1. Force on a particle inside a uniform solid Earth

For a point at distance $x$ from the center of a uniformly dense Earth, only the mass enclosed within radius $x$ contributes to gravity.

Hence, $M_x = M\left(\frac{x^3}{R^3}\right)$

So the gravitational force on mass $m$ is $F = -\frac{G M_x m}{x^2} = -\frac{G M m}{R^3}x$

Using $g = \frac{GM}{R^2} \implies GM = gR^2$ we get $F = -\frac{g}{R}mx$

Therefore, $m\frac{d^2x}{dt^2} = -\frac{g}{R}mx$

or $\frac{d^2x}{dt^2} + \frac{g}{R}x = 0$

This is the equation of simple harmonic motion with $\omega = \sqrt{\frac{g}{R}}$

So the time period is $T = 2\pi\sqrt{\frac{R}{g}}$

2. Substitute the given values

Given: $R = 6400\text{ km} = 6.4\times 10^6\text{ m}$ $g = 10\text{ m s}^{-2}$

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

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

Hence, $T = 2\pi(800) = 1600\pi \text{ s}$

Using $\pi \approx 3.14$, $T \approx 1600\times 3.14 = 5024\text{ s}$

Convert into minutes: $\frac{5024}{60} \approx 83.7\text{ min}$

That is approximately $84\text{ min} = 1\text{ hour }24\text{ minutes}$

3. Check options

• A: $12$ hours $\times$
• B: $1$ hour $24$ minutes $\checkmark$
• C: $24$ hours $\times$
• D: $1$ hour $40$ minutes $\times$

So the correct option is B.

4. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.