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Current Question (ID: 19700)

Question:
$\text{In the series sequence of two engines } E_1 \text{ and } E_2 \text{ as shown.}$ $T_1 = 600 \text{ K and } T_2 = 300 \text{ K.}$ $\text{It is given that both the engines working on the Carnot principle have the same efficiency, then temperature } T \text{ at which exhaust of } E_1 \text{ is fed into } E_2 \text{ is equal to:}$
Options:
  • 1. $300\sqrt{2} \text{ K}$
  • 2. $200\sqrt{2} \text{ K}$
  • 3. $300\sqrt{3} \text{ K}$
  • 4. $200\sqrt{3} \text{ K}$
Solution:
$\eta_1 = 1 - \frac{T}{600}$ $\eta_2 = 1 - \frac{300}{T}$ $\text{Given: } \eta_1 = \eta_2$ $\Rightarrow \frac{T}{600} = \frac{300}{T}$ $\Rightarrow T = \sqrt{180000} \text{ K} = 300\sqrt{2} \text{ K}$ $\Rightarrow n = 2$

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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
}