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 of the circular cardboard from which a square disc of maximum possible area is cut. We are given that the moment of inertia of the square is \( \frac{Ma^2}{6} \).

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( M \) be the mass of the circular cardboard.
- Let \( r \) be the radius of the circular cardboard.
- Let \( b \) be the side length of the square cut from the circular cardboard.
- The area of the square is \( A = b^2 \).

2. **Moment of Inertia of the Square:**
- The moment of inertia of the square about an axis through its center and perpendicular to its plane is given by:
\[ I = \frac{Mb^2}{6} \]
- According to the problem, this is equal to \( \frac{Ma^2}{6} \). Therefore, we can equate:
\[ \frac{Mb^2}{6} = \frac{Ma^2}{6} \]
- Canceling \( \frac{M}{6} \) from both sides gives:
\[ b^2 = a^2 \]

3. **Finding the Maximum Size of the Square:**
- The maximum size of the square that can be cut from the circular cardboard occurs when the square is inscribed in the circle.
- The diagonal of the square is equal to the diameter of the circle:
\[ \text{Diagonal of square} = b\sqrt{2} = 2r \]
- Therefore, we can express \( b \) in terms of \( r \):
\[ b = \frac{2r}{\sqrt{2}} = r\sqrt{2} \]

4. **Substituting \( b \) into the Moment of Inertia Equation:**
- Substitute \( b = r\sqrt{2} \) into the equation \( b^2 = a^2 \):
\[ (r\sqrt{2})^2 = a^2 \]
- This simplifies to:
\[ 2r^2 = a^2 \]
- Rearranging gives:
\[ r^2 = \frac{a^2}{2} \]

5. **Finding the Radius \( r \):**
- Taking the square root of both sides, we find:
\[ r = \frac{a}{\sqrt{2}} \]

### Final Result:
The radius of the circular cardboard is:
\[ r = \frac{a}{\sqrt{2}} \]