Question

Three particles of each mass \( m \) are kept at the three vertices of an equilateral triangle. If \( I_1 \) is the moment of inertia of the system of the particles about an axis along one side of the triangle and \( I_2 \) is the moment of inertia of the system of the particles about an axis along the perpendicular bisector of a side, then \( I_1 : I_2 \) is

• A. \( \sqrt{3} : 2 \)
• B. \( \sqrt{3} : 4 \)
• C. \( 3 : 4 \)
• D. \( 3 : 2 \)

Hint:

In moment of inertia problems with symmetric objects, carefully consider the axis about which the calculation is done.

Answer 1

The Correct Option is D

For the moment of inertia about an axis along one side of the triangle \( I_1 \), the distance of each particle from the axis is proportional to the side length of the triangle. For the moment of inertia about the perpendicular bisector \( I_2 \), the distance is based on the distance from the center of mass. After calculating both moments of inertia, we find the ratio \( I_1 : I_2 = 3 : 2 \). Thus, the correct answer is: \[ \boxed{3 : 2} \]