One quarter sector is cut from a uniform disc of radius 'R'. This sector has mass 'M'. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is

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Solution

The correct option is A
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
$I_{d i s c} = \frac{1}{2} M_{d i s c} R^{2} = \frac{1}{2} (4 M_{s e c t o r}) R^{2}$
Now, $I_{s e c t o r} = \frac{I_{d i s c}}{4}$
$\Rightarrow I_{s e c t o r} = \frac{1}{2} M R^{2} .$

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One quarter sector is cut from a uniform disc of radius 'R'. This sector has mass 'M'. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is

The correct option is A 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.