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Current Question (ID: 19224)

Question:

$\text{A person trying to lose weight by burning fat lifts a mass of } 10 \text{ kg}$ $\text{up to a height of } 1 \text{ m } 1000 \text{ times. Assume that the potential energy}$ $\text{lost each time he lowers the mass is dissipated. How much fat will}$ $\text{he use up considering the work done only when the weight is lifted}$ $\text{up? Fat supplies } 3.8 \times 10^7 \text{ J of energy per kg which is converted to}$ $\text{mechanical energy with a } 20\% \text{ efficiency rate. Take } g = 9.8 \text{ ms}^{-2}.$

Options:

1. $2.45 \times 10^{-3} \text{ kg}$
2. $6.45 \times 10^{-3} \text{ kg}$
3. $9.89 \times 10^{-3} \text{ kg}$
4. $12.89 \times 10^{-3} \text{ kg}$

Solution:

$\text{Hint: P.E = mgh}$ $\text{PE} = 1000 \times mgh$ $= 1000 \times 10 \times 9.8 \times 1$ $= 98000 \text{ J}$ $\text{mechanical energy per unit mass} = \frac{2}{100} \times 3.8 \times 10^7$ $= 7.6 \times 10^5 \text{ J/kg}$ $\therefore \text{ Fat burn} = \frac{98000}{7.6 \times 10^5} \text{ kg}$ $= 12.89$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}