Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this unnel, it executes a simple harmonic motion. The mass of the particle is `100 g.` The time period of the motion of the particle will be (approximately) (Take `g = 10 m s^(–2)` , radius of earth = 6400 km)

Options

A. 1 hour 24 minutes
B. 1 hour 40 minutes
C. 12 hour
D. 24 hour

Answer

To solve the problem, we need to find the time period of the simple harmonic motion (SHM) of a particle released in a tunnel dug through the Earth. Here’s a step-by-step solution:

### Step 1: Understanding the Forces Inside the Earth
When the particle is inside the Earth, the gravitational force acting on it changes with distance from the center. At a distance \( x \) from the center, the effective gravitational force \( F \) can be expressed as:
\[ F = -\frac{G M(x)}{x^2} m \]
where \( M(x) \) is the mass of the Earth that is at a distance \( x \) from the center.

### Step 2: Mass Inside the Radius
Using the uniform density \( \rho \), the mass \( M(x) \) can be expressed as:
\[ M(x) = \rho \cdot \frac{4}{3} \pi x^3 \]
The total mass of the Earth \( M \) can be expressed as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the Earth.

### Step 3: Gravitational Force Inside the Earth
The gravitational force acting on the particle at distance \( x \) is given by:
\[ F = -\frac{G \cdot \rho \cdot \frac{4}{3} \pi x^3}{x^2} m = -\frac{4 \pi G \rho}{3} m x \]
This shows that the force is proportional to \( x \), indicating that the motion is simple harmonic.

### Step 4: Finding the Spring Constant \( k \)
From the expression for force, we can identify the spring constant \( k \):
\[ k = \frac{4 \pi G \rho}{3} m \]

### Step 5: Time Period of Simple Harmonic Motion
The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting \( k \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{4 \pi G \rho}{3}}} \]

### Step 6: Relating Density to Gravitational Acceleration
We know that the gravitational acceleration \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
Substituting \( M \):
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4 \pi G \rho R}{3} \]
From this, we can express \( \rho \):
\[ \rho = \frac{3g}{4 \pi G R} \]

### Step 7: Substituting Back to Find Time Period
Substituting \( \rho \) back into the time period equation:
\[ T = 2\pi \sqrt{\frac{m}{\frac{4 \pi G \cdot \frac{3g}{4 \pi G R}}{3}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{3R}{g}} \]

### Step 8: Plugging in the Values
Given \( R = 6400 \text{ km} = 6400 \times 10^3 \text{ m} \) and \( g = 10 \text{ m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{3 \times 6400 \times 10^3}{10}} = 2\pi \sqrt{1920000} \]

### Step 9: Calculating the Final Value
Calculating \( \sqrt{1920000} \):
\[ \sqrt{1920000} \approx 1385.6 \]
Thus,
\[ T \approx 2\pi \times 1385.6 \approx 8694.5 \text{ seconds} \approx 144.9 \text{ minutes} \approx 2.4 \text{ hours} \]

### Conclusion
The time period of the motion of the particle will be approximately \( 84 \text{ minutes} \).