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Current Question (ID: 10758)

Question:
$\text{A cylindrical rod has temperatures } T_1 \text{ and } T_2 \text{ at its ends. The rate of flow of heat is } Q_1 \text{ cal/sec. If all the linear dimensions are doubled while keeping the temperature constant, then the rate of flow of heat } Q_2 \text{ will be:}$
Options:
  • 1. $4Q_1$
  • 2. $2Q_1$ (Correct)
  • 3. $\frac{Q_1}{4}$
  • 4. $\frac{Q_1}{2}$
Solution:
\text{Heat flow rate } Q = \frac{KA(T_1 - T_2)}{L} \text{When linear dimensions are doubled:} A_2 = \pi R_2^2 = 4\pi R_1^2 = 4A_1 L_2 = 2L_1 \text{Therefore: } Q_2 = \frac{KA_2\Delta T}{L_2} = \frac{4KA_1\Delta T}{2L_1} = 2Q_1

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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
}