One quarter sector disc of radius 'R', has mass 'M'. Its moment of inertia about the axis shown in figure is.
A
12MR2
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B
14MR2
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C
18MR2
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D
√2MR2
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Solution
The correct option is A12MR2 If we assume complete disc to be present, then it would have a mass 4 times the mass of the sector. Then, moment of inertia of the complete disc is Idisc=12MdiscR2=12(4Msector)R2
Now, Isector=Idisc4 ⇒Isector=12MR2.