Question

If a straight tunnel is bored connecting two points on the surface of the earth, then the time taken by a particle released from any point of the tunnel, to come to instantaneously rest again in the tunnel is

• A. independent of the point of the tunnel from which the particle is released.
• B. depends on the perpendicular distance of the tunnel from the centre of the earth.
• C. dependent on the density (assumed to be uniform) of the earth.
• D. is equal to the time period of the satellite of the earth, the altitude of which is equal to the radius of the earth.

Answer 1

Answer: (A), (C)

dependent on the density (assumed to be uniform) of the earth.

We know that time period,
$T=2\pi \sqrt{\frac{R}{g}}$

$g=\frac{GM}{R^2}=\frac{G}{R^2}\times \frac{4}{3}\pi R^3\times \rho$

$\Rightarrow g=\frac{4}{3}\pi GR\rho$

So, $T=2\pi \sqrt{\frac{R}{4/3\pi GR\rho }}=2\pi \sqrt{\frac{3}{4\pi G\rho }}$

Now, the time interval between two successive positions of instant rest of the particle is $t=T/2$.

We can see that,

$\left(A\right)t$ is independent of point of tunnel from which the particle is released.

$\left(B\right)t$ does not depend on the perpendicular distance of the tunnel from the centre of the earth.

$\left(C\right)t$ is dependent on the density $\left(\rho \right)$.

$\left(D\right)t$ is not equal to the time period of the satellite of the earth, the altitude of which is equal to the radius of the earth, as its time period is given by

$T=\frac{2\pi \left(2R\right)}{v}=\frac{2\pi \left(2R\right)}{\sqrt{GM/2R}}=2\sqrt{2}T$

Therefore, the correct options are $\left(A\right)$ and $\left(C\right)$.