An equilateral triangle ABC is cut from a thin solid sheet of wood .(see figure ) D,E and F are the mid-points of its sides as shown and G is the centre of the triangle.The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane to the triangle is `I_(0)` If the smaller triangle DEF is removed from ABC , the moment of inertia of the remaining figure about the same axis is I. Then :
A
`I=9/16I+0`
B
`I=15/16I_0`
C
`I=3/4I_0`
D
`I=I_0/4`
Text Solution
Verified by Experts
The correct Answer is:
B
Since, moment of inertia is proportional to square of side length and mass. I of `triangleDEF` about its centroid will be `1/4`th because of side length and `1/4` th because of mass. So `I(DEF)=I_0/16` So, for the remaining portion `I=I_0-I_0/4. 1.4=(15I_0)/16`
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
