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Current Question (ID: 20591)

Question:
$\text{The frequencies at which the current amplitude in an } LCR \text{ series circuit becomes } \frac{1}{\sqrt{2}} \text{ times its maximum value, are } 212 \text{ rad s}^{-1} \text{ and } 232 \text{ rad s}^{-1}. \text{ The value of resistance in the circuit is } R = 5 \, \Omega. \text{ The self-inductance in the circuit is:}$
Options:
  • 1. $510 \, \text{mH}$
  • 2. $110 \, \text{mH}$
  • 3. $250 \, \text{mH}$
  • 4. $340 \, \text{mH}$
Solution:
$\Delta \omega = \omega_2 - \omega_1 = \frac{R}{L}$

Import JSON File

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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
}