| 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 \[\text{3}.\text{8}\times \text{1}0\text{7J}\]of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms-2:-[JEE MAIN - I 3-4-2016] |
A) \[\text{12}.\text{89}\times \text{1}{{0}^{\text{3}}}\text{kg}\]
B) \[\text{2}.\text{45}\times \text{1}{{0}^{\text{3}}}\text{kg}\]
C) \[\text{6}.\text{45}\times \text{1}{{0}^{\text{3}}}\text{kg}\]
D) \[\text{9}.\text{89}\times \text{1}{{0}^{\text{3}}}\text{kg}\]
Correct Answer: A
Solution :
| [a] Work done against gravity = (mgh) 1000 in lifting 1000 times |
| \[=\text{1}0\times \text{9}.\text{8}\times \text{1}{{0}^{\text{3}}}\] |
| \[=\text{9}.\text{8}\times \text{1}{{0}^{\text{4}}}\]Joule |
| 20% efficiency is to converts fat into energy. |
| \[[\text{2}0%\text{of3}.\text{8}\times \text{1}0\text{7J}]\times (m)=\text{9}.\text{8}\times \text{1}{{0}^{\text{4}}}\] |
| (Where m is mass)\[\text{m}=\text{12}.\text{89}\times \text{1}{{0}^{\text{3}}}\text{kg}\] |
You need to login to perform this action.
You will be redirected in
3 sec
