T is the time period of simple pendulum on the earth's surface. Its time period becomes `x T` when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of `x` will be :

Options

A. 45017
B. 4
C. 44958
D. 2

Answer

To solve the problem, we need to determine how the time period of a simple pendulum changes when it is taken to a height equal to the Earth's radius.

### Step-by-Step Solution:

1. **Understand the Time Period of a Simple Pendulum**:
The time period \( T \) of a simple pendulum at the Earth's surface 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 at the Earth's surface.

2. **Determine the New Gravity at Height \( R \)**:
When the pendulum is taken to a height \( h = R \) (where \( R \) is the radius of the Earth), the new acceleration due to gravity \( g' \) can be calculated using the formula:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Substituting \( h = R \):
\[ g' = \frac{g}{(1 + 1)^2} = \frac{g}{4} \]

3. **Calculate the New Time Period \( T' \)**:
The time period at the new height \( T' \) is given by:
\[ T' = 2\pi \sqrt{\frac{l}{g'}} \]
Substituting \( g' = \frac{g}{4} \):
\[ T' = 2\pi \sqrt{\frac{l}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4l}{g}} = 2 \times 2\pi \sqrt{\frac{l}{g}} = 2T \]

4. **Determine the Value of \( x \)**:
We have found that:
\[ T' = 2T \]
This means that the new time period \( T' \) is \( xT \) where \( x = 2 \).

### Final Answer:
Thus, the value of \( x \) is:
\[ \boxed{2} \]