Three dense point size bodies of same mass are attached at three vertices of a light equilateral triangular frame . Identify the increasing order of their moment of inertia about following axis .
I) About an axis `bot^(r )` to plane and passing through a corner
II) About an axis `bot^(r )` to plane passing through centre
III) About `bot^(r)` bisector of any side .

Options

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

Answer

To solve the problem of determining the increasing order of the moment of inertia of three point masses attached to the vertices of an equilateral triangle about different axes, we will analyze each case step by step.

### Step 1: Define the System
We have three point masses \( m \) located at the vertices of an equilateral triangle with side length \( 2a \). We will denote the moment of inertia for each axis as \( I_1 \), \( I_2 \), and \( I_3 \) respectively.

### Step 2: Moment of Inertia about an Axis Perpendicular to the Plane and Passing Through a Corner (Case I)
- For this case, we choose one vertex as the axis of rotation. The moment of inertia \( I_1 \) can be calculated as follows:

\[ I_1 = m \cdot 0^2 + m \cdot (2a)^2 + m \cdot (2a)^2 \]

Here, the first term is zero because the point mass at the axis has a distance of zero from the axis. The other two point masses are at a distance of \( 2a \) from the axis.

\[ I_1 = 0 + m \cdot (2a)^2 + m \cdot (2a)^2 = 0 + 4ma^2 + 4ma^2 = 8ma^2 \]

### Step 3: Moment of Inertia about an Axis Perpendicular to the Plane and Passing Through the Center (Case II)
- For this case, we find the distance of each mass from the center of the triangle. The distance \( d \) from the center to each vertex can be derived using geometry. The center divides the height of the triangle into a ratio of 2:1.

The height \( h \) of the triangle is given by:

\[ h = \sqrt{(2a)^2 - a^2} = \sqrt{3a^2} = a\sqrt{3} \]

The distance from the centroid to each vertex is:

\[ d = \frac{2}{3}h = \frac{2}{3}(a\sqrt{3}) = \frac{2a\sqrt{3}}{3} \]

Now, we calculate \( I_2 \):

\[ I_2 = 3m \cdot d^2 = 3m \cdot \left(\frac{2a\sqrt{3}}{3}\right)^2 \]

\[ I_2 = 3m \cdot \frac{4a^2 \cdot 3}{9} = \frac{4ma^2}{3} \]

### Step 4: Moment of Inertia about the Perpendicular Bisector of Any Side (Case III)
- For this case, we consider the bisector of one side of the triangle. The distance from the bisector to the two vertices is \( a \) and to the opposite vertex is \( h \) where \( h = \sqrt{3}a \).

Thus, the moment of inertia \( I_3 \) is:

\[ I_3 = m \cdot a^2 + m \cdot a^2 + m \cdot (h^2) \]

\[ I_3 = 2ma^2 + m \cdot (3a^2) = 2ma^2 + 3ma^2 = 5ma^2 \]

### Step 5: Compare the Moments of Inertia
Now we have:
- \( I_1 = 8ma^2 \)
- \( I_2 = 4ma^2 \)
- \( I_3 = 5ma^2 \)

### Step 6: Increasing Order of Moment of Inertia
To find the increasing order:
- \( I_2 < I_3 < I_1 \)
- Thus, the increasing order is \( I_2, I_3, I_1 \).

### Final Answer
The increasing order of the moment of inertia about the specified axes is:
**II < III < I**