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(A) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMmr}{R^3}$.In the tunnel

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$2 \pi \sqrt{\frac{R}{g}}$

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(A) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMmr}{R^3}$.In the tunnel,the component of this force along the tunnel is $F_t = F \cos 	heta$,where $	heta$ is the angle between the position vector $r$ and the direction perpendicular to the tunnel.From the geometry of the tunnel,the perpendicular distance from the center is $d = R/2$. If $x$ is the displacement of the particle from the midpoint of the chord,then $r \cos 	heta = x$.Thus,the restoring force is $F_t = -\left(\frac{GMm}{R^3}ight) r \cos 	heta = -\left(\frac{GMm}{R^3}ight) x$.Since $g = \frac{GM}{R^2}$,we can write $F_t = -\left(\frac{mg}{R}ight) x$.The acceleration is $a = \frac{F_t}{m} = -\left(\frac{g}{R}ight) x$.This is the equation of simple harmonic motion $a = -\omega^2 x$,where $\omega^2 = \frac{g}{R}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$.

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	$(s)$ $\frac{GM_Em}{2r}$

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:

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