$A$ spherical body of radius $R$ consists of a fluid of constant density $\rho$ and is in equilibrium under its own gravity. If $P(r)$ is the pressure at a distance $r$ from the center $(r < R)$,then the correct option$(s)$ is(are):
$(A) P(r=0) = P_c$ (maximum pressure at center)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$

$A$ particle of mass $m$ moves in a circular orbit under the central potential field,$U(r) = -\frac{C}{r}$,where $C$ is a positive constant. The correct radius-velocity graph of the particle's motion is:

$Assertion$ : The Earth is slowing down and as a result the Moon is coming nearer to it.
$Reason$ : The angular momentum of the Earth-Moon system is not conserved.

$A$ planet has a core of density $3\rho$ and an outer crust of density $\rho$. There is a small tunnel through the core. $A$ small particle of mass $m$ is released from end $A$. What is the time required to reach end $B$?

$A$ binary star system consists of two stars $A$ (mass $M_A = 2.2 M_S$) and $B$ (mass $M_B = 11 M_S$),where $M_S$ is the mass of the Sun. They are separated by a distance $d$ and rotate about their common centre of mass,which is stationary. What is the ratio of the angular momentum of star $A$ to the angular momentum of star $B$ about the centre of mass?

Consider a spherical gaseous cloud of mass density $\rho(r)$ in free space,where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time,the particle number density $n(r) = \rho(r) / m$ is:
[$G$ is the universal gravitational constant]