Q 11-06-176JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
Shown in the figure is a hollow ice-cream cone (it is open at the top). If its mass is $M$, the radius of its top is $R$ and its height is $H$, then its moment of inertia about its axis is:
Answer: (A) $\dfrac{MR^2}{2}$
Cut the cone surface into thin rings. A ring at slant distance $s$ from the tip has radius $r = \dfrac{Rs}{L}$ ($L$ = slant height) and mass proportional to its area $2\pi r\,ds$, i.e. $dm = \dfrac{2M s\,ds}{L^2}$.
$$I = \int r^2\,dm = \int_0^L \frac{R^2 s^2}{L^2}\cdot\frac{2Ms}{L^2}\,ds = \frac{2MR^2}{L^4}\cdot\frac{L^4}{4} = \frac{MR^2}{2}$$
(The result does not depend on $H$.)
Solution by Sreeraj P, M.Sc Physics