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
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
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In moment of inertia problems with symmetric objects, carefully consider the axis about which the calculation is done.
Updated On: May 15, 2025
- \( \sqrt{3} : 2 \)
- \( \sqrt{3} : 4 \)
- \( 3 : 4 \)
- \( 3 : 2 \)
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The Correct Option is D
Solution and Explanation
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:
\[
\boxed{3 : 2}
\]
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