The moment of inertia of a thin sheet of mass M of the given shape about the specified axis is ( axis and sheet both are in same plane )
A. `3/2 Ma^2`
B. `3/4 Ma^2`
C. `7/12 Ma^2`
D. `1/3 Ma^2`

Options

A. a. `3/2 Ma^2`
B. b. `3/4 Ma^2
C. `7/12 Ma^2`
D. `1/3 Ma^2`

Answer

To solve the problem of finding the moment of inertia of a thin sheet of mass \( M \) about a specified axis in the same plane as the sheet, we will follow these steps:

### Step 1: Understand the Geometry
We have a thin sheet in the shape of a square with side length \( a \). The axis of rotation is a diagonal of the square, which makes an angle of 45 degrees with the horizontal.

### Step 2: Moment of Inertia About the Center
The moment of inertia of a square sheet about an axis through its center and parallel to one of its sides is given by:

\[ I_O = \frac{1}{12} M a^2 \]

where \( M \) is the mass of the sheet and \( a \) is the side length of the square.

### Step 3: Apply the Parallel Axis Theorem
Since we want to find the moment of inertia about the diagonal axis (let's denote it as \( I_P \)), we can use the parallel axis theorem:

\[ I_P = I_O + M d^2 \]

where \( d \) is the distance from the center of mass to the new axis of rotation.

### Step 4: Calculate the Distance \( d \)
The distance \( d \) from the center of the square to the diagonal can be calculated as follows. The diagonal of the square divides it into two equal triangles, and the distance from the center to the diagonal can be found using the formula for the distance from a point to a line. For a square of side \( a \), the distance \( d \) from the center to the diagonal is:

\[ d = \frac{a}{\sqrt{2}} \]

### Step 5: Substitute \( d \) into the Equation
Now, substituting \( d \) into the parallel axis theorem:

\[ I_P = \frac{1}{12} M a^2 + M \left( \frac{a}{\sqrt{2}} \right)^2 \]

Calculating \( \left( \frac{a}{\sqrt{2}} \right)^2 \):

\[ \left( \frac{a}{\sqrt{2}} \right)^2 = \frac{a^2}{2} \]

Thus, substituting this back into the equation gives:

\[ I_P = \frac{1}{12} M a^2 + M \cdot \frac{a^2}{2} \]

### Step 6: Combine the Terms
Now, we need a common denominator to combine the terms:

\[ I_P = \frac{1}{12} M a^2 + \frac{6}{12} M a^2 = \frac{7}{12} M a^2 \]

### Conclusion
Thus, the moment of inertia of the thin sheet about the specified axis is:

\[ I_P = \frac{7}{12} M a^2 \]

The correct answer is option C: \( \frac{7}{12} M a^2 \).

---