The time period of a simple pendulum of length 1 m is _______ second. (take `pi^(2)=10,g=10ms^(-2)`)

Explanation

To find the time period of a simple pendulum with a length of 1 meter, we can use the formula for the time period \( T \) of a simple pendulum, which is given by:

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

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

### Step 1: Identify the values
We are given:
- Length \( L = 1 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- \( \pi^2 = 10 \)

### Step 2: Substitute the values into the formula
Substituting the values into the formula, we have:

\[ T = 2\pi \sqrt{\frac{1}{10}} \]

### Step 3: Simplify the expression
Now, we can simplify the expression inside the square root:

\[ T = 2\pi \sqrt{0.1} \]

### Step 4: Calculate \( \sqrt{0.1} \)
Since \( \sqrt{0.1} \) can be expressed as \( \frac{1}{\sqrt{10}} \), we can rewrite the equation as:

\[ T = 2\pi \cdot \frac{1}{\sqrt{10}} \]

### Step 5: Substitute \( \pi^2 \) value
Given \( \pi^2 = 10 \), we can find \( \pi \):

\[ \pi = \sqrt{10} \]

Substituting this back into our equation gives:

\[ T = 2\sqrt{10} \cdot \frac{1}{\sqrt{10}} = 2 \]

### Step 6: Conclusion
Thus, the time period \( T \) of the simple pendulum of length 1 meter is:

\[ \boxed{2 \text{ seconds}} \]

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