Import Question JSON

Current Question (ID: 18012)

Question:

$\text{The magnetic substances that have lesser magnetic moment than expected is known as -}$

Options:

1. $\text{Paramagnetic.}$
2. $\text{Ferrimagnetic.}$
3. $\text{Antiferromagnetic.}$
4. $\text{None of the above.}$

Solution:

$\text{The substances that have lesser magnetic moments than expected are called ferrimagnetic substances.}$ $\text{Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances are } \text{O}_2, \text{Fe}^{+3} \text{ and } \text{Cr}^{+3}. \text{Paramagnetic substances get magnetised in a magnetic field in the same direction but lose magnetism when the magnetic field is removed.}$ $\text{Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism. Examples include } \text{Fe}_3\text{O}_4 \text{ (magnetite), ferrites such as } \text{MgFe}_2\text{O}_4 \text{ and } \text{ZnFe}_2\text{O}_4. \text{Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances. On heating, these substances become paramagnetic.}$ $\text{Antiferromagnetism: Antiferromagnetic substances have domain structures similar to ferromagnetic substances, but are oppositely oriented. The oppositely oriented domains cancel out each other's magnetic moments.}$

Import JSON File

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
}