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Current Question (ID: 16910)

Question:

$\text{The magnetic moment of a magnet } (10 \text{ cm} \times 4 \text{ cm} \times 1 \text{ cm}) \text{ is } 4 \text{ Am}^2. \text{ Its intensity of magnetisation is:}$

Options:

1. $10^3 \text{ A/m}$
2. $10^2 \text{ A/m}$
3. $10^5 \text{ A/m}$
4. $10^4 \text{ A/m}$

Solution:

$\text{Hint: } I = \frac{M}{V}$ $\text{Step: Find the intensity of magnetisation.}$ $\text{The intensity of magnetization } I \text{ of the given magnet, we can use the formula:}$ $\Rightarrow I = \frac{M}{V}$ $\text{Substitute the values into the intensity of magnetization we get;}$ $I = \frac{4 \text{ A} - \text{m}^2}{4 \times 10^{-5} \text{ m}^3}$ $\Rightarrow I = \frac{4}{4 \times 10^{-5}} = 1 \times 10^5 \text{ A/m}$ $\text{Hence, option (3) is the correct answer.}$

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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
}