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JEE Main 2016 · Conservation of Energy

Conservation of Energy question from JEE Main 2016

Q31JEE Main 2016Medium

Conservation of Energy

A person trying to lose weight by burning fat lifts a mass of 10 kg10\text{ kg} upto a height of 1 m 10001\text{ 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 3.8×107 J3.8 \times 10^{7}\text{ J} of energy per kg\text{kg} which is converted to mechanical energy with a 20%20\% efficiency rate. Take g=9.8 ms−2g = 9.8\text{ ms}^{-2}:

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Answer: (b)

  1. Work in lifting: W=1000 mgh=1000×10×9.8×1=9.8×104 JW = 1000\,mgh = 1000 \times 10 \times 9.8 \times 1 = 9.8 \times 10^4\text{ J}.
  2. Only 20% of fat energy becomes work: energy from fat =9.8×1040.2=4.9×105 J= \frac{9.8 \times 10^4}{0.2} = 4.9 \times 10^5\text{ J}.
  3. Fat used =4.9×1053.8×107=1.289×10−2 kg= \frac{4.9 \times 10^5}{3.8 \times 10^7} = 1.289 \times 10^{-2}\text{ kg}
  4. So the answer is (b) 12.89×10−3 kg12.89 \times 10^{-3}\text{ kg}.
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