Question

Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if it is projected into the tunnel with a speed of $\sqrt{gR}$

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

Answer 1

The correct option is B

$\frac{\pi }{2}\frac{\sqrt{g}}{R}$

Let M be the total mass of the earth.

At any position x,

$\therefore \frac{M^{\prime }}{M}=\frac{\rho \times \left(\frac{4}{3}\right)\pi \times x^3}{\rho \times \left(\frac{4}{3}\right)\pi \times R^3}=\frac{x^3}{R^3}\Rightarrow M^{\prime }=\frac{M{x}^3}{R^3}$

So force on the particle is given by,

$\therefore F_x=\frac{G{M}^{\prime }m}{x^2}=\frac{GMm}{R^3}x$ ....(1)

So, acceleration of the mass 'M' at that position is given by,

$a_x=\frac{GM}{R^2}x\Rightarrow \frac{a_x}{x}=w^2=\frac{GM}{R^3}=\frac{g}{R}$ $(\therefore g=\frac{GM}{R^2})$

So, $T=2\pi \sqrt{\frac{R}{g}}=$Time period of oscillation.

a) Now, using velocity- displacement equation.

$V=\omega \sqrt{(A^2-R^2)}$ [Where, A=amplitude]

Given when, $y=R,v=\sqrt{gR},\omega =\sqrt{\frac{g}{R}}$

$\Rightarrow \sqrt{gR}=\sqrt{\frac{g}{R}}\sqrt{(A^2-R^2)}$ [because $\omega =\sqrt{\frac{g}{R}}$]

$\Rightarrow R^2=A^2-R^2\Rightarrow A=\sqrt{2}R$

[Now, the phase of the particle at the point P is greater than $\frac{\pi }{2}$ but less than $\pi$ and at Q is greater

than $\pi$ but less than $\frac{3\pi }{2}$. Let the times taken by the particle to reach the positions P and Q be $t_1\&t_2$

respectively, then using displacement time equaion]

$y=rsin\omega t$

We have, $R=\sqrt{2}Rsin\omega t_1$ $\Rightarrow \omega t_1=\frac{3\pi }{4}$

$\&-R=\sqrt{2}Rsin\omega t_2$ $\Rightarrow \omega t_2=\frac{5\pi }{4}$

So, $\omega (t_2-t_1)=\frac{\pi }{2}\Rightarrow t_2-t_1=\frac{\pi }{2\omega }=\frac{\pi }{2\sqrt{\left(\frac{R}{g}\right)}}sec$