$A$ particle of mass $m$ is dropped from a height $R$ equal to the radius of the earth above the tunnel dug through the earth as shown in the figure. Hence the correct statement is

Two bodies,each of mass $M$,are kept fixed with a separation $2L$. $A$ particle of mass $m$ is projected from the mid-point of the line joining their centres,perpendicular to the line. The gravitational constant is $G$. The correct statement$(s)$ is/are:
$(a)$ The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $4 \sqrt{\frac{GM}{L}}$
$(b)$ The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \sqrt{\frac{GM}{L}}$
$(c)$ The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{2GM}{L}}$
$(d)$ The energy of the mass $m$ remains constant

Two particles of equal mass $m$ each move in a circle of radius $r$ under the action of their mutual gravitational attraction. The time period of each particle is

Select the correct choice$(s)$:

Match List-$I$ with List-$II$:

$(a)$ Gravitational constant $(G)$	$(i)$ $[L^{2}T^{-2}]$
$(b)$ Gravitational potential energy	$(ii)$ $[M^{-1}L^{3}T^{-2}]$
$(c)$ Gravitational potential	$(iii)$ $[LT^{-2}]$
$(d)$ Gravitational intensity	$(iv)$ $[ML^{2}T^{-2}]$

Choose the correct answer from the options given below:

$A$ 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)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$