Two smooth tunnels are dug from one side of earth's surface to the other side, one along a diameter and other along the chord .Now two particle are dropped from one end of each of the tunnels. Both time particles oscillate simple harmonically along the tunnels. Let `T_(1)` and `T_(2)` be the time particle and v1and v2 be the maximum speed along the diameter and along the chord respectively. Then:

Options

A. a. `T_(1)=T_(2)`
B. b. `T_(1) gt T_(2)`
C. c. `v_(1)=v_(2)`
D. d. `v_(1) gtv_(2)`

Answer

To solve the problem, we need to analyze the motion of particles in two different tunnels: one dug along the diameter of the Earth and the other along a chord.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two tunnels: one along the diameter (Tunnel 1) and one along a chord (Tunnel 2).
- Both tunnels allow particles to oscillate in simple harmonic motion (SHM) when dropped.

2. **Time Period of Oscillation**:
- The time period \( T \) of a particle in SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{r}{g}} \]
where \( r \) is the effective radius and \( g \) is the acceleration due to gravity.
- For Tunnel 1 (along the diameter), the effective radius is equal to the radius of the Earth, \( R \).
- For Tunnel 2 (along the chord), the effective radius is less than \( R \) because it does not extend all the way to the center of the Earth.
- Therefore, we can write:
\[ T_1 = 2\pi \sqrt{\frac{R}{g}} \quad \text{(for Tunnel 1)} \]
\[ T_2 = 2\pi \sqrt{\frac{r_2}{g}} \quad \text{(for Tunnel 2, where \( r_2 < R \))} \]

3. **Comparing Time Periods**:
- Since \( r_2 < R \), it follows that:
\[ T_2 < T_1 \]
- Thus, we conclude:
\[ T_1 > T_2 \]

4. **Maximum Speed of the Particles**:
- The maximum speed \( v \) of a particle in SHM can be derived from the potential energy change. The maximum speed is given by:
\[ v = \sqrt{2g h} \]
where \( h \) is the maximum height (or depth) from which the particle is dropped.
- For Tunnel 1, the maximum height is \( R \) (the full radius of the Earth).
- For Tunnel 2, the maximum height is less than \( R \) because the tunnel does not reach the center.
- Therefore, we can express:
\[ v_1 = \sqrt{2gR} \quad \text{(for Tunnel 1)} \]
\[ v_2 = \sqrt{2g h_2} \quad \text{(for Tunnel 2, where \( h_2 < R \))} \]

5. **Comparing Maximum Speeds**:
- Since \( h_2 < R \), it follows that:
\[ v_2 < v_1 \]
- Thus, we conclude:
\[ v_1 > v_2 \]

### Final Results:
- Time Period Comparison: \( T_1 > T_2 \)
- Maximum Speed Comparison: \( v_1 > v_2 \)