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Current Question (ID: 17026)

Question:
$\text{Two coils have a mutual inductance of } 5 \text{ mH.}$ $\text{The current changes in the first coil according to the equation } I = I_0 \cos \omega t,$ $\text{where } I_0 = 10 \text{ A and } \omega = 100\pi \text{ rad/s.}$ $\text{The maximum value of emf induced in the second coil is:}$
Options:
  • 1. $5\pi \text{ V}$
  • 2. $2\pi \text{ V}$
  • 3. $4\pi \text{ V}$
  • 4. $\pi \text{ V}$
Solution:
$\text{Emf induced in the second will be-}$ $e = -M \frac{di}{dt} = -5 \times 10^{-3} \times (-I_0 \sin \omega t) \times \omega$ $= 5 \times 10^{-3} \times 100\pi \times (10 \sin \omega t) \text{ Volt}$ $\Rightarrow e_{\text{max}} = 5 \times 10^{-3} \times 100\pi \times 10 = 5\pi \text{ Volt}$

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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
}