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ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I0. If part ADE is removed, the moment of inertia of the remaining part about the same axis is NI016 where N is an integer. Value of N is:

 

PhysicsRotational MotionJEE MainJEE Main 2020 (04 Sep Shift 1)
Solution:
1035 Upvotes Verified Answer
The correct answer is: 11

Let mass of lamina =m, and length of side =l, then moment of inertila of lamina about an axis passing through G and perpendicular to the plane.

I0ml2

I0=kml2

Let moment of inertila of DEF=I1 about G

then, I1=I016

then,IADE=IBDF=IEFC=I2

3I2+I1=I0

3I2+I016=I0

 I2=5I016

So, Moment of inertila of DECB=2I2+I125I016+I016

=11I016

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