JEE Main 2019 — Rotational Motion Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :
Choose an option
Show full solutionCorrect option: A
Correct answer
A16 cm
Step-by-step explanation
Consider an element of radius x and thickness dx
Mass of element, dm =
Here, = mass per unit area =
Moment of inertia of element, dI = (dm)x2
I =
=
=
= .....(i)
Moment of inertia of thin cylinder of same mass,
I = m ......(ii)
m =
= 250
r0 16 cm
Mass of element, dm =
Here, = mass per unit area =
Moment of inertia of element, dI = (dm)x2
I =
=
=
= .....(i)
Moment of inertia of thin cylinder of same mass,
I = m ......(ii)
m =
= 250
r0 16 cm
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Rotational Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, 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.