Two smooth tunnels are dug from one side of earth's surface to the other side, one along a diameter are dropped from one end of each of the tunnels. Both particles oscillate simple harmonically along the tunnels. Let `T_(1)` and `T_(2)` be the time period of particles along the diameter and along the chord respectively. Then:

Options

A. `T_(1) = T_(2)`
B. `T_(1) gt T_(2)`
C. `v_(1)=v_(2)`
D. `v_(1) gt v_(2)`

Answer

To solve the problem of two particles oscillating in smooth tunnels dug through the Earth, we need to analyze the time periods of the oscillations in both the diameter and the chord tunnels.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two tunnels: one along the diameter of the Earth (Tunnel 1) and another along a chord (Tunnel 2).
- When a particle is dropped into either tunnel, it will oscillate back and forth due to gravitational forces.

2. **Forces Acting on the Particles**:
- The gravitational force acting on the particle inside the tunnel can be derived from Newton's law of gravitation.
- For a particle at a distance \( r \) from the center of the Earth, the force is given by:
\[ F = -\frac{G M m}{r^2} \]
- However, only the mass of the Earth that is within radius \( r \) contributes to the gravitational force on the particle.

3. **Deriving the Equation of Motion**:
- Using the concept of uniform density, the mass \( m' \) of the Earth within radius \( r \) can be expressed as:
\[ m' = \rho \cdot \frac{4}{3} \pi r^3 \]
- The gravitational force acting on the particle can be expressed as:
\[ F = -\frac{G m' m}{r^2} = -\frac{G \cdot \rho \cdot \frac{4}{3} \pi r^3 m}{r^2} = -\frac{4 \pi G \rho m}{3} r \]
- This shows that the motion is simple harmonic, where the acceleration \( a \) is proportional to the displacement \( r \).

4. **Finding the Time Period**:
- The time period \( T \) of simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Here, \( k \) is the effective spring constant, which can be derived from the force equation:
\[ k = \frac{4 \pi G \rho m}{3} \]
- Substituting for \( k \) gives:
\[ T = 2\pi \sqrt{\frac{3}{4 \pi G \rho}} \]
- Since the density \( \rho \) is constant for both tunnels, the time period for both tunnels (diameter and chord) will be equal:
\[ T_1 = T_2 \]

5. **Comparing Maximum Velocities**:
- The maximum velocity of the particle at the center of the Earth can be derived using energy conservation.
- For Tunnel 1 (diameter), the maximum velocity \( V_1 \) can be derived as:
\[ V_1 = \sqrt{\frac{GM}{R}} \]
- For Tunnel 2 (chord), the maximum velocity \( V_2 \) will be less than \( V_1 \) due to the shorter effective distance traveled in the chord tunnel:
\[ V_2 < V_1 \]
- Thus, we conclude that:
\[ V_1 > V_2 \]

### Conclusion:
- The time periods for both tunnels are equal: \( T_1 = T_2 \).
- The maximum velocity of the particle in the diameter tunnel is greater than that in the chord tunnel: \( V_1 > V_2 \).

### Final Answers:
- **Correct Options**:
- A: \( T_1 = T_2 \)
- D: \( V_1 > V_2 \)