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Current Question (ID: 20381)

Question:
$\text{The ratio of magnetic field due to coil at centre and at a distance of } R \text{ from the centre on the axis passing through the centre and perpendicular to the plane of ring is } \sqrt{x} : 1 \text{ (R is the radius of coil), find the value of } x.$
Options:
  • 1. $1$
  • 2. $8$
  • 3. $\sqrt{8}$
  • 4. $2$
Solution:
$\text{Hint: } B_C = \frac{\mu_0 i}{2R}$ $B_P = \frac{\mu_0}{4\pi} \times \frac{2 \times i \times \pi R^2}{(R^2 + R^2)^{3/2}} = \frac{\mu_0}{2R} \frac{i}{2\sqrt{2}} = \left(\frac{\mu_0 i}{4\sqrt{2}R}\right)$ $\frac{B_C}{B_P} = \frac{4\sqrt{2}}{2} = \sqrt{8} : 1$

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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
}