JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Medium
- MCQ
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 \(=6400\,km\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1\) hour \(24\) minutes
Step-by-step Solution
Detailed explanation
Let at some time particle is at a distance \(x\) from centre of Earth, then at that position field
\(E=\frac{G M}{R^3} x\)
\(\therefore \text { Acceleration of particle }\)
\(\vec{a}=-\frac{ GM }{ R ^3} \overrightarrow{ x }\)
\(\Rightarrow \omega=\sqrt{\frac{ GM }{ R ^3}}=\sqrt{\frac{ g }{ R }}\)
\(\text { Now } T =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{ R }{ g }}\)
\(\Rightarrow T =2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}}\)
\(=2 \times 3.14 \times 800\,sec \approx 1 \text { hour } 24 \text { minutes }\)
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