`T=2pisqrt(l/g)` is the time period of a simple pendulum, then the unit of `4pi^(2)l/T^(2)` in SI system is ______ .

Options

A. `ms^(-1)`
B. `s^(-2)`
C. `ms^(-2)`
D. `s^(-1)`

Explanation

To find the unit of \( \frac{4\pi^2 l}{T^2} \) in the SI system, we will follow these steps:

### Step 1: Understand the variables
- \( l \) is the length of the pendulum, which is measured in meters (m).
- \( T \) is the time period of the pendulum, which is measured in seconds (s).

### Step 2: Write down the formula
We need to find the unit of the expression:
\[ \frac{4\pi^2 l}{T^2} \]

### Step 3: Identify the units
- The unit of \( l \) (length) is \( \text{m} \).
- The unit of \( T \) (time) is \( \text{s} \).

### Step 4: Substitute the units into the expression
Substituting the units into the expression gives us:
\[ \frac{4\pi^2 \cdot \text{m}}{\text{s}^2} \]

### Step 5: Simplify the expression
Since \( 4\pi^2 \) is a constant and does not have any units, we can ignore it for the unit calculation. Therefore, the unit simplifies to:
\[ \frac{\text{m}}{\text{s}^2} \]

### Step 6: Conclusion
Thus, the unit of \( \frac{4\pi^2 l}{T^2} \) in the SI system is:
\[ \text{m/s}^2 \]