JEE Main 2017PhysicsWork Power EnergyHardMCQ

JEE Main 2017 — Work Power Energy Question with Solution

JEE Main 2017 (02 Apr)

Question

A body of mass m=10−2 kg is moving in a medium and experiences a frictional force F=−k v 2 . Its initial speed is v0=10 m s−1. After 10 s its kinetic energy is 18mv02, then value of k will be:-

Choose an option

Show full solutionCorrect option: D
Correct answer
D10−4 kg m-1

Step-by-step explanation

After 10 sec, 12mv2=18mv02
⇒v=v02=5 m s-1
Acceleration,  a=− k v 2 m
⇒  dv dt =− k v 2 m
⇒ ∫105dvv2=−km ∫010dt
-15+110=-k10-2×10
k=10−4 kg m-1

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About this question

This is a previous-year question from JEE Main 2017, covering the Work Power Energy 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.