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

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

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(D) The gravitational force on a particle of mass $m$ at a distance $r$ from the center of the Earth is given by $F = -\frac{GMmr}{R^3}$.Along the tunnel,only the component of this force directed towards the center of the tunnel (the equilibrium position) acts as a restoring force.Let $x$ be the displacement of the particle from the center of the tunnel. The distance from the Earth's center is $r = \sqrt{x^2 + (R/2)^2}$.The component of the force along the tunnel is $F_{restoring} = F \cos 	heta$,where $\cos 	heta = \frac{x}{r}$.Thus,$F_{restoring} = -\left(\frac{GMmr}{R^3}ight) \left(\frac{x}{r}ight) = -\frac{GMm}{R^3} x$.Since $g = \frac{GM}{R^2}$,we have $\frac{GM}{R^3} = \frac{g}{R}$.Therefore,$F_{restoring} = -\left(\frac{mg}{R}ight) x$.Comparing this with the $SHM$ equation $F = -m \omega^2 x$,we get $\omega^2 = \frac{g}{R}$,which implies $\omega = \sqrt{\frac{g}{R}}$.The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R}{g}}$.

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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