Frequency f of a simple pendulum depends on its length l and acceleration g due to gravity according to the following equation `f=1/(2pi)sqrt(g/l)`
Graph between which of the following quantities is a straight line?

Options

A. f on the odinate and l on the abscissa
B. f on the ordinate and `sqrt(l)` on the abscissa
C. `f^(2)` on the ordinate and l on the abscissa
D. `f^(2)` on the ordinate and `1//l` on the adscissa

Answer

To determine which graph represents a straight line based on the relationship between the frequency \( f \) of a simple pendulum, its length \( l \), and the acceleration due to gravity \( g \), we start with the given equation:

\[ f = \frac{1}{2\pi} \sqrt{\frac{g}{l}} \]

### Step 1: Rearranging the Equation
To analyze the relationship, we can manipulate this equation. First, we can square both sides to eliminate the square root:

\[ f^2 = \left(\frac{1}{2\pi}\right)^2 \cdot \frac{g}{l} \]

This simplifies to:

\[ f^2 = \frac{g}{4\pi^2} \cdot \frac{1}{l} \]

### Step 2: Identifying the Linear Relationship
Now, we can express this equation in a form that resembles the equation of a straight line, \( y = mx + c \). Here, we can identify:

- \( y \) as \( f^2 \)
- \( m \) as \( \frac{g}{4\pi^2} \)
- \( x \) as \( \frac{1}{l} \)
- \( c \) as 0 (since there is no constant term)

Thus, we have:

\[ f^2 = \left(\frac{g}{4\pi^2}\right) \cdot \frac{1}{l} \]

This indicates that if we plot \( f^2 \) on the y-axis (ordinate) and \( \frac{1}{l} \) on the x-axis (abscissa), the graph will be a straight line.

### Step 3: Conclusion
From our analysis, we conclude that the graph between \( f^2 \) (ordinate) and \( \frac{1}{l} \) (abscissa) is a straight line. Therefore, the correct option is:

**Option D: \( f^2 \) vs \( \frac{1}{l} \)**

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