A particle takes a time `t_(1)` to move down a straight tunnel from the surface of earth to its centre. If gravity were to remain constant this time would be `t_(2)` calculate the ratio `(t_(1))/(t_(2))`

Options

A. `(pi)/(2sqrt(2))`
B. `(pi)/(sqrt(2))`
C. `sqrt(2)pi`
D. `(pi)/(sqrt(2))`

Explanation

To solve the problem, we need to calculate the time taken by a particle to move from the surface of the Earth to its center through a straight tunnel (denoted as \( t_1 \)) and compare it to the time taken if gravity were to remain constant (denoted as \( t_2 \)).

### Step-by-step Solution:

1. **Understanding the Motion in the Tunnel**:
- When a particle moves towards the center of the Earth, the gravitational force acting on it changes with distance. The force is maximum at the surface and decreases linearly to zero at the center.
- The gravitational acceleration \( g' \) at a distance \( r \) from the center is given by:
\[ g' = \frac{G M(r)}{r^2} \]
where \( M(r) \) is the mass enclosed within radius \( r \).

2. **Using the Concept of Simple Harmonic Motion (SHM)**:
- The motion of the particle can be modeled as simple harmonic motion. The time period \( T \) for one complete oscillation is given by:
\[ T = 2\pi \sqrt{\frac{R}{g}} \]
- However, since we are only interested in the time taken to reach the center, we take a quarter of the period:
\[ t_1 = \frac{T}{4} = \frac{1}{4} \times 2\pi \sqrt{\frac{R}{g}} = \frac{\pi}{2} \sqrt{\frac{R}{g}} \]

3. **Calculating \( t_2 \) with Constant Gravity**:
- If gravity were to remain constant throughout the tunnel, the particle would experience a constant acceleration \( g \). The distance to the center of the Earth is \( R \).
- Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \) with initial velocity \( u = 0 \), we have:
\[ R = \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ t_2^2 = \frac{2R}{g} \implies t_2 = \sqrt{\frac{2R}{g}} \]

4. **Finding the Ratio \( \frac{t_1}{t_2} \)**:
- Now we can find the ratio of \( t_1 \) to \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{\frac{\pi}{2} \sqrt{\frac{R}{g}}}{\sqrt{\frac{2R}{g}}} \]
- Simplifying this expression:
\[ \frac{t_1}{t_2} = \frac{\frac{\pi}{2}}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]

### Final Result:
Thus, the ratio \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = \frac{\pi}{2\sqrt{2}} \]