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JEE Main 2019 (12 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (12 Jan, Shift 2)

Q160JEE Main 2019 · 12 JAN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly?

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Answer: (c)

Parallel axis theorem: $I(x) = \frac{2}{5}MR^2 + Mx^2$ .
At $x = 0$ , $I = \frac{2}{5}MR^2 \neq 0$ , and $I$ grows as $x^2$ : a parabola starting above the origin.
So the answer is (c).

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