Question

Six particles each of mass 100 g are placed at the six vertices of a regular hexagon of side $10\sqrt{3}$ m. The moment of inertia of the system of particles about an axis passing through any two diagonally opposite vertices of the hexagon is

• A. 75 $kg m^2$
• B. 90 $kg m^2$
• C. 25 $kg m^2$
• D. 50 $kg m^2$

Hint:

For a regular hexagon of side $a$, the perpendicular distance from a vertex to the longest diagonal is given by $\frac{\sqrt{3}}{2}a$.

Answer 1

The Correct Option is B

Step 1: Identify the Geometry:
We have a regular hexagon with side length $a = 10\sqrt{3}$ m. Mass of each particle $m = 100 \text{ g} = 0.1 \text{ kg}$. The axis of rotation passes through two diagonally opposite vertices (e.g., A and D).
Step 2: Determine Perpendicular Distances ($r$):
Moment of Inertia $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.

• Vertices on Axis (2 particles): The distance $r = 0$. Contribution $= 0$.
• Vertices off Axis (4 particles): In a regular hexagon, the distance of the vertices (say B, C, E, F) from the main diagonal (AD) is equal to the altitude of the equilateral triangle formed by the center and two adjacent vertices. Distance $d = a \sin(60^\circ) = a \frac{\sqrt{3}}{2}$.

Step 3: Calculate Moment of Inertia:
Total $I$ comes from the 4 particles at distance $d$. \[ I = 4 \times m \times d^2 \] \[ I = 4m \left( \frac{a\sqrt{3}}{2} \right)^2 \] \[ I = 4m \left( \frac{3a^2}{4} \right) = 3ma^2 \]
Step 4: Substitute Values:
$m = 0.1$ kg. $a = 10\sqrt{3}$ m $\implies a^2 = 100 \times 3 = 300 \, m^2$. \[ I = 3 \times 0.1 \times 300 \] \[ I = 0.3 \times 300 = 90 \, kg m^2 \]
Step 5: Final Answer:
The moment of inertia is 90 $kg m^2$.