Import Question JSON

Current Question (ID: 16875)

Question:

$\text{The magnetic moment of a bar magnet of length } L \text{ and area of cross-section } A \text{ is } M. \text{ If the magnet is cut into four identical parts each of length } L \text{ and area of cross-section } \frac{A}{4}, \text{ then the magnetic moment of each part is:}$

Options:

1. $\frac{M}{4}$
2. $\frac{M}{2}$
3. $M$
4. $4M$

Solution:

$\text{Hint: The magnetic moment } M = m \times L$ $\text{Step: Find the magnetic moment of the magnet of each part.}$ $\text{The magnetic moment } (M) \text{ of a bar magnet is given by;} M = m \times L$ $\text{When the magnet is cut into smaller parts, the pole strength is affected by the area of the cross-section, as the pole strength is proportional to the area.}$ $\text{If the magnet is cut into four identical parts, each with a cross-sectional area of } \frac{A}{4},$ $\text{the pole strength } m' \text{ of each part becomes } \frac{m}{4}, \text{ because the pole strength is directly proportional to the area.}$ $\text{Since the length } L \text{ of each part remains the same, the magnetic moment of each part is given by;} M' = m' \times L \Rightarrow M' = \frac{m}{4} \times L = \frac{M}{4}$ $\text{Thus, the magnetic moment of each part is } \frac{M}{4}. \text{ Hence, option (1) is the correct answer.}$

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
}