A person trying to lose weight by burning fat filts a mass of `10 kg` upto a being of `1m 1000` time . Assume that the potential energy lost each time be lower the mass is dissipated . How much far will be use up considering the work done only when the weight is lifted up ? Far supplies `3.8 xx 10^(7) J` of energy per kg wich is canverted to mechanical energy with `x 20 %` efficiency rate Take ` = 9.8 ms^(-2)`

Options

A. `2.45xx10^(-3)kg`
B. `6.45xx10^(-3)kg`
C. `9.89xx10^(-3)kg`
D. `12.89xx10^(-3)kg`

Answer

D

Explanation

Here, `m=10kg, h=1m, n=1000`
Work done in lifting mass `=nxxmgh`
`=1000xx10xx9.8xx1` ….(i)
If `M` is mass of fat burnt, then energy gained
`=Mxx3.8xx10^(7)xx(20)/(100)` ...(ii)
If there is no loss of energy, then from (i) and (ii)
`Mxx3.8xx10^(7)xx(20)/(100)=1000xx10xx9.8`
`M=(9.8xx10^(4)xx5)/(3.8xx10^(7))=12.89xx10^(-3)kg`