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Current Question (ID: 20585)

Question:
$\text{The rms value of conduction current in a parallel plate capacitor is } 6.9 \, \mu\text{A. The capacity of this capacitor, if it is connected to } 230 \, \text{V AC supply with an angular frequency of } 600 \, \text{rad/s, will be:}$
Options:
  • 1. $5 \, \text{pF}$
  • 2. $50 \, \text{pF}$
  • 3. $100 \, \text{pF}$
  • 4. $200 \, \text{pF}$
Solution:
$\text{Hint: } I = \frac{V}{X_C}$ $I_{\text{rms}} = \frac{V_{\text{rms}}}{\frac{1}{\omega C}}$ $C = \frac{I_{\text{rms}}}{V_{\text{rms}} \times \omega}$ $C = \frac{6.9 \times 10^{-6}}{230 \times 600}$ $C = 50 \, \text{pF}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}