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Current Question (ID: 16402)

Question:

$\text{A particle of mass } m \text{ carrying charge } -q_1 \text{ is moving around a charge } +q_2$ $\text{along a circular path of radius } r. \text{ The period of revolution of the charge } -q_1 \text{ is:}$

Options:

1. $\sqrt{\frac{16\pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$
2. $\sqrt{\frac{8\pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$
3. $\sqrt{\frac{q_1 q_2}{16\pi^3 \varepsilon_0 m r^3}}$
4. $\text{zero}$

Solution:

$\text{The net force acting on } -q_1 \text{ is:}$ $\vec{F}_{q_1} = \frac{k q_1 q_2}{r^2} \quad (\text{attractive force}) \quad \left[ k = \frac{1}{4\pi \varepsilon_0} \right]$ $\frac{1}{4\pi \varepsilon_0} \times \frac{q_1 q_2}{r^2} = \frac{mv^2}{r} = mr\omega^2 = \text{centripetal force}$ $\omega^2 = \frac{q_1 q_2}{4\pi \varepsilon_0 m r^3}$ $\left[ \omega = \frac{2\pi}{T} \right]$ $\frac{4\pi^2}{T^2} = \frac{q_1 q_2}{4\pi \varepsilon_0 m r^3}$ $T = \sqrt{\frac{16\pi^3 \varepsilon_0 m r^3}{q_1 q_2}}$

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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
}