If R is the radius of the earth and the acceleration due to gravity on the surface of earth is `g = pi^2 m//s^2`, then the length of the second's pendulum at a height `h = 2R` from the surface of earth will be,:

Options

A. `2/9 m`
B. `1/9 m`
C. `4/9 m`
D. `8/9 m`

Answer

To find the length of the second's pendulum at a height \( h = 2R \) from the surface of the Earth, we can follow these steps:

### Step 1: Understand the Length of a Second's Pendulum
The length \( L \) of a second's pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( T \) is the time period of the pendulum, and \( g \) is the acceleration due to gravity. For a second's pendulum, \( T = 2 \) seconds, hence:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]

### Step 2: Rearranging the Formula
We can rearrange the formula to find the length \( L \):
\[ 1 = \pi \sqrt{\frac{L}{g}} \quad \Rightarrow \quad \sqrt{\frac{L}{g}} = \frac{1}{\pi} \]
Squaring both sides gives:
\[ \frac{L}{g} = \frac{1}{\pi^2} \quad \Rightarrow \quad L = \frac{g}{\pi^2} \]

### Step 3: Calculate \( g \) at Height \( h = 2R \)
The acceleration due to gravity at a height \( h \) above the surface of the Earth can be calculated using the formula:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Substituting \( h = 2R \):
\[ g' = \frac{g}{(1 + \frac{2R}{R})^2} = \frac{g}{(1 + 2)^2} = \frac{g}{9} \]

### Step 4: Substitute \( g' \) into the Length Formula
Now we can substitute \( g' \) into the length formula:
\[ L' = \frac{g'}{\pi^2} = \frac{\frac{g}{9}}{\pi^2} = \frac{g}{9\pi^2} \]

### Step 5: Substitute the Value of \( g \)
Given that \( g = \pi^2 \, m/s^2 \):
\[ L' = \frac{\pi^2}{9\pi^2} = \frac{1}{9} \, m \]

### Final Answer
Thus, the length of the second's pendulum at a height \( h = 2R \) from the surface of the Earth is:
\[ L' = \frac{1}{9} \, m \]

---