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Current Question (ID: 11101)

Question:
$\text{A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are:}$
Options:
  • 1. $80°\text{C}, 37°\text{C}$
  • 2. $95°\text{C}, 28°\text{C}$
  • 3. $90°\text{C}, 37°\text{C}$
  • 4. $99°\text{C}, 37°\text{C}$ (Correct)
Solution:
$\text{Initially } \eta = \left(1 - \frac{T_2}{T_1}\right) = \frac{W}{Q} = \frac{1}{6} \ldots(i)$ $\text{Finally } \eta' = \left(1 - \frac{T_2'}{T_1}\right) = \left(1 - \frac{(T_2-62)}{T_1}\right) = 1 - \frac{T_2}{T_1} + \frac{62}{T_1}$ $= \eta + \frac{62}{T_1} \ldots(ii)$ $\text{It is given that } \eta' = 2\eta. \text{ Hence solving equation (i) and (ii)}$ $\Rightarrow T_1 = 372 \text{ K} = 99°\text{C and } T_2 = 310\text{K} = 37°\text{C}$

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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
}