Question

If a tunnel is cut at any orientation through earth, then at what time will a ball released from one end reach the other end (neglect the rotation of the earth)?
A. $84.6$ minutes
B. $42.3$ minutes
C. $8$ minutes
D. depends on orientation

Answer 1

When a body is taken at a depth from the surface of the Earth then the acceleration due to gravity also changes. Hence, we have to find the acceleration due to gravity at that depth. Then, by using the equation and the equation of time period related with displacement and acceleration we will find the answer.

Formula used:
$g = \dfrac{{GM}}{{{R^2}}} - - - - \left( 1 \right)$
Where $G =$ Universal Gravitational Constant, $M =$ Mass of the Earth and $R =$ radius of the Earth.

Complete step by step answer:

We know that the formula of acceleration due to gravity at the surface of Earth stands as,
$g = \dfrac{{GM}}{{{R^2}}} - - - - \left( 1 \right)$
Where $G =$ Universal Gravitational Constant, $M =$ Mass of the Earth and $R =$ radius of the Earth

If we take the body at a depth $d$ from the surface of the Earth, the mass that it encompasses will be $M'$. Let us consider Earth as spherical with mass $M$ the volume that it encloses is $\dfrac{4}{3}\pi {R^3} - - - - \left( 2 \right)$
With mass $M'$ the volume that it encloses is $\dfrac{4}{3}\pi {(R - d)^3} - - - - \left( 3 \right)$
Dividing $\left( 2 \right)$ by $\left( 3 \right)$ we get,
$\dfrac{M}{{M'}} = \dfrac{{\dfrac{4}{3}\pi {R^3}}}{{\dfrac{4}{3}\pi {{(R - d)}^3}}} = \dfrac{{{R^3}}}{{{{\left( {R - d} \right)}^3}}}$
Hence, we find the relation,
$M' = \dfrac{{{{\left( {R - d} \right)}^3}}}{{{R^3}}} \times M$
By using relation $\left( 1 \right)$ we get the acceleration due to gravity at depth $d$ as,
$g' = \dfrac{{GM'}}{{{{(R - d)}^2}}} = \dfrac{G}{{{{(R - d)}^2}}} \times \dfrac{{{{\left( {R - d} \right)}^3}}}{{{R^3}}} \times M = \dfrac{{GM}}{{{R^3}}} \times \left( {R - d} \right)$

Simplifying the equation we get,
$g' = \dfrac{{GM}}{{{R^2}}} \times \left( {1 - \dfrac{d}{R}} \right)$
Substituting the value of $g$ from equation $\left( 1 \right)$ we get,
$g' = g\left( {1 - \dfrac{d}{R}} \right)$
$\Rightarrow g' = g\left( {\dfrac{{R - d}}{R}} \right) - - - - - \left( 4 \right)$
Let us consider $R - d$ as $x$ and the mass of the ball as $m$.
Thus, $g' = g\left( {\dfrac{x}{R}} \right) - - - - \left( 5 \right)$
The weight that is acting here is equal to the force $F$,
$F = mg'$
$ma = mg\left( {\dfrac{x}{R}} \right)$ where $a$ is the acceleration of the body.
$\Rightarrow a = g\left( {\dfrac{x}{R}} \right)$
Arranging the equation in terms of displacement and acceleration,
$\dfrac{x}{a} = \dfrac{R}{g} - - - - \left( 6 \right)$

The relation between time period $T$ with acceleration and displacement is,
$T = 2\pi \sqrt {\dfrac{x}{a}} = 2\pi \sqrt {\dfrac{R}{g}}$
This time period is the total time taken by the body to return back to the same position from where it initially started. Thus, the time period $t$ in which the body only reaches the end of the tunnel is $t = \dfrac{T}{2}$.Therefore, we get,
$t = \dfrac{T}{2} = \dfrac{{2\pi \sqrt {\dfrac{R}{g}} }}{2} \\\ \Rightarrow t= \pi \sqrt {\dfrac{R}{g}} \\\ \Rightarrow t= 3.14 \times \sqrt {\dfrac{{6.4 \times {{10}^6}}}{{9.8}}}$
$\Rightarrow t \approx 2537{\text{ seconds}}$
$\therefore t \approx 42.3{\text{ minutes}}$

Therefore, the correct option is B.

Note: It must be noted that we must consider the amount of mass of the earth whose radius would be the distance between the centre of the Earth and the position of the body in which it is placed. Weight of a body is the product of the acceleration due to gravity at that place to the mass of the body. Hence, the mass of a body never changes while the weight changes from place to place. The time period is defined as the total time taken by the body to complete one oscillation, in other words to return back to its mean position.