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Current Question (ID: 16898)

Question:

$\text{Two bar magnets are held together tightly in a vibration magnetometer.}$ $\text{When their like poles are together, they make 20 oscillations per minute}$ $\text{and when their unlike poles are together, they make 8 oscillations per minute.}$ $\text{The ratio of the magnetic dipole moments of two bar magnets is:}$

Options:

1. $29 : 21$
2. $6 : 15$
3. $1 : 6$
4. $25 : 4$

Solution:

$\text{Hint: The frequency of vibration is given by;} \; f = \frac{1}{2\pi} \sqrt{\frac{mB}{I}}$ $\text{Step 1: Calculate the net magnetic moment.}$ $\text{case 1: } m_A = m_1 + m_2.$ $\text{case 2: } m_B = m_1 - m_2.$ $\text{Step 2: Calculate the ratio of the magnetic moment.}$ $\frac{f_1}{f_2} = \sqrt{\frac{m_A}{m_B}}$ $\left(\frac{20}{8}\right)^2 = \frac{m_1 + m_2}{m_1 - m_2}$ $\frac{25}{4} = \frac{m_1 + m_2}{m_1 - m_2}$ $\text{Use componendo and dividendo we get;} \; \frac{25 + 4}{25 - 4} = \frac{2m_1}{2m_2}$ $\Rightarrow \frac{m_1}{m_2} = \frac{29}{21}$ $\text{Hence, option (1) is the correct answer.}$

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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
}