A simple pendulum has a time period `T_(1)` when on the earth's surface and `T_(2)` when taken to a height R above the earth's surface, where R is the radius of the earth. The value of `(T_(2))/(T_(1))` is

Options

A. 1
B. `sqrt(2)`
C. 4
D. 2

Explanation

To solve the problem, we need to find the ratio of the time periods of a simple pendulum at the Earth's surface and at a height equal to the radius of the Earth.

### Step-by-Step Solution:

1. **Understanding the Time Period of a 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.

2. **Time Period at Earth's Surface**:
Let \( T_1 \) be the time period of the pendulum at the Earth's surface. Thus,
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]

3. **Time Period at Height \( R \)**:
When the pendulum is taken to a height \( R \) (which is equal to the radius of the Earth), the acceleration due to gravity \( g' \) at that height can be calculated using the formula:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Here, \( h = R \), so:
\[ g' = \frac{g}{(1 + 1)^2} = \frac{g}{4} \]

4. **Calculating \( T_2 \)**:
Now, let \( T_2 \) be the time period of the pendulum at height \( R \):
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T_2 = 2\pi \sqrt{\frac{L}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \cdot 2 \sqrt{\frac{L}{g}} = 4\pi \sqrt{\frac{L}{g}} \]

5. **Finding the Ratio \( \frac{T_2}{T_1} \)**:
Now, we can find the ratio of the two time periods:
\[ \frac{T_2}{T_1} = \frac{4\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{L}{g}}} = \frac{4}{2} = 2 \]

### Final Answer:
Thus, the value of \( \frac{T_2}{T_1} \) is \( 2 \).

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