Question

A tunnel is made across the earth of radius $R$ , passing through its centre. A ball is dropped from a height $h$ in the tunnel, height $h$ is very small. The motion will be periodic with time period:

• A. $2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{h}{g}}$
• B. $2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$
• C. $2\pi \sqrt{\frac{R}{g}}+\sqrt{\frac{h}{g}}$
• D. $2\pi \sqrt{\frac{R}{g}}+\sqrt{\frac{2h}{g}}$

Answer 1

Answer: (B)

When the ball is in the tunnel at distance $x$ from the centre of the earth, then gravitational force acting on ball is
$F=\frac{Gm}{x^2}\times \left(\frac{4}{3}\pi x^3\rho \right)=G\times \left(\frac{4}{3}\pi \rho \right)mx$
Mass of the earth, $M=\frac{4}{3}\pi R^3\rho$
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
$T^{{}^{\prime }}=2\pi \sqrt{\frac{inertiafactor}{springfactor}}$
$=2\pi \sqrt{\frac{m}{gm/R}}$
$=2\pi \sqrt{\frac{R}{g}}$
Time spent by ball outside the tunnel on both the sides will be $=4\sqrt{2\text{h}/\text{g}}$
Therefore, total time period of oscillation of ball is
$=2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$