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Current Question (ID: 19696)

Question:

$\text{Let } \eta_1 \text{ is the efficiency of an engine at } T_1 = 447 \, ^\circ\text{C} \text{ and } T_2 = 147 \, ^\circ\text{C} \text{ while } \eta_2 \text{ is the efficiency at } T_1 = 947 \, ^\circ\text{C} \text{ and } T_2 = 47 \, ^\circ\text{C} . \text{ The ratio } \frac{\eta_1}{\eta_2} \text{ will be:}$

Options:

1. $0.41$
2. $0.56$
3. $0.73$
4. $0.70$

Solution:

$\text{Hint: } \eta = 1 - \frac{T_L}{T_H}$ $\text{Step 1: Find the efficiency of both engines.}$ $\eta_1 = 1 - \frac{147 + 273}{447 + 273} = 1 - \frac{420}{720}$ $\Rightarrow \eta_1 = \frac{300}{720}$ $\eta_2 = 1 - \frac{47 + 273}{947 + 273} = 1 - \frac{320}{1220}$ $\Rightarrow \eta_2 = \frac{900}{1220}$ $\text{Step 2: Find the ratio of efficiencies.}$ $\frac{\eta_1}{\eta_2} = \frac{300}{720} \times \frac{1220}{900} = \frac{122}{72 \times 3}$ $\Rightarrow \frac{\eta_1}{\eta_2} = 0.56$

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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
}