Statement-1: The time period of a simple pendulum on a satellite orbiting the earth is infinity
Because
Statement-2: Effective value of g in orbiting satellite is zero and `T = 2pi sqrt((l)/(g))`

Options

A. Statement-1 is True, Statement-2 is true, Statement -2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1
C. Statement-1 is true, Statement -2 is false
D. Statement -1 is false, statement-2 is true

Answer

To solve the question, we need to analyze both statements and understand the relationship between them.

### Step-by-Step Solution:

1. **Understanding Statement-1**:
- The statement claims that the time period of a simple pendulum on a satellite orbiting the Earth is infinity.
- In a gravitational field, the time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
- Here, \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Understanding Statement-2**:
- This statement asserts that the effective value of \( g \) in an orbiting satellite is zero.
- In a satellite in orbit, the satellite and everything inside it are in free fall. This creates a condition of weightlessness, effectively making the acceleration due to gravity \( g \) equal to zero for the pendulum inside the satellite.

3. **Analyzing the Time Period Formula**:
- If we substitute \( g = 0 \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{l}{0}} \]
- This results in \( T \) approaching infinity, as division by zero is undefined and indicates that the time period becomes infinitely large.

4. **Conclusion**:
- Since the effective value of \( g \) is zero in the satellite, the time period of the pendulum indeed becomes infinite.
- Therefore, both statements are true, and Statement-2 provides the correct explanation for Statement-1.

### Final Answer:
- **Statement 1 is true. Statement 2 is true, and Statement 2 is the correct explanation for Statement 1.**