Question 10 of 223

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 $100g$. The time period of the motion of the particle will be (approximately)
(take $g=10{ms}^{-2}$, radius of earth $=6400km$)

A$24$ hoursB$1$ hour $24$ minutesC$1$ hour $40$ minutesD$12$ hours

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Answer: (B) $1$ hour $24$ minutes

The gravitational force acting on a particle of mass $m$ at a distance$r(\lt R)$ from the centre of Earth is
$F=-G\dfrac{{M}^{'}m}{{r}^{2}}$ ....(1)
where ${M}^{'}$ is the mass of portion of Earth enclosed by a sphere of radius $r$ concentric with Earth. Here $-$ sign indicates radially inward direction of force.
Taking the density of Earth to be uniform
$\dfrac{{M}^{'}}{4{\pi r}^{3}}=\dfrac{M}{4{\pi R}^{3}}$
$\Rightarrow {M}^{'}=\dfrac{M{r}^{3}}{{R}^{3}}$ ....(2)
Substituting ${M}^{'}$ from equation 2 into equation 1:
$F=-\left(\dfrac{GMm}{{R}^{3}}\right)r$
Which is similar to the equation of motion $F=-kx$ with $k=G\dfrac{Mm}{{R}^{3}}=\dfrac{mg}{R}(∵g=G\dfrac{M}{{R}^{2}})$
For such motion, the time period is given by:
$T=2\pi \sqrt{\dfrac{m}{k}}$
Substituting k from above,
$T=2\pi \sqrt{\dfrac{R}{g}}$
Putting the values, we get
$T=2\times 3.14\sqrt{\dfrac{6400\times {10}^{3}}{10}}=2\times 3.14\times 800\approx 1hour24minutes$