JEE MAIN - Physics (2016 (Offline) - No. 24)
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 $$3.8 \times {10^7}J$$ of energy per $$kg$$ which is converted to mechanical energy with a $$20\% $$ efficiency rate. Take $$g = 9.8\,m{s^{ - 2}}$$ :
$$9.89 \times {10^{ - 3}}\,\,kg$$
$$12.89 \times {10^{ - 3}}\,kg$$
$$2.45 \times {10^{ - 3}}\,\,kg$$
$$6.45 \times {10^{ - 3}}\,\,kg$$
Shpjegimi
Assume the amount of fat is used = x kg
So total Mechanical energy available through fat
= $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$
And work done through lifting up
= 10 $$ \times $$ 9.8 $$ \times $$ 1000 = 98000 J
$$ \Rightarrow $$ $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$ = 98000
$$ \Rightarrow $$ $$x$$ = 12.89 $$ \times $$ 10-3 kg
So total Mechanical energy available through fat
= $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$
And work done through lifting up
= 10 $$ \times $$ 9.8 $$ \times $$ 1000 = 98000 J
$$ \Rightarrow $$ $$x \times 3.8 \times {10^7} \times {{20} \over {100}}$$ = 98000
$$ \Rightarrow $$ $$x$$ = 12.89 $$ \times $$ 10-3 kg