Question

A straight tunnel is dug into the Earth as shown in the figure at a distance $d$ from its center. A ball of mass $\mathrm{m}$ is dropped from one of its ends. The time it takes to reach the other end is approximately equal to

• A. $42\min$
• B. $84\min$
• C. $84\left(\frac{\mathrm{d}}{\mathrm{R}}\right)\min$
• D. $42\left(\frac{\mathrm{d}}{\mathrm{R}}\right)\min$

Answer 1

The correct option is A

$42\min$

Step 1: Given information

Consider the figure given below

Step 2: Calculations required

The gravitational force on mass $m$ at distance $r$ is,

$F=\frac{GMmr}{R^3}$

Where, $G$ is the gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth, and $r$ is the distance from the center O.

From the figure given below,

$$\begin{gathered} \frac{x}{r}=\sin \theta \\ \Rightarrow r=\frac{x}{\sin \theta } \end{gathered}$$

Now,

$$\begin{gathered} F_x=-Fsin\theta \\ {F}_x=\frac{-GMmr}{R^3}\frac{x}{r} \\ {F}_x=\frac{-GMmx}{R^3} \end{gathered}$$

Since $F_x\propto \left(-x\right)$, its motion is simple harmonic in nature. Further,

$$\begin{gathered} m{a}_x=\frac{GMmx}{R^3} \\ {a}_x=\frac{GMx}{R^3} \end{gathered}$$

Hence the time period of oscillation is,

$T=2\pi \sqrt{\frac{x}{g}}$

Where, $\mathrm{T}$ is the time period.

$$\begin{gathered} T=2\pi \frac{\sqrt{x}}{\sqrt{a_x}} \\ T=2\pi \frac{\sqrt{x}}{\sqrt{\frac{GMx}{R^3}}} \\ =2\pi \sqrt{\frac{R^3}{GM}}\left[\because g=\frac{GM}{R^2}\right] \\ =2\pi \sqrt{\frac{R}{g}} \\ =2\pi \sqrt{\frac{6.4\times {10}^6}{9.8}} \\ =5077.5\mathrm{seconds} \\ =\frac{5077.5}{60}\min \\ =84\min \end{gathered}$$

The time take for a particle to go from one end to another is,

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{T}}{2} \\ =\frac{84}{2} \\ =42\min \end{gathered}$$

Therefore, option (A) is correct.