Three point masses,each of mass $m$,are kept at the corners of an equilateral triangle of side $L$. The system rotates about the center of the triangle. The period of rotation $T$ is directly proportional to:

$A$ particle of mass $m_0$ is projected from the mid-point of the line joining two fixed particles each of mass $m.$ If the distance of separation between the fixed particles is $l,$ the minimum velocity of projection of the particle so as to escape to infinity is equal to

$A$ small point mass $m$ is placed at a distance $2R$ from the centre $O$ of a big uniform solid sphere of mass $M$ and radius $R$. The gravitational force on $m$ due to $M$ is $F_1$. $A$ spherical part of radius $R/3$ is removed from the big sphere as shown in the figure and the gravitational force on $m$ due to the remaining part of $M$ is found to be $F_2$. The value of the ratio $F_1: F_2$ is

$A$ mass $m$,travelling at speed $V_0$ in a straight line from far away,is deflected when it passes near a black hole of mass $M$ which is at a perpendicular distance $R$ from the original line of flight. $a$,the distance of closest approach between the mass and the black hole,is given by the relation:

Match the column $-I$ with column $-II$ for a satellite in circular orbit:

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

(where $M_E$ is the mass of the earth,$m$ is the mass of the satellite,and $r$ is the radius of the orbit)

If the mass of the Earth remains constant and its radius is reduced to $\frac{1}{n}$ of its original value,what will be the duration of $1$ day?