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Current Question (ID: 16868)

Question:

$\text{A bar magnet having a magnetic moment of } 2.0 \times 10^5 \text{ JT}^{-1} \text{ is placed along the direction of the uniform magnetic field of magnitude, } B = 14 \times 10^{-5} \text{ T.}$ $\text{The work done in rotating the magnet slowly through } 60^\circ \text{ from the direction of the field is:}$

Options:

1. $14 \text{ J}$
2. $8.4 \text{ J}$
3. $4 \text{ J}$
4. $1.4 \text{ J}$

Solution:

$\text{Hint: } W = MB (\cos \theta_1 - \cos \theta_2)$ $\text{Step: Find the work done in rotating the bar magnet.}$ $\text{The work done in rotating the magnetic field is given by;}$ $W = MB (\cos \theta_1 - \cos \theta_2)$ $\Rightarrow W = MB (\cos 0^\circ - \cos 60^\circ) \quad [\theta_1 = 0^\circ, \theta_2 = 60^\circ]$ $\Rightarrow W = 28 \left(1 - \frac{1}{2}\right) = 14 \text{ J}$ $\text{Hence, option (1) is the correct answer.}$

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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
}