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Current Question (ID: 16892)

Question:
$\text{A short bar magnet of magnetic moment } 0.4 \text{ J/T is placed in a uniform magnetic field of } 0.16 \text{ T. The magnet is in stable equilibrium when the potential energy is:}$ $1. \ 0.064 \text{ J}$ $2. \ \text{zero}$ $3. \ -0.082 \text{ J}$ $4. \ -0.064 \text{ J}$
Options:
  • 1. $0.064 \text{ J}$
  • 2. $\text{zero}$
  • 3. $-0.082 \text{ J}$
  • 4. $-0.064 \text{ J}$
Solution:
$\text{Hint: } U = -mB \cos \theta$ $\text{Step: Find potential energy in stable equilibrium conditions.}$ $\text{In stable equilibrium conditions } \theta \text{ becomes zero.}$ $U_{\min} = -mB \cos \theta$ $\Rightarrow U_{\min} = -0.4 \times 0.16 \cos(0^\circ)$ $\Rightarrow U_{\min} = -0.064 \text{ J}$ $\text{Hence, option (4) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}