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Current Question (ID: 20346)

Question:
$\text{A proton, an electron, and a helium nucleus have the same energy.}$ $\text{They are in circular orbits in a plane due to a magnetic field}$ $\text{perpendicular to the plane. Let } r_p, r_e \text{ and } r_{\text{He}} \text{ be their respective radii, then:}$ $1.\ r_e < r_p < r_{\text{He}}$ $2.\ r_e > r_p = r_{\text{He}}$ $3.\ r_e < r_p = r_{\text{He}}$ $4.\ r_e > r_p > r_{\text{He}}$
Options:
  • 1. $r_e < r_p < r_{\text{He}}$
  • 2. $r_e > r_p = r_{\text{He}}$
  • 3. $r_e < r_p = r_{\text{He}}$
  • 4. $r_e > r_p > r_{\text{He}}$
Solution:
$\text{Hint: } r = \frac{\sqrt{2mE}}{Bq}$ $r = \frac{\sqrt{2mE}}{Bq} \quad E = \text{same}$ $r \propto \frac{\sqrt{m}}{q}$ $\text{proton } \frac{\sqrt{m_p}}{q_p} = \frac{\sqrt{m_\alpha}}{q_\alpha}, \quad \text{He}^{+2}$ $\therefore \ r_p = r_{\text{He}}$ $\text{For } \frac{\sqrt{m_e}}{q_e} < \frac{\sqrt{m_p}}{q_p} \quad \text{proton}$ $\therefore \ r_e < r_p = r_{\text{He}}$

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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
}