Question

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) $\perp$ bisector of any side (IV) $\perp$ to the plane and passing through any corner

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

Answer 1

Answer: (B)

B

Answer 2

Let the mass of each point body be 'm' and the side length of the square be 'a'.

1. Moment of inertia about an axis passing through any side ($I_I$):
   Choose side AB. Masses at A and B are on the axis (distance = 0). Masses at C and D are at a perpendicular distance 'a' from the axis.
   $I_I = m(0)^2 + m(0)^2 + m(a)^2 + m(a)^2 = 2ma^2$.

2. Moment of inertia about an axis passing through opposite corners ($I_{II}$):
   Choose diagonal AC. Masses at A and C are on the axis (distance = 0). Masses at B and D are at a perpendicular distance of $a/\sqrt{2}$ from the diagonal.
   $I_{II} = m(0)^2 + m(a/\sqrt{2})^2 + m(0)^2 + m(a/\sqrt{2})^2 = ma^2/2 + ma^2/2 = ma^2$.

3. Moment of inertia about an axis perpendicular to the plane and passing through the midpoint of any side ($I_{III}$):
   Choose the midpoint M of side AB. The axis passes through M and is perpendicular to the plane.
   Masses at A and B are at a distance $a/2$ from M.
   Masses at C and D are at a distance $\sqrt{(a/2)^2 + a^2} = a\sqrt{5}/2$ from M.
   $I_{III} = m(a/2)^2 + m(a/2)^2 + m(a\sqrt{5}/2)^2 + m(a\sqrt{5}/2)^2$
   $I_{III} = ma^2/4 + ma^2/4 + 5ma^2/4 + 5ma^2/4 = 12ma^2/4 = 3ma^2$.

4. Moment of inertia about an axis perpendicular to the plane and passing through any corner ($I_{IV}$):
   Choose corner A. The axis passes through A and is perpendicular to the plane.
   Mass at A is on the axis (distance = 0).
   Masses at B and D are at a distance 'a' from A.
   Mass at C is at a distance $a\sqrt{2}$ (diagonal length) from A.
   $I_{IV} = m(0)^2 + m(a)^2 + m(a\sqrt{2})^2 + m(a)^2 = ma^2 + 2ma^2 + ma^2 = 4ma^2$.

Comparing the moments of inertia: $I_I = 2ma^2$
$I_{II} = ma^2$
$I_{III} = 3ma^2$
$I_{IV} = 4ma^2$

Decreasing order: $I_{IV} > I_{III} > I_I > I_{II}$
$4ma^2 > 3ma^2 > 2ma^2 > ma^2$

This corresponds to the order: (IV), (III), (I), (II).

The final answer is $\boxed{B}$