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Current Question (ID: 20339)

Question:
$\text{An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii } r_e, r_p, r_\alpha \text{ respectively in a uniform magnetic field } B. \text{ The relation between } r_e, r_p, r_\alpha \text{ is:}$ $1. \ r_e > r_p = r_\alpha$ $2. \ r_e < r_p = r_\alpha$ $3. \ r_e < r_p < r_\alpha$ $4. \ r_e < r_\alpha < r_p$
Options:
  • 1. $r_e > r_p = r_\alpha$
  • 2. $r_e < r_p = r_\alpha$
  • 3. $r_e < r_p < r_\alpha$
  • 4. $r_e < r_\alpha < r_p$
Solution:
$\text{Hint: } r = \frac{\sqrt{2mKE}}{qB}$ $e^-, \ \text{p}^+, \ \alpha$ $r = \frac{\sqrt{2mKE}}{qB}$ $r \propto \frac{\sqrt{m}}{q}$ $\text{For electron, } \frac{\sqrt{m}}{q} = \frac{\sqrt{m_e}}{1e}$ $\text{For proton, } \frac{\sqrt{m}}{q} = \frac{\sqrt{m_p}}{1e}$ $\text{For } \alpha \ \frac{\sqrt{m}}{q} = \frac{\sqrt{4u}}{2e}$ $= \frac{2\sqrt{1u}}{2e}$ $r_e < r_p = r_\alpha$

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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
}