Find the length of seconds pendulum at a place where `g = pi^(2) m//s^(2)`.

Explanation

To find the length of a seconds pendulum at a place where \( g = \pi^2 \, \text{m/s}^2 \), we can use the formula for the time period of a simple pendulum:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

Where:
- \( T \) is the time period of the pendulum,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 1: Set the time period for a seconds pendulum
A seconds pendulum has a time period of \( T = 2 \) seconds.

### Step 2: Substitute the values into the formula
We can substitute \( T = 2 \) seconds and \( g = \pi^2 \, \text{m/s}^2 \) into the formula:

\[ 2 = 2\pi \sqrt{\frac{L}{\pi^2}} \]

### Step 3: Simplify the equation
Dividing both sides by \( 2 \):

\[ 1 = \pi \sqrt{\frac{L}{\pi^2}} \]

### Step 4: Isolate the square root
Now, we can isolate the square root:

\[ \frac{1}{\pi} = \sqrt{\frac{L}{\pi^2}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:

\[ \left(\frac{1}{\pi}\right)^2 = \frac{L}{\pi^2} \]

This simplifies to:

\[ \frac{1}{\pi^2} = \frac{L}{\pi^2} \]

### Step 6: Solve for \( L \)
Multiplying both sides by \( \pi^2 \):

\[ L = 1 \, \text{meter} \]

### Conclusion
The length of the seconds pendulum at the specified location is \( 1 \, \text{meter} \).

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