JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 12)
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 :
_6th_September_Morning_Slot_en_12_1.png)
$${{M\left( {{R^2} + {H^2}} \right)} \over 3}$$
$${{M{R^2}} \over 2}$$
$${{M{R^2}} \over 3}$$
$${{M{H^2}} \over 3}$$
Explanation
Moment of inertia of this cone will same as
circular disk of mass (M) and radius R.
_6th_September_Morning_Slot_en_12_2.png)
I = $${{M{R^2}} \over 2}$$
I = $${{M{R^2}} \over 2}$$
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