The acceleration due to gravity at height `R` above the surface of the earth is `g/4`. The periodic time of simple pendulum in an artifical satellie at this height will be :-

Options

A. `T = 2pisqrt((2l)/(g))`
B. `T = 2pisqrt((l)/(2g))`
C. zero
D. infinity

Explanation

To solve the problem, we need to determine the periodic time of a simple pendulum in an artificial satellite at a height where the acceleration due to gravity is \( g/4 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are given that the acceleration due to gravity at a height \( R \) above the Earth's surface is \( g/4 \).
- We need to find the periodic time of a simple pendulum at this height.

2. **Formula for Period of a Simple Pendulum**:
- The formula for the period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

3. **Identifying the Effective Gravity**:
- At the height \( R \), the effective acceleration due to gravity is given as \( g/4 \).
- Thus, we can substitute \( g \) in the formula with \( g/4 \).

4. **Substituting into the Period Formula**:
- Substituting \( g = g/4 \) into the period formula, we have:
\[ T = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \cdot 2 \sqrt{\frac{L}{g}} = 4\pi \sqrt{\frac{L}{g}} \]

5. **Considering the Condition of an Artificial Satellite**:
- In an artificial satellite, the effective gravity \( g \) is effectively zero because the satellite is in free fall. Therefore, the pendulum cannot oscillate in the usual sense.
- This leads to the conclusion that the time period \( T \) approaches infinity because the pendulum does not experience any restoring force.

6. **Final Conclusion**:
- The periodic time of a simple pendulum in an artificial satellite at this height is effectively infinite.

### Answer:
The periodic time of the simple pendulum in an artificial satellite at this height will be infinite.