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QJEEM20Ph3bf0: JEE Main 2020 Physics - Class 11 JEEM20Ph3bf0 Solution: Shown in the figure is a hollow icecream cone (it is open at…

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 : JEE Main 2020 (Online) 6th September Morning Slot Physics - Rotational Motion Question 118 English

Multiple Choice Options

  • A M(R2+H2)3{{M\left( {{R^2} + {H^2}} \right)} \over 3}
  • B MR22{{M{R^2}} \over 2}
  • C MR23{{M{R^2}} \over 3}
  • D MH23{{M{H^2}} \over 3}

Step-by-Step Verified Solution

Moment of inertia of this cone will same as circular disk of mass (M) and radius R.
JEE Main 2020 (Online) 6th September Morning Slot Physics - Rotational Motion Question 118 English Explanation

I = MR22{{M{R^2}} \over 2}

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