JEE Main 2016 · 3 Apr · Shift 1 · Physics Q4

← JEE Main 2016 · 3 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 4 of 90

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 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 $3.8\times10^7$ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms $^{-2}$ :

2.45\times10^{-3}

6.45\times10^{-3}

9.89\times10^{-3}

12.89\times10^{-3}

Answer and worked solution

Official answer: D

Work done in lifting: $1000\times10\times9.8\times1=9.8\times10^4$ J. At 20% efficiency the fat must supply $4.9\times10^5$ J, i.e. $\frac{4.9\times10^5}{3.8\times10^7}=12.89\times10^{-3}$ kg.

Topics:Power Potential energy and conservative forces

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 4