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 $=6400\,km$ )
JEE MAIN 2023, Medium
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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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