JEE Main · 2016 · 03 Apr 2016 · Offline · Work, Power & Energy

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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 \times 1 0^{7} J

of energy per kg which is converted to mechanical energy with a

20%

efficiency rate. Take

g = 9.8 ms^{- 2}

(A) $6.45 \times 1 0^{- 3} kg$
(B) $9.89 \times 1 0^{- 3} kg$
(C) $12.89 \times 1 0^{- 3} kg$ ✓ Correct
(D) $2.45 \times 1 0^{- 3} kg$

Answer: (C)

Solution

The useful work is only the lifting work

m g h

done

n

times:

W = 10 \times 9.8 \times 1 \times 1000 = 9.8 \times 1 0^{4}

J. Each kilogram of fat yields

3.8 \times 1 0^{7}

J, but only 20% converts to mechanical work, so

x

kg supplies

x \times 3.8 \times 1 0^{7} \times 0.20

J. Equating supply to demand,

x \times 7.6 \times 1 0^{6} = 9.8 \times 1 0^{4}

, giving

x = \frac{9.8 \times 1 0 ^{4}}{7.6 \times 1 0 ^{6}} \approx 12.89 \times 1 0^{- 3}

kg.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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JEE Main · 2016 · 03 Apr 2016 · Offline · Work, Power & Energy

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