Explanation

(b) When the ball of mass $m$ falls from a height$h$, it reaches the surface of earth in time$t = \sqrt{2h/g}$. Its velocity is$v = \sqrt{2gh}$. It then moves in to the tunnel and reaches on the other side of earth and goes again upto a height $h$ from that side of earth. The ball again returns back and thus executes periodic motion. Outside the earth ball crosses distance $h$ four times.
When the ball is in the tunnel at distance $x$ from the centre of the earth, then gravitational force acting on ball is
$F = \frac{Gm}{x^{2}} \times \left( \frac{4}{3}\pi x^{2}\rho \right) = G \times \left( \frac{4}{3}\pi\rho \right)mx$
Mass of the earth, $M = \frac{4}{3}\pi R^{2}\rho$
or $\frac{4}{3}\pi\rho = \frac{M}{R^{3}}$
$\therefore\ \ \ \ F = \frac{GMmx}{R^{3}}\ \ ie,\ F \propto x$
As this $F$ is directed towards the centre of earth $ie$, the mean position so the ball will execute periodic motion about the centre of earth
Here inertia factor=mass of ball$= m$
Spring factor$= \frac{GMm}{R^{3}} = \frac{gm}{R}$
$\therefore$ time period of oscillation of ball in the tunnel is
$T^{'} = 2\pi\sqrt{\frac{inertia\ factor}{spring\ factor}}$
$= 2\pi\sqrt{\frac{m}{\frac{gm}{R}} = 2\pi\sqrt{\frac{R}{g}}}$
Time spent by ball outside the tunnel on both the sides will be $= 4\sqrt{2h/g}$
Therefore, total time period of oscillation of ball is
$= 2\pi\sqrt{\frac{R}{g}} + 4\sqrt{\frac{2h}{g}}$