JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2016
- Medium
- MCQ
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 \(g=9.8\) \(ms^{-2}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(12.89 \times 10^{-3}\) \( kg\)
Step-by-step Solution
Detailed explanation
\(n = \frac{W}{{input}} = \frac{{mgh \times 1000}}{{input}} = \frac{{10 \times 9.8 \times 1 \times 1000}}{{input}}\)
Input \(=\frac{{98000}}{{0.2}} = 49 \times {10^4}\,J\)
Fat used \(=\frac{{49 \times {{10}^4}}}{{3.8 \times {{10}^7}}} = 12.89 \times {10^{ - 3}}\,kg.\)
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