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Current Question (ID: 16792)

Question:
$\text{A proton and an } \alpha\text{-particle enter a uniform magnetic field perpendicularly at the same speed. If a proton takes } 25 \, \mu s \text{ to make } 5 \text{ revolutions, then the periodic time for the } \alpha\text{-particle will be:}$ $1. \, 50 \, \mu s$ $2. \, 25 \, \mu s$ $3. \, 10 \, \mu s$ $4. \, 5 \, \mu s$
Options:
  • 1. $50 \, \mu s$
  • 2. $25 \, \mu s$
  • 3. $10 \, \mu s$
  • 4. $5 \, \mu s$
Solution:
$\text{Time period of proton } T_p = \frac{25}{5} = 5 \, \mu \text{ sec}$ $\text{By using } \frac{T}{T_p} = \frac{m_\alpha}{m_p} \times \frac{q_p}{q_\alpha} = \frac{4m_p}{m_p} \times \frac{q_p}{2q_p}$ $\Rightarrow T_\alpha = 2T_p = 10 \, \mu \text{ sec}$

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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
}