JEE Main 2019 — Rotational Motion Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: A
Correct answer
A16 rad/s2
Step-by-step explanation
40 + f = m(R) . . . .(i)
40 R f R = mR2
40 f = mR . . . .(ii)
From (i) and (ii)
= = 16
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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.