JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(\frac{ MR ^{2}}{2}\)
- B \(\frac{ MH ^{2}}{3}\)
- C \(\frac{ MR ^{2}}{3}\)
- D \(\frac{ M \left( R ^{2}+ H ^{2}\right)}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{ MR ^{2}}{2}\)
Step-by-step Solution
Detailed explanation
Area \(=\pi R \ell=\pi R \left(\sqrt{ H ^{2}+ R ^{2}}\right)\) Area of element \(d A =2 \pi rd \ell==2 \pi r \frac{ dh }{\cos \theta}\) mass of element \(dm =\frac{ M }{\pi R \sqrt{ H ^{2}+ R ^{2}}} \times \frac{2 \pi rdh }{\cos \theta}\)…
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