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 know that the acceleration due to gravity at a height \( R \) above the surface of the Earth is \( g/4 \). The question asks for the time period of a simple pendulum at this height.

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

3. **Effective Gravity in an Artificial Satellite**: In an artificial satellite, the effective acceleration due to gravity \( g_{\text{effective}} \) is not the same as the gravitational acceleration at that height. Instead, due to the satellite's orbital motion, the effective gravity becomes zero. This is because the satellite is in free fall, creating a condition of weightlessness.

4. **Substituting Values**: Since the effective gravity \( g_{\text{effective}} = 0 \) in the satellite, we can substitute this into the time period formula:
\[ T = 2\pi \sqrt{\frac{l}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{l}{0}} \]
This expression indicates that the time period \( T \) approaches infinity because division by zero is undefined.

5. **Conclusion**: Therefore, the time period of a simple pendulum in an artificial satellite at the given height is infinite.

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