In the above question, the curve between log I and log R will be :

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the relationship between the moment of inertia \( I \) of a solid cylinder and its radius \( R \). The moment of inertia about the axis through the center of mass for a solid cylinder is given by the formula:

\[ I = \frac{1}{2} m R^2 \]

Where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the cylinder,
- \( R \) is the radius of the cylinder.

### Step 1: Express the Moment of Inertia in Terms of Logarithms
We can take the logarithm of both sides of the equation to analyze the relationship between \( \log I \) and \( \log R \):

\[ \log I = \log\left(\frac{1}{2} m R^2\right) \]

Using the properties of logarithms, we can separate the terms:

\[ \log I = \log\left(\frac{1}{2}\right) + \log m + \log R^2 \]

Further simplifying, we have:

\[ \log I = \log\left(\frac{1}{2}\right) + \log m + 2 \log R \]

### Step 2: Rearranging the Equation
We can rearrange the equation to express it in the form of a linear equation \( y = mx + c \):

\[ \log I = 2 \log R + \left(\log m + \log\left(\frac{1}{2}\right)\right) \]

Here, we can identify:
- \( y = \log I \)
- \( x = \log R \)
- The slope \( m = 2 \)
- The y-intercept \( c = \log m + \log\left(\frac{1}{2}\right) \)

### Step 3: Analyzing the Graph
From the equation \( \log I = 2 \log R + c \), we can see that if we plot \( \log I \) on the y-axis and \( \log R \) on the x-axis, the graph will be a straight line with:
- A slope of 2,
- A y-intercept of \( c \).

### Conclusion
The curve between \( \log I \) and \( \log R \) will be a straight line with a slope of 2. This indicates that as the radius \( R \) increases, the moment of inertia \( I \) increases quadratically.