Import Question JSON

Current Question (ID: 16947)

Question:
$\text{A circular loop of radius } R \text{ carrying current } i \text{ lies in the } x-y \text{ plane.}$ $\text{If the centre of the loop coincides with the origin, then the total magnetic flux passing through the } x-y \text{ plane will be:}$
Options:
  • 1. $\text{directly proportional to } i.$
  • 2. $\text{directly proportional to } R.$
  • 3. $\text{directly proportional to } R^2.$
  • 4. $\text{Zero.}$
Solution:
$(4) \text{ The circular loop behaves as a magnetic dipole whose one surface will be } N\text{-pole and another will be } S\text{-pole.}$ $\text{Therefore magnetic lines of force emerge from } N \text{ will meet at } S. \text{ Hence total magnetic flux through } x-y \text{ plane is zero.}$

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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
}