JEE Main 2016PhysicsWork Power EnergyHardMCQ

JEE Main 2016 — Work Power Energy Question with Solution

JEE Main 2016 (03 Apr)

Question

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms-2 :

Choose an option

Show full solutionCorrect option: B
Correct answer
B12.89×10-3 kg

Step-by-step explanation

m=10kg, h=1m, 1000 times

Work done against gravity mgh

=10×1×9.8=98 J

In lifting 1000 times

=1000×98

=9.8×104 J

Fat burn with 20% efficiency =3.8×107 J×0.2

=7.6×106 J per kg

m=9.8×1047.6×106=1.289×10-2

=12.89×10-3 kg

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Work Power Energy 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 2016, 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.