JEE Main 2020PhysicsLaws of MotionMediumMCQ

JEE Main 2020 — Laws of Motion Question with Solution

JEE Main 2020 (07 Jan Shift 1)

Question

A 60HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : 1 HP=746 W, g=10 m s-2

Choose an option

Show full solutionCorrect option: B
Correct answer
B1.9 m s-1

Step-by-step explanation

4000×V+mg×V=P
60×7464000+20000=V
V=1.9 m s-1

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About this question

This is a previous-year question from JEE Main 2020, covering the Laws of 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.