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Current Question (ID: 20545)

Question:

$\text{In a series } LCR \text{ resonant circuit, the quality factor } (Q) \text{ at resonance is measured as } 100.$ $\text{The inductance } (L) \text{ is increased two-fold } (L' = 2L), \text{ and the resistance } (R) \text{ is decreased two-fold } (R' = \frac{R}{2}), \text{ while the capacitance } (C) \text{ remains unchanged.}$ $\text{Assuming the new quality factor is calculated at the new resonance frequency of the modified circuit, then the new quality factor will be:}$

Options:

1. $173.25$
2. $282.84$
3. $453.97$
4. $621.24$

Solution:

$\text{Hint: } Q = \frac{\omega L}{R} = \frac{1}{\sqrt{LC}} \times \frac{L}{R}.$ $Q = \frac{\omega L}{R} = \frac{1}{\sqrt{LC}} \frac{L}{R} = \frac{\sqrt{L}}{R\sqrt{C}}$ $Q' = \frac{\sqrt{2L}}{\left(\frac{R}{2}\right)\sqrt{C}} = 2\sqrt{2}Q = 2\sqrt{2}\left(100\right)$ $= 282.84$

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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
}