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Current Question (ID: 20493)

Question:
$\text{A uniform magnetic field } B \text{ exists in a direction perpendicular to the plane of a square loop made of a metal wire.}$ $\text{The wire has a diameter of } 4 \text{ mm and a total length of } 30 \text{ cm.}$ $\text{The magnetic field changes with time at a steady rate } \frac{dB}{dt} = 0.032 \text{ T s}^{-1}. \text{ The induced current in the loop is close to: (the resistivity of the metal wire is } 1.23 \times 10^{-8} \, \Omega \text{m})$
Options:
  • 1. $0.34 \, \text{A}$
  • 2. $0.53 \, \text{A}$
  • 3. $0.61 \, \text{A}$
  • 4. $0.43 \, \text{A}$
Solution:
$\text{Hint: } \varepsilon = -\frac{d\phi}{dt}$ $q_i = \frac{d(Ba^2)}{dt} = a^2 \frac{dB}{dt}$ $i = \frac{q}{R} = \frac{a^2 \frac{dB}{dt}}{\frac{p(40)}{\pi^2}}$

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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
}