Shown in the figure is a hollow icecream cone (it is open at the… | JEE Main 2020 (6 Sep, Shift 1)

Q152JEE Main 2020 · 6 SEP · SHIFT 1Difficult

Moment of Inertia & Radius of Gyration

Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is:

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Answer: (b)

Cut the cone surface into rings: a ring at slant distance $s$ has radius $r = \frac{R}{l}s$ and mass $dm = \frac{2M}{l^2}s\,ds$ .
$I = \int_0^l \frac{R^2 s^2}{l^2} \cdot \frac{2M}{l^2}s\,ds = \frac{2MR^2}{l^4} \cdot \frac{l^4}{4} = \frac{MR^2}{2}$
Independent of $H$ , so the answer is (b) $\frac{MR^2}{2}$ .

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