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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)
$$P (r = 0) = 0$$
$$\frac{P \left(r = \frac{3R}{4}\right)}{P \left(r = \frac{2R}{3}\right)} = \frac{63}{80}$$
$$\frac{P \left(r = \frac{3R}{5}\right)}{P \left(r = \frac{2R}{5}\right)} = \frac{16}{21}$$
$$\frac{P \left(r = \frac{R}{2}\right)}{P \left(r = \frac{R}{3}\right)} = \frac{20}{27}$$
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