If a tunnel is cut at any orientation through the Earth,a ball released from one end will reach the other end in time ........ $\text{min}$ (neglect Earth's rotation).

Assume that a tunnel is dug along a chord of the Earth,at a perpendicular distance $(R / 2)$ from the Earth's centre,where $R$ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel,it will execute a simple harmonic motion with a time period:

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)

As shown in the figure,two spherical cavities are made in a uniform solid sphere of radius $R$. The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the $X$-axis. The mass of the solid sphere before the cavities were created was $M$. The gravitational force on a point mass $m$ at a distance $d$ away from the centre of the solid sphere is

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only,then the distance covered by the smaller body before collision is (in $R$)

Three identical spheres of mass $m$ are placed at the vertices of an equilateral triangle of side length $a$. When released,they interact only through gravitational force and collide after a time $T = 4 \text{ s}$. If the sides of the triangle are increased to length $2a$ and the masses of the spheres are made $2m$,then they will collide after . . . . . . seconds.