

**D: {{P(r = R/2)} \over {P(r = R/3)}} = {{20} \over {27}}$`** {{P(R/2)} \over {P(R/3)}} = {{C(R^2 - (R/2)^2)} \over {C(R^2 - (R/3)^2)}} = {{R^2(1 - 1/4)} \over {R^2(1 - 1/9)}} = {{3/4} \over {8/9}} = {3 \over 4} \times {9 \over 8} = {27 \over 32} $$ The calculated ratio is 27/32, not 20/27. So, option D is incorrect.

Conclusion: The correct options are B and C.