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Current Question (ID: 17309)

Question:

$\text{For a } LCR \text{ series circuit with an AC source of angular frequency } \omega:$ $1. \text{ circuit will be capacitive if } \omega > \frac{1}{\sqrt{LC}}$ $2. \text{ circuit will be inductive if } \omega = \frac{1}{\sqrt{LC}}$ $3. \text{ power factor of circuit will be unity if capacitive reactance equals inductive reactance}$ $4. \text{ current will be leading voltage if } \omega > \frac{1}{\sqrt{LC}}$

Options:

1. $\text{circuit will be capacitive if } \omega > \frac{1}{\sqrt{LC}}$
2. $\text{circuit will be inductive if } \omega = \frac{1}{\sqrt{LC}}$
3. $\text{power factor of circuit will be unity if capacitive reactance equals inductive reactance}$
4. $\text{current will be leading voltage if } \omega > \frac{1}{\sqrt{LC}}$

Solution:

$\text{At resonance}$ $X_L = X_C$ $\omega L = \frac{1}{\omega C}$ $\omega = \frac{1}{\sqrt{LC}} = \omega_o$ $\text{If } \omega = \omega_o \Rightarrow X_L = X_C$ $\text{If } \omega = \omega_o, \text{ circuit will be at resonance.}$ $\text{Power factor is unity, when } X_L = X_C.$

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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
}