Four point size dense bodies of same mass are attached at four corners of a light square frame . Identify the decreasing order of their moments of inertia about following axes .
I) Passing through any side
II) Passing through opposite corners III) `bot^(r)` bisector of any side
(IV) `bot^(r)` to the plane and passing through any corner

Options

A. III,IV,I,II
B. IV,III,I,II
C. III,II,IV,I
D. IV,III,II,I

Answer

To determine the decreasing order of the moments of inertia of four point masses located at the corners of a square frame about various axes, we will calculate the moments of inertia for each case step by step.

### Given:
- Four point masses, each of mass \( m \), are located at the corners of a square frame.
- The side length of the square frame is assumed to be \( 2a \).

### Moments of Inertia Calculations:

#### I) Moment of Inertia about an axis passing through any side (I1):
- For this axis, only the two masses that are not on the axis contribute to the moment of inertia.
- The distance of each of these two masses from the axis is \( a \) (half the side length).
- Thus, the moment of inertia \( I_1 \) is given by:
\[ I_1 = m \cdot a^2 + m \cdot a^2 = 2m \cdot a^2 \]

#### II) Moment of Inertia about an axis passing through opposite corners (I2):
- For this axis, the two masses at the corners through which the axis passes do not contribute (distance = 0).
- The distance of the other two masses from the axis is the diagonal distance.
- The diagonal length \( d \) is calculated as:
\[ d = \sqrt{(2a)^2 + (2a)^2} = \sqrt{8a^2} = 2\sqrt{2}a \]
- The distance of each mass from the axis is \( \frac{d}{2} = \sqrt{2}a \).
- Thus, the moment of inertia \( I_2 \) is given by:
\[ I_2 = m \left(\sqrt{2}a\right)^2 + m \left(\sqrt{2}a\right)^2 = 2m \cdot 2a^2 = 4m \cdot a^2 \]

#### III) Moment of Inertia about the bisector of any side (I3):
- The axis bisects the side, so the distance of the two masses on the opposite corners is \( \sqrt{(a^2 + a^2)} = \sqrt{2}a \).
- Thus, the moment of inertia \( I_3 \) is given by:
\[ I_3 = m \left(\sqrt{2}a\right)^2 + m \left(\sqrt{2}a\right)^2 = 2m \cdot 2a^2 = 4m \cdot a^2 \]

#### IV) Moment of Inertia about an axis perpendicular to the plane and passing through any corner (I4):
- The distance from the axis to each of the other three masses is \( 2a \) (the height from the corner to the plane).
- Thus, the moment of inertia \( I_4 \) is given by:
\[ I_4 = m(2a)^2 + m(2a)^2 + m(2a)^2 = 3m \cdot 4a^2 = 12m \cdot a^2 \]

### Summary of Moments of Inertia:
- \( I_1 = 2ma^2 \)
- \( I_2 = 4ma^2 \)
- \( I_3 = 4ma^2 \)
- \( I_4 = 12ma^2 \)

### Decreasing Order:
Now we can arrange the moments of inertia in decreasing order:
1. \( I_4 = 12ma^2 \)
2. \( I_2 = 4ma^2 \)
3. \( I_3 = 4ma^2 \)
4. \( I_1 = 2ma^2 \)

Thus, the decreasing order of the moments of inertia is:
\[ I_4 > I_2 = I_3 > I_1 \]