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Question

Three identical thin rods each of mass $$m$$ and length $$L$$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:


A
2mL23
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B
3mL22
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C
4mL23
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D
3mL24
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Solution

The correct option is A $$\dfrac{3mL^{2}}{2}$$
Sum of MI of the two rods, at vertex of which the axis passes, is $$2\times mL^2/3=2mL^2/3$$
For the third rod,  the distance between center of the rod and the axis is $$d=\sqrt3L/2$$
Using parallel axis theorem, its MI is $$mL^2/12+3mL^2/4=5mL^2/6$$
So total MI of the system is $$2mL^2/3+5mL^2/6=3mL^2/2$$

409174_20180_ans_c20c632c3dc5405695425eef5e729f71.png

Physics

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