Given that `T` stands for time and `l` stands for the length of simple pendulum . If `g` is the acceleration due to gravity , then which of the following statements about the relation `T^(2) = ( l// g)` is correct?

Options

A. It is correct both dimensionally as well as numerically.
B. It is neither dimensionally correct nor numerically.
C. It is dimensionally correct but not numerically.
D. It is numerically correct but not dimensionally.

Explanation

To solve the problem, we need to analyze the given relation \( T^2 = \frac{l}{g} \) where \( T \) is the time period of a simple pendulum, \( l \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- \( T \) represents the time period of the pendulum.
- \( l \) represents the length of the pendulum.
- \( g \) represents the acceleration due to gravity.

2. **Analyzing the Given Relation**:
- The relation given is \( T^2 = \frac{l}{g} \).
- This suggests that the square of the time period is proportional to the length of the pendulum divided by the acceleration due to gravity.

3. **Dimensional Analysis**:
- We need to check the dimensions of both sides of the equation to see if they are consistent.
- The dimension of time \( T \) is represented as \( [T] \).
- The dimension of length \( l \) is represented as \( [L] \).
- The dimension of acceleration due to gravity \( g \) is \( [L][T^{-2}] \) (length per time squared).

4. **Calculating Dimensions**:
- The left-hand side (LHS) of the equation is \( T^2 \), which has the dimension \( [T^2] \).
- The right-hand side (RHS) is \( \frac{l}{g} \):
\[ \text{Dimension of } \frac{l}{g} = \frac{[L]}{[L][T^{-2}]} = [T^2] \]
- Both sides have the same dimensions, which means the equation is dimensionally consistent.

5. **Numerical Analysis**:
- The correct formula for the time period of a simple pendulum is actually:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
- Squaring both sides gives:
\[ T^2 = 4\pi^2 \frac{l}{g} \]
- This indicates that the original relation \( T^2 = \frac{l}{g} \) is not numerically correct because it lacks the factor \( 4\pi^2 \).

6. **Conclusion**:
- The relation \( T^2 = \frac{l}{g} \) is dimensionally correct but not numerically correct.
- Therefore, the correct statement about the relation is that it is dimensionally correct but not numerically correct.

### Final Answer:
The relation \( T^2 = \frac{l}{g} \) is dimensionally correct but not numerically correct.