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

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$2.45 \times 10^{- 3} k g$

$6.45 \times 10^{- 3} k g$

$9.89 \times 10^{- 3} k g$

$12.89 \times 10^{- 3} k g$

answer is D.

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Detailed Solution

Here, $m = 10 kg, h = 1 m, g = 9.8 m s^{- 2}, n = 1000$

Energy of fat $= 3.8 \times 10^{7} J k g^{- 1}$

Efficiency, $η = 20 % = \frac{1}{5}$

Net work done by the man in lifting the mass

$= n \times (gain in potential energy of the mass)$

$n (m g h) = 1000 \times 10 \times 9.8 \times 1 = 98000 J$

$η = \frac{N e t w o r k d o n e b y t h e m a n}{E n e r g y i n t h e f a t}$

$\frac{1}{5} = \frac{98000}{m \times 3.8 \times 10^{7}} o r m = \frac{98000 \times 5}{3.8 \times 10^{7}}$

Therefore $m = 12.89 \times 10^{- 3} kg$

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