The time period of a simple pendulum of infinite length is (R=radius of earth).

Options

A. infinite
B. `2pisqrt((R)/(g))`
C. `2pisqrt((g)/(R))`
D. `(1)/(2pi)sqrt((R)/(g))`

Explanation

To find the time period of a simple pendulum of infinite length, we can follow these steps:

### Step 1: Understand the formula for the 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:
- \( T \) is the time period,
- \( l \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Define the length of the pendulum
In this case, the length of the pendulum is said to be infinite. However, we can relate this to the radius of the Earth \( R \). For the purpose of this problem, we consider the effective length to be equal to the radius of the Earth:
\[ l = R \]

### Step 3: Substitute the effective length into the formula
Substituting \( l = R \) into the time period formula, we get:
\[ T = 2\pi \sqrt{\frac{R}{g}} \]

### Step 4: Use the known values for \( R \) and \( g \)
The radius of the Earth \( R \) is approximately \( 6400 \) km or \( 6.4 \times 10^6 \) m, and the acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

### Step 5: Calculate the time period
Now, substituting these values into the formula:
\[ T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}} \]

Calculating the value inside the square root:
\[ \frac{6.4 \times 10^6}{9.8} \approx 65384.6 \]

Now, taking the square root:
\[ \sqrt{65384.6} \approx 255.6 \]

Finally, calculating \( T \):
\[ T \approx 2\pi \times 255.6 \approx 1604.4 \, \text{seconds} \]

### Step 6: Convert seconds to minutes
To convert seconds to minutes:
\[ T \approx \frac{1604.4}{60} \approx 26.74 \, \text{minutes} \]

### Final Answer
Thus, the time period of a simple pendulum of infinite length is approximately \( 26.74 \) minutes.

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