Q.

Moment of inertia of a thin ring of mass m and radius r about an axis passing through its centre and perpendicular to its plane is I. Three such identical rings are joined together as shown in the figure. Then moment of inertia of the system about axis AB is

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a

534I

b

8+233I

c

3+432I

d

15+432I

answer is D.

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Detailed Solution

I=mR2 (R=Radius)M.O.I. of a ring about a diameter =12mR2=I2d = Distance of centre of ring 3 from axis AB =R+2Rsin60o=3+1R∴IAB=mR22+mR2 x 2+mR22+m3+12R2=72mR2+22+3 x mR2=15+432 x I
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