A particle would take a time t to move down a straight tunnel from the surface of earth (supposed to be a homogeneous sphere) to its centre. If gravity were to remain constant this time would be t'. The ratio of t /t' will be

The correct option is π/2√2.

To find the ratio of the times, we will calculate the time taken by the particle in both scenarios: under varying gravity (simple harmonic motion) and under constant gravity.

Case 1: Under varying gravity (t)
For a homogeneous sphere of radius $R$ and acceleration due to gravity at the surface $g_0$, the acceleration due to gravity at a distance $r$ from the centre is given by:
$g\left(r\right)=\frac{g_0}{R}r$
Since the force is directed towards the centre, the acceleration of the particle is:
$a=-\frac{g_0}{R}r$
This equation represents Simple Harmonic Motion (SHM) of the form $a=-{\omega }^{2}r$, where the angular frequency is:
$\omega =\sqrt{\frac{g_0}{R}}$
The time $t$ taken to reach the centre from the surface (which is one-quarter of a full SHM cycle) is:
$t=\frac{T}{4}=\frac{\pi }{2\omega }=\frac{\pi }{2}\sqrt{\frac{R}{g_0}}$

Case 2: Under constant gravity (t')
If the acceleration due to gravity remains constant at its surface value $g_0$, the particle moves under uniform acceleration.
Using the second equation of motion for a particle starting from rest:
$R=\frac{1}{2}{g}_0{(t\text{'})}^2$
Solving for $t\text{'}$:
$t\text{'}=\sqrt{\frac{2R}{g_0}}$

Calculating the Ratio t/t'
Now, we find the ratio of the two times:
$\frac{t}{t\text{'}}=\frac{\frac{\pi }{2}\sqrt{\frac{R}{g_0}}}{\sqrt{2}\sqrt{\frac{R}{g_0}}}=\frac{\pi }{2\sqrt{2}}$