An equilateral triangle ABC is cut from a thin solid sheet of wood. [JEE Main 11-Jan-2019 Morning] |
(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 of 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) \[I=\frac{15}{16}{{I}_{0}}\]
B) \[I=\frac{9}{16}{{I}_{0}}\]
C) \[I=\frac{3}{4}{{I}_{0}}\]
D) \[I=\frac{{{I}_{0}}}{4}\]
Correct Answer: A
Solution :
[a] Moment of inertia at triangular lamina, where K= constant of proportionality |
Now moment of inertia of small lamina, |
So, moment of inertia of remaining part, |
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