Import Question JSON

Current Question (ID: 11096)

Question:

$\text{In a Carnot engine, when } T_2 = 0°\text{C and } T_1 = 200°\text{C, its efficiency is } \eta_1 \text{ and when } T_1 = 0°\text{C and } T_2 = -200°\text{C, its efficiency is } \eta_2\text{. What is the value of } \frac{\eta_1}{\eta_2}\text{?}$

Options:

1. $0.577$ (Correct)
2. $0.733$
3. $0.638$
4. $\text{cannot be calculated}$

Solution:

$\text{Hint: Find efficiency of the heat engine}$ $\text{Step 1:}$ $\eta = \frac{T_1 - T_2}{T_1}$ $\eta_1 = \frac{200 - 0}{200 + 273} = \frac{200}{473}$ $\eta_2 = \frac{0 - (-200)}{0 + 273} = \frac{200}{273}$ $\frac{\eta_1}{\eta_2} = \frac{273}{473} = 0.577$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}