Answer

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**Hint:**Here efficiency is given 20% which means:

\[Efficiency{\text{ }}\left( n \right){\text{ }} = {\text{ }}output/input.\] (Where output is work done)

Therefore, the fat used when weight is lifted by considering the input will be related with work done and fat supplied i.e,

\[\Rightarrow Fat{\text{ }}used{\text{ }} = {\text{ }}Input/fat{\text{ }}supplied\]

\[\Rightarrow {{20}{l}}{ = {\text{ }}5 \times {{10}^5}{\text{ }}J/3.8 \times {{10}^7}{\text{ }}J} \\

\Rightarrow { = {\text{ }}12.89 \times {{10}^{ - 3}}{\text{ }}kg.}

\]

**Complete step by step answer:**

Here, we know that

\[

\Rightarrow {{20}{l}}{m{\text{ }} = {\text{ }}10kg,} \\

\Rightarrow {h{\text{ }} = {\text{ }}1m,} \\

\Rightarrow {n{\text{ }} = {\text{ }}1000}

\]

Work done in lifting mass = $n \times mgh$

$ = 1000 \times 10 \times 9.8 \times 1 - - (1)$

If $m$ is mass of fat burnt, then energy gained

$ = m \times 3.8 \times {10^7} \times \dfrac{{20}}{{100}} - - (2)$

If there is no loss of energy, then from (1) and (2)

$

\Rightarrow m \times 3.8 \times {10^7} \times \dfrac{{20}}{{100}} = 1000 \times 10 \times 9.8 \\

\therefore m = \dfrac{{9.8 \times {{10}^4} \times 5}}{{3.8 \times {{10}^7}}} = 12.89 \times {10^{ - 3}}kg \\

$

From the above calculations, we can conclude that $12.89 \times {10^{ - 3}}kg$ is the correct answer.

**Hence,option (C) is the right answer.**

**Note:**Here in the given question, the work done is considered when the weight is lifted and not when he lowers the mass because, while lifting the weight energy is consumed due to which the work is done against the gravitational force.

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