Import Question JSON

Current Question (ID: 16916)

Question:

$\text{Which of the following is not dimensionless?}$ $\text{(where symbols stand for their usual meanings in magnetism)}$

Options:

1. $\frac{I}{H}$
2. $\frac{B}{\mu_0 H}$
3. $\mu_r$
4. $\frac{\mu_r B}{H}$

Solution:

$\text{Hint: } \frac{\mu_r B}{H} \text{ has the dimensions of permeability } [\text{H/m}], \text{ so it is not dimensionless.}$ $\text{Explanation: Let's analyze each option for dimensional consistency:}$ $\frac{I}{H}: (I) \text{ (magnetization) and } (H) \text{ (magnetic field intensity) both have the same dimensions: } [\text{A/m}]. \text{ Therefore, } \frac{I}{H} \text{ is dimensionless.}$ $\frac{B}{\mu_0 H}: (B) \text{ (magnetic flux density) has the dimension } (\text{T}) \text{ (Tesla) or } [\text{N/(A .m)}]. \mu_0 H \text{ has the same dimension as } B. \text{ Hence, } \frac{B}{\mu_0 H} \text{ is dimensionless.}$ $\mu_r \text{ (relative permeability): This is inherently dimensionless, as it represents a ratio of two permeabilities } (\mu/\mu_0).$ $\frac{\mu_r B}{H}: \mu_r \text{ is dimensionless, but } B/H \text{ has the dimensions of permeability } (\mu), \text{ which has dimensions of } [\text{H/m}] \text{ (Henry per meter). Therefore, } \frac{\mu_r B}{H} \text{ is not dimensionless; it has the dimensions of permeability.}$ $\text{Hence, option (4) is the correct answer.}$

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
}