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Current Question (ID: 16896)

Question:

$\text{Magnets } A \text{ and } B \text{ are geometrically similar but the magnetic moment of } A \text{ is twice that of } B. \text{ If } T_1 \text{ and } T_2 \text{ be the time periods of the oscillation when their like poles and unlike poles are kept together respectively, then } \frac{T_1}{T_2} \text{ will be:}$

Options:

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

Solution:

$T_{\text{sum}} = 2\pi \sqrt{\frac{(I_1 + I_2)}{(M_1 + M_2) B_H}}$ $T_{\text{diff}} = 2\pi \sqrt{\frac{(I_1 + I_2)}{(M_1 - M_2) B_H}}$ $\Rightarrow \frac{T_s}{T_d} = \frac{T_1}{T_2} = \sqrt{\frac{M_1 - M_2}{M_1 + M_2}} = \sqrt{\frac{2M - M}{2M + M}} = \frac{1}{\sqrt{3}}$

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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
}