JEE Physics 2023 Question with Solution

Given: Earth is assumed to be a uniform solid sphere, a tunnel is dug along its diameter, and the particle mass is $100 \, \text{g}$.

Find: The time period of oscillation inside the Earth.

At a distance $x$ from the centre of the Earth, the gravitational field inside a uniform sphere is proportional to displacement:

Hence the acceleration of the particle is directed toward the centre:

This is the condition for simple harmonic motion, so

Using $g = \frac{GM}{R^2}$,

Therefore,

Now substitute $R = 6400 \times 10^3 \, \text{m}$ and $g = 10 \, \text{m/s}^2$:

So,

which is approximately $1 \, \text{hour} \; 24 \, \text{minutes}$.

Therefore, the correct option is B.

Given: Motion of a particle through a diameter tunnel inside a uniform Earth.

Find: The approximate time period.

For SHM inside the Earth, the effective force is proportional to displacement:

and therefore

So the time period is

Using

we get

Now take

and

Then

This is approximately $1 \, \text{hour} \; 24 \, \text{minutes}$.

The particle mass does not affect the time period. Although one the solution marks option A, the worked numerical result clearly matches $1 \, \text{hour} \; 24 \, \text{minutes}$, so the correct option is B.