Questions

From a disc of radius* R* and mass *M*, a circular hole of diameter *R*, whose rim passes through the centre

is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? [NEET 2016]

60 mins Expert Faculty Ask Questions

a

13MR2/32

b

11MR2/32

c

9 MR2/32

d

15 MR2/32

detailed solution

Correct option is A

Considering the information given in the question, let us draw the figure.If the above figure is considered, then moment of inertia of disc will be given asI=Iremaining +I(R/2)⇒ Iremaining =I−I(R/2)Putting the values, we get=MR22−M4R222+M4R22=MR22−MR232+MR216=MR22−MR2+2MR232=MR22−3MR232=16MR2−3MR232Iremaining =13MR232

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The moment of inertia of a thin uniform rod of mass *M* and length *L* about an axis passing through its mid-point and perpendicular to its length is *I*_{0} , Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

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