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Current Question (ID: 16908)

Question:

$\text{When a magnetic material is subjected to a very small magnetising force } H, \text{ the intensity of magnetisation is proportional to:}$

Options:

1. $\frac{1}{\sqrt{H}}$
2. $\sqrt{H}$
3. $H$
4. $H^2$

Solution:

$\text{Hint: The intensity of the magnetisation, } I = \chi H.$ $\text{Explanation: The magnetic susceptibility } (\chi) \text{ is a dimensionless quantity}$ $\text{that indicates how much a material will become magnetized in an applied}$ $\text{magnetic field.}$ $\text{It is defined as: } \chi = \frac{I}{H}$ $\text{Rearranging this gives: } I = \chi \cdot H$ $\text{For small magnetizing fields, the magnetic susceptibility } (\chi) \text{ is typically}$ $\text{considered to be a constant for a given material.}$ $\text{Therefore, we can express the relationship between } (I) \text{ and } (H) \text{ as:}$ $I \propto H$ $\text{This indicates that the intensity of magnetization } (I) \text{ is directly proportional}$ $\text{to the magnetizing field } (H).$ $\text{Hence, option (3) is the correct answer.}$

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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
}