Two identical balls are dropped from the surface of earth, on isdropped in atunnel along the diameter of the earth and other isdroppped in tunnel along a chord.
STATEMENT-1 `:` Both balls will execute S.H.M. with same time period.
and
STATEMENT -2 `:` Both balls cross their mean position ( i.e., centre of earth ) with same speed

Answer

To solve the problem, we need to analyze the two statements regarding the behavior of two identical balls dropped from the surface of the Earth: one dropped through a tunnel along the diameter and the other through a tunnel along a chord.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical balls dropped from the surface of the Earth.
- One ball is dropped through a tunnel that goes straight through the center of the Earth (along the diameter).
- The other ball is dropped through a tunnel that is along a chord of the Earth.

2. **Analyzing the First Ball (Diameter)**:
- When the ball is dropped through the tunnel along the diameter, it will experience a gravitational force that varies with distance from the center of the Earth.
- The gravitational force acting on the ball at a distance \( r \) from the center is given by:
\[ F = -\frac{G M m}{r^2} \]
However, for a solid sphere, the force can be expressed as:
\[ F = -\frac{G M m}{R^3} r \]
where \( R \) is the radius of the Earth.
- This force leads to simple harmonic motion (SHM) about the center of the Earth.

3. **Calculating the Time Period for the First Ball**:
- The acceleration \( a \) of the ball can be expressed as:
\[ a = -\frac{G M}{R^3} r \]
- This shows that the motion is SHM with angular frequency \( \omega \) given by:
\[ \omega = \sqrt{\frac{G M}{R^3}} \]
- The time period \( T \) of the SHM is:
\[ T = 2\pi \sqrt{\frac{R^3}{G M}} \]

4. **Analyzing the Second Ball (Chord)**:
- For the ball dropped through the chord, the analysis is similar. The gravitational force acting on it also leads to SHM.
- The force acting on the ball at a distance \( x \) from the center can be derived similarly, and it will also lead to SHM.
- The time period for this ball is also:
\[ T = 2\pi \sqrt{\frac{R^3}{G M}} \]
- Therefore, both balls have the same time period.

5. **Crossing the Mean Position**:
- When the first ball (through the diameter) reaches the center, all its potential energy has converted to kinetic energy, resulting in maximum speed.
- For the second ball (through the chord), when it reaches the center, it still has some potential energy due to its position along the chord.
- Thus, the speed of the second ball at the center will be less than that of the first ball.

### Conclusion:
- **Statement 1**: Both balls will execute SHM with the same time period. **(True)**
- **Statement 2**: Both balls cross their mean position with the same speed. **(False)**

Thus, the correct option is that Statement 1 is true and Statement 2 is false.