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Current Question (ID: 20543)

Question:

$\text{A resonance circuit having inductance and resistance } 2 \times 10^{-4} \text{ H}$ $\text{and } 6.28 \ \Omega \text{ respectively oscillates at } 10 \ \text{MHz frequency.}$ $\text{The value of quality factor of this resonator is: } [\pi = 3.14]$

Options:

1. $1000$
2. $2000$
3. $3000$
4. $4000$

Solution:

$\text{Hint: } Q = \omega_0 \frac{L}{R}$ $\text{Given: } L = 2 \times 10^{-4} \text{ H}$ $R = 6.28 \ \Omega$ $f = 10 \ \text{MHz} = 10^7 \ \text{Hz}$ $\text{Since quality factor,}$ $Q = \omega_0 \frac{L}{R} = 2 \pi f \frac{L}{R}$ $\therefore Q = 2 \pi \times 10^7 \times \frac{2 \times 10^{-4}}{6.28}$ $Q = 2 \times 10^3 = 2000$ $\therefore \text{Ans. is } 2000$

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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
}