Question 65 of 280

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

A$\dfrac{{MR}^{2}}{2}$B$\dfrac{M\left({R}^{2}+{H}^{2}\right)}{4}$C$\dfrac{{MH}^{2}}{3}$D$\dfrac{{MR}^{2}}{3}$

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Answer: (A) $\dfrac{{MR}^{2}}{2}$

It can be assumed as several parts of discs having different radius, so

$I=\int dl=\int dm\left({r}^{2}\right)$
From diagram
$\dfrac{r}{y}=\tan \text{θ}=\dfrac{R}{H}$
$r=\dfrac{R}{H}\cdot y$
$dm=\text{ρ}\left(\text{π}{r}^{2}\right)dy$
From equation (i), (ii) and (iii)
$I=\dfrac{M{R}^{2}}{2}$

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.