For a circular cardboard of uniform thickness and mass M, a square disc of the maximum possible are is cut. If the moment of inertia of the square with the moment of inertial of the square with the axis of rotation at the centre and perpendicular to the plane of the disc is `(Ma^2)/6`, the radius of the circular cardboard is

Options

A. `sqrt(2)a`
B. `a/(sqrt(2))`
C. `2a`
D. `2sqrt(2)a`

Answer

To solve the problem, we need to find the radius \( R \) of the circular cardboard from which the maximum possible square disc is cut. We are given that the moment of inertia of the square is \( \frac{M a^2}{6} \), where \( a \) is the side length of the square.

### Step-by-Step Solution:

1. **Understand the Moment of Inertia of the Square:**
The moment of inertia \( I \) of a square plate about an axis through its center and perpendicular to its plane is given by:
\[ I = \frac{M' a^2}{6} \]
where \( M' \) is the mass of the square and \( a \) is the side length of the square.

2. **Relate the Mass of the Square to the Mass of the Circular Cardboard:**
Since the square is cut from the circular cardboard, we can denote the mass of the square as \( M' \) and the mass of the circular cardboard as \( M \). The mass of the square \( M' \) can be expressed in terms of the mass density and area, but we will focus on the relationship with the radius.

3. **Determine the Side Length of the Square:**
The maximum size of the square that can be cut from the circular cardboard is when its diagonal is equal to the diameter of the circle. The diagonal \( d \) of the square can be expressed as:
\[ d = a\sqrt{2} \]
The diameter of the circular cardboard is \( 2R \). Therefore, we have:
\[ a\sqrt{2} = 2R \]
From this, we can solve for \( a \):
\[ a = \frac{2R}{\sqrt{2}} = R\sqrt{2} \]

4. **Substituting the Side Length into the Moment of Inertia Equation:**
Now substituting \( a = R\sqrt{2} \) into the moment of inertia equation:
\[ I = \frac{M' (R\sqrt{2})^2}{6} = \frac{M' \cdot 2R^2}{6} = \frac{M' R^2}{3} \]
We know from the problem statement that \( I = \frac{M a^2}{6} \). Thus, we equate:
\[ \frac{M' R^2}{3} = \frac{M (R\sqrt{2})^2}{6} \]
Simplifying the right side:
\[ \frac{M (2R^2)}{6} = \frac{M R^2}{3} \]

5. **Equating the Two Expressions:**
Now we can set the two expressions equal to each other:
\[ \frac{M' R^2}{3} = \frac{M R^2}{3} \]
Since \( R^2 \) is common and non-zero, we can cancel it out:
\[ M' = M \]

6. **Finding the Radius:**
Since \( M' \) is the mass of the square and \( M \) is the mass of the circular cardboard, we have established that the maximum square that can be cut has a relationship with the radius \( R \).

From the earlier equation \( a = R\sqrt{2} \) and substituting back, we can find:
\[ R = \frac{a}{\sqrt{2}} \]

### Final Answer:
Thus, the radius \( R \) of the circular cardboard is:
\[ R = \frac{a}{\sqrt{2}} \]