Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

The correct option is 6 I.

To find the moment of inertia of a uniform circular disc about an axis perpendicular to its plane and passing through a point on its rim, we can use the theorems of moment of inertia step-by-step.

Step 1: Understand the given moment of inertia
Let the mass of the uniform circular disc be $M$ and its radius be $R$.
The moment of inertia of a circular disc about a diameter (which lies in the plane of the disc) is given by:

$I_{\text{diameter}}=\frac{1}{4}M{R}^2$

According to the question, this moment of inertia is $I$:

$I=\frac{1}{4}M{R}^2\Rightarrow M{R}^2=4I$

Step 2: Find the moment of inertia about a central perpendicular axis
Using the perpendicular axis theorem, the moment of inertia of the disc about an axis passing through its center of mass ($C$) and perpendicular to its plane ($I_c$) is twice the moment of inertia about its diameter:

$I_c=\frac{1}{2}M{R}^2$

Step 3: Find the moment of inertia about the target axis using the parallel axis theorem
We need to find the moment of inertia about an axis perpendicular to the plane of the disc and passing through a point on its rim.
The distance between the center of mass and the point on the rim is the radius $R$.
According to the parallel axis theorem:

$I_{\text{rim}}=I_c+M{R}^2$

Substituting the value of $I_c$ into the equation:

$I_{\text{rim}}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Step 4: Express the final result in terms of $I$
Now, substitute $M{R}^2=4I$ into the expression for $I_{\text{rim}}$:

$I_{\text{rim}}=\frac{3}{2}\left(4I\right)=6I$

Thus, the moment of inertia of the disc about the specified axis is 6 I.