If the length of a simple pendulum is equal to the radius of the earth, its time period will be

Options

A. `2pisqrt(R//g)`
B. `2pi sqrt (R//2g)`
C. `2 pi sqrt (2R//g)`
D. infinite

Explanation

To find the time period of a simple pendulum when its length is equal to the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Time Period**:
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.

2. **Identify the Condition**:
In this case, we are given that the length of the pendulum \( l \) is equal to the radius of the Earth \( r \):
\[ l = r \]

3. **Adjust the Formula for Large Lengths**:
When the length of the pendulum is comparable to the radius of the Earth, we need to use a modified formula for the time period:
\[ T = 2\pi \sqrt{\frac{1}{\frac{1}{l} + \frac{1}{r}}} \]
Here, we substitute \( l = r \).

4. **Substitute the Values**:
Substituting \( l = r \) into the modified formula:
\[ T = 2\pi \sqrt{\frac{1}{\frac{1}{r} + \frac{1}{r}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{1}{\frac{2}{r}}} \]

5. **Simplify the Expression**:
Continuing with the simplification:
\[ T = 2\pi \sqrt{\frac{r}{2}} \]
Thus, we can express the time period as:
\[ T = 2\pi \sqrt{\frac{r}{2g}} \]
where \( g \) is the acceleration due to gravity.

6. **Final Result**:
Therefore, the time period of the pendulum when its length is equal to the radius of the Earth is:
\[ T = 2\pi \sqrt{\frac{r}{2g}} \]

### Answer:
The time period will be \( T = 2\pi \sqrt{\frac{r}{2g}} \).

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