A second's pendulum is placed in a space laboratory orbiting around the earth at a height `3R`, where R is the radius of the earth. The time period of the pendulum is

Options

A. zero
B. `2sqrt3s`
C. `4s`
D. infinite

Explanation

To solve the problem of finding the time period of a second's pendulum placed in a space laboratory orbiting around the Earth at a height of \(3R\) (where \(R\) is the radius of the Earth), we can follow these steps:

### Step 1: Understand the Concept of a Second's Pendulum
A second's pendulum is defined as a pendulum that has a time period of 2 seconds (1 second for each swing). 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.

### Step 2: Determine the Effective Gravity in Orbit
In the given scenario, the pendulum is placed in a space laboratory orbiting the Earth at a height of \(3R\). In orbit, the effective acceleration due to gravity \(g_{\text{effective}}\) is not the same as on the surface of the Earth. In fact, when in free fall (as in orbit), the effective gravitational force acting on the pendulum is essentially zero because both the pendulum and the laboratory are in free fall together.

### Step 3: Calculate the Time Period
Since the effective gravity \(g_{\text{effective}} = 0\), 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 results in an undefined situation, indicating that the time period approaches infinity.

### Conclusion
Thus, the time period of the pendulum in the space laboratory orbiting at a height of \(3R\) is infinite.

**Final Answer: The time period of the pendulum is infinite.**

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