Given that T stands for time period nd 1 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 ?
(I) It is correct both dimensionally as well unmerically
(II) It is neither dimensionally correct nor numerically
(III) It is dimensionally correct but not numerically
(IV) It is numerically correct but not dimensionally

Options

A. Only I is correct
B. Only II is correct
C. Only III is correct
D. Only IV is correct

Answer

To solve the problem regarding the relation \( T^2 = \frac{l}{g} \), we will analyze the dimensional correctness and numerical correctness of the equation step by step.

### Step 1: Understand the Variables
- \( T \) is the time period of the pendulum.
- \( l \) is the length of the pendulum.
- \( g \) is the acceleration due to gravity.

### Step 2: Analyze the Dimensions
1. **Dimension of Time Period \( T \)**:
- The dimension of time is represented as \( [T] \).

2. **Dimension of Length \( l \)**:
- The dimension of length is represented as \( [L] \).

3. **Dimension of Acceleration due to Gravity \( g \)**:
- Acceleration has dimensions of length per time squared, which can be expressed as:
\[ [g] = \frac{[L]}{[T]^2} = [L][T]^{-2} \]

### Step 3: Substitute Dimensions into the Equation
Now let's substitute the dimensions into the equation \( T^2 = \frac{l}{g} \).

- The left-hand side (LHS) has dimensions:
\[ [T^2] = [T]^2 \]

- The right-hand side (RHS) can be expressed as:
\[ \frac{[l]}{[g]} = \frac{[L]}{[L][T]^{-2}} = [L] \cdot [L]^{-1} \cdot [T]^2 = [T]^2 \]

### Step 4: Compare Dimensions
Since both the LHS and RHS have the same dimensions:
\[ [T^2] = [T^2] \]
This confirms that the equation is dimensionally correct.

### Step 5: Check Numerical Correctness
The original equation \( T^2 = \frac{l}{g} \) does not include the constant \( 2\pi \) that is present in the actual formula for the time period of a simple pendulum:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Thus, while the relationship \( T^2 = \frac{l}{g} \) is dimensionally correct, it is not numerically correct because it lacks the factor of \( 2\pi \).

### Conclusion
Based on the analysis:
- The equation \( T^2 = \frac{l}{g} \) is dimensionally correct but not numerically correct.

Therefore, the correct answer is:
**(III) It is dimensionally correct but not numerically.**

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