An equilateral triangle $\mathrm{ABC}$ 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 $\mathrm{G}$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $\mathrm{G}$ and perpendicular to the plane of the triangle is $\mathrm{I}_{0}$. If the smaller triangle $\mathrm{DEF}$ is removed from $\mathrm{ABC}$, the moment of inertia of the remaining figure about the same axis is I. Then :

Question:

An equilateral triangle $\mathrm{ABC}$ 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 $\mathrm{G}$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $\mathrm{G}$ and perpendicular to the plane of the triangle is $\mathrm{I}_{0}$. If the smaller triangle $\mathrm{DEF}$ is removed from $\mathrm{ABC}$, the moment of inertia of the remaining figure about the same axis is I. Then :

  1. $\mathrm{I}=\frac{15}{16} \mathrm{I}_{0}$

  2. $\mathrm{I}=\frac{3}{4} \mathrm{I}_{0}$

  3. $\mathrm{I}=\frac{9}{16} \mathrm{I}_{0}$

  4. $\mathrm{I}=\frac{\mathrm{I}_{0}}{4}$

JEE Main Previous Year Single Correct Question of JEE Main from Physics System of Particles and Rotational Motion chapter.

JEE Main Previous Year 11 Jan. 2019 I


Correct Option: 1

Solution:

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