Shown in the fig. is a hollow ice-cream 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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$\frac{M H^{2}}{3}$

$\frac{M R^{2}}{2}$

$\frac{M R^{2}}{3}$

$\frac{M (R^{2} + H^{2})}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} F r o m t h e v e r t e x, a l o n g c u r v e d s u r f a c e g o t o y a n d t a k e a s t r i p o f t h i c k n e s s d y \\ a n d r a d i u s x . \\ M a s s o f t h i s s t r i p = d m = \frac{m a s s o n C S A}{C S A} \times a r e a o f s t r i p \\ \Rightarrow d m = \frac{M}{π R \sqrt{R^{2} + H^{2}}} \times 2 π x d x \\ A l s o, f r o m s i m i l a r t r i a n g l e s, \frac{x}{y} = \frac{R}{\sqrt{R^{2} + H^{2}}} \end{array}$

$\begin{array}{l} I =_{}^{} d m x^{2} = \frac{M}{π R \sqrt{R^{2} + H^{2}}}_{}^{} 2 π x^{3} d y \\ I = \frac{2 M}{R \sqrt{R^{2} + H^{2}}} \times \frac{R^{3}}{{(R^{2} + H^{2})}^{\frac{3}{2}}} \times {[\frac{y^{4}}{4}]}_{0}^{\sqrt{R^{2} + H^{2}}} = \frac{M R^{2}}{2} \end{array}$