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

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{R}{g}} \\ = 2 π \sqrt{\frac{6400 \times 10^{3}}{10}} \\ = 2 π \times 8 \times 10^{2} = 16 π \times 10^{2} s = 5024 s = 83.73 \min \\ = 1 hr : 23.73 \min \\ ≃ 1 hr : 24 \min \end{array}$

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