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JEE Main 2019 (11 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (11 Jan, Shift 1)

Q167JEE Main 2019 · 11 JAN · SHIFT 1Difficult

Moment of Inertia & Radius of Gyration

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 of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:

Figure

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Answer: (b)

  1. DEF has side a2\frac{a}{2}, mass M4\frac{M}{4} and the same centre G.
  2. I∝Ma2⇒IDEF=14×14I0=I016I \propto Ma^2 \Rightarrow I_{DEF} = \frac{1}{4} \times \frac{1}{4}I_0 = \frac{I_0}{16}
  3. I=I0−I016=1516I0I = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0, so the answer is (b).
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