An equilateral triangle ABC is cut from a thin solid sheet of wood. 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 $\mathrm { I } _ { 0 }$. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is $I$. Then (1) $I = \frac { 15 } { 16 } I _ { 0 }$ (2) $\mathrm { I } = \frac { 3 } { 4 } \mathrm { I } _ { 0 }$ (3) $I = \frac { 9 } { 16 } I _ { 0 }$ (4) $\mathrm { I } = \frac { \mathrm { I } _ { 0 } } { 4 }$
An equilateral triangle ABC is cut from a thin solid sheet of wood. 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 $\mathrm { I } _ { 0 }$. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is $I$. Then\\
(1) $I = \frac { 15 } { 16 } I _ { 0 }$\\
(2) $\mathrm { I } = \frac { 3 } { 4 } \mathrm { I } _ { 0 }$\\
(3) $I = \frac { 9 } { 16 } I _ { 0 }$\\
(4) $\mathrm { I } = \frac { \mathrm { I } _ { 0 } } { 4 }$