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Question

(B) For a fluid sphere of constant density $\rho$ in equilibrium under its own gravity,the pressure $P(r)$ at a distance $r$ from the center is given

Vedclass · Accepted Answer

$(B, C)$

Vedclass · Answer

(B) For a fluid sphere of constant density $ho$ in equilibrium under its own gravity,the pressure $P(r)$ at a distance $r$ from the center is given by the hydrostatic equilibrium equation: $\frac{dP}{dr} = -ho g(r)$.The gravitational field at distance $r$ is $g(r) = \frac{G M(r)}{r^2} = \frac{G (\frac{4}{3}\pi r^3 ho)}{r^2} = \frac{4}{3}\pi G ho r$.Substituting this into the equilibrium equation: $\frac{dP}{dr} = -\frac{4}{3}\pi G ho^2 r$.Integrating from $r$ to $R$ (where $P(R) = 0$): $\int_{P(r)}^{0} dP = -\int_{r}^{R} \frac{4}{3}\pi G ho^2 r dr$.$0 - P(r) = -\frac{4}{3}\pi G ho^2 [\frac{r^2}{2}]_r^R = -\frac{2}{3}\pi G ho^2 (R^2 - r^2)$.Thus,$P(r) = \frac{2}{3}\pi G ho^2 R^2 (1 - \frac{r^2}{R^2}) = P_c (1 - \frac{r^2}{R^2})$.Checking the ratios:$(B) \frac{P(3R/4)}{P(2R/3)} = \frac{1 - (3/4)^2}{1 - (2/3)^2} = \frac{1 - 9/16}{1 - 4/9} = \frac{7/16}{5/9} = \frac{63}{80}$. (Correct)$(C) \frac{P(3R/5)}{P(2R/5)} = \frac{1 - 9/25}{1 - 4/25} = \frac{16/25}{21/25} = \frac{16}{21}$. (Correct)$(D) \frac{P(R/2)}{P(R/3)} = \frac{1 - 1/4}{1 - 1/9} = \frac{3/4}{8/9} = \frac{27}{32} 
eq \frac{20}{27}$. (Incorrect)Therefore,the correct options are $(B)$ and $(C)$.

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