JEE Main 2017 — Rotational Motion Question with Solution
From: JEE Main 2017 (Online) 9th April Morning Slot
Question
A circular hole of radius is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
Mass of removed disc = Radius of removed disc =
Moment of inertia of removed disc about it's own axis (O')
= =
Moment of inertia of removed disc about O,
= Icm + mx2
= +
=
Moment of inertia of complete disc about point 'O',
I =
Moment of Inertia of remaining disc
=
=
Moment of inertia of removed disc about it's own axis (O')
= =
Moment of inertia of removed disc about O,
= Icm + mx2
= +
=
Moment of inertia of complete disc about point 'O',
I =
Moment of Inertia of remaining disc
=
=
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This is a previous-year question from JEE Main 2017, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.