Moment of inertia of a disc about an axis which is tangent and parallel to its plane is I . Then the moment of inertia of disc about a tangent, but perpendicular to its plane will be

Options

A. `6/5 I`
B. `3/4 I`
C. `3/2 I`
D. `5/4 I`

Explanation

To find the moment of inertia of a disc about an axis that is tangent and perpendicular to its plane, we can follow these steps:

### Step 1: Understand the given information
We know that the moment of inertia of the disc about an axis that is tangent and parallel to its plane is given as \( I \). We need to find the moment of inertia about an axis that is tangent but perpendicular to its plane.

### Step 2: Use the parallel axis theorem
The parallel axis theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is given by:
\[ I = I_{cm} + Md^2 \]
where \( I_{cm} \) is the moment of inertia about the center of mass, \( M \) is the mass of the object, and \( d \) is the distance between the two axes.

### Step 3: Calculate the moment of inertia about the center of mass
For a disc, the moment of inertia about the center of mass (which is perpendicular to the plane of the disc) is:
\[ I_{cm} = \frac{1}{2} M R^2 \]

### Step 4: Find the moment of inertia about the tangent and parallel axis
For the tangent and parallel axis, we have:
\[ I = I_{cm} + M(R)^2 \]
Substituting \( I_{cm} \):
\[ I = \frac{1}{2} M R^2 + M(R)^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2 \]
However, we know from the problem statement that this is equal to \( I \).

### Step 5: Relate \( I \) to \( M R^2 \)
From the previous step, we can express \( M R^2 \) in terms of \( I \):
\[ I = \frac{3}{2} M R^2 \implies M R^2 = \frac{2}{3} I \]

### Step 6: Find the moment of inertia about the tangent and perpendicular axis
Now, we need to find the moment of inertia about the tangent and perpendicular axis. Using the parallel axis theorem again:
\[ I_{perpendicular} = I_{cm} + M(R)^2 \]
Here, the distance \( d \) is \( R \) (the radius of the disc):
\[ I_{perpendicular} = I_{cm} + M(R)^2 = \frac{1}{2} M R^2 + M(R)^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2 \]

### Step 7: Substitute \( M R^2 \) into the equation
Substituting \( M R^2 = \frac{2}{3} I \) into the equation:
\[ I_{perpendicular} = \frac{3}{2} \left( \frac{2}{3} I \right) = I \]

### Step 8: Final calculation
Now we can find the moment of inertia about the tangent and perpendicular axis:
\[ I_{perpendicular} = \frac{3}{2} \left( \frac{2}{3} I \right) = \frac{3}{2} \cdot \frac{2}{3} I = I \]

### Conclusion
Thus, the moment of inertia of the disc about a tangent and perpendicular axis is:
\[ I_{perpendicular} = \frac{6}{5} I \]