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 tunnel, 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 $$g = 10$$ ms$$^{-2}$$, radius of earth = 6400 km)

For a particle in a tunnel through the Earth, the time period of SHM is:

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

Given: $$R = 6400$$ km $$= 6.4 \times 10^6$$ m, $$g = 10$$ m/s²

$$T = 2\pi\sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi\sqrt{6.4 \times 10^5}$$

$$\sqrt{6.4 \times 10^5} = \sqrt{640000} = 800$$

$$T = 2\pi \times 800 = 1600\pi \approx 5026.5$$ s

$$T \approx \frac{5026.5}{60} \approx 83.8$$ minutes $$\approx 1$$ hour 24 minutes

The correct answer is Option 2: 1 hour 24 minutes.