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Current Question (ID: 16886)

Question:
$\text{Two magnets } A \text{ and } B \text{ are identical and these are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. A magnetic needle is placed between the magnets at point } P \text{ which gets deflected through an angle } \theta \text{ under the influence of magnets. The ratio of distance } d_1 \text{ and } d_2 \text{ will be:}$
Options:
  • 1. $(2 \tan \theta)^{\frac{1}{3}}$
  • 2. $(2 \tan \theta)^{-\frac{1}{3}}$
  • 3. $(2 \cot \theta)^{\frac{1}{3}}$
  • 4. $(2 \cot \theta)^{-\frac{1}{3}}$
Solution:
$\text{In equilibrium, } B_1 = B_2 \tan \theta$ $\Rightarrow \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_1^3} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d_2^3} \tan \theta$ $\Rightarrow \frac{d_1}{d_2} = (2 \cot \theta)^{\frac{1}{3}}$

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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
}