question_answer

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,