Question

If a tunnel is cut at any orientation through earth, then find the time taken by a ball released from one end to reach the other end. (neglect earth rotation and take radius of earth as $R=64\times {10}^5\text{m}$ and $g=10{\text{m/sec}}^2$)

• A.
  $84.6\text{minutes}$
• B.
  $41.9\text{minutes}$
• C.
  $8\text{minutes}$
• D.
  depend on orientation

Answer 1

The correct option is B

$41.9\text{minutes}$
As we know that when the particle of mass $m$ is at a distance $x(x<R)$ from the center inside the earth, the gravitational force on it be equal to $$F=-\frac{GMmx}{R^3}$$ Now, the acceleration of the particle at this, point is,

$\Rightarrow a=\frac{F}{m}=-\frac{GM}{R^3}x$

Thus, $a\propto (-x)$, this is the case of SHM.

So, comparing with the standard equation of SHM, $a=-{\omega }^{2}x$ we have, angular frequency of oscillation,

$\Rightarrow \omega =\sqrt{\frac{GM}{R^3}}$

As, $g=\frac{GM}{R^2}$ ,

$\therefore \omega =\sqrt{\frac{g}{R}}$

Time period is given by,

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R}{g}}$

Substituting, $R=64\times {10}^5\text{m}$ and $g=10{\text{m/s}}^2$, we get

$\Rightarrow T=2\pi \sqrt{\frac{64\times {10}^5}{10}}$

$\Rightarrow T=1600\pi \text{sec}$

$\Rightarrow T=\frac{1600\pi }{60}=83.8\text{min}$

So, time taken by ball from one end to other end is $\left(\frac{T}{2}\right)=41.9\text{min}$.

Hence, option (b) is the correct answer.