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Current Question (ID: 17322)

Question:

$\text{In a series } LC \text{ circuit, if } L = 10^{-3} \text{ H and } C = 3 \times 10^{-7} \text{ F is connected to a } (100 \text{ V, } 50 \text{ Hz}) \text{ AC source, the impedance of the circuit is:}$

Options:

1. $\frac{10^5}{3\pi} - 10\pi$
2. $0.1\pi - 3 \times 10^{-5}\pi$
3. $\frac{10^5}{3\pi} - \frac{\pi}{10}$
4. $\text{None of these}$

Solution:

$\text{Hint: Impedance, } Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\text{Step 1: Find angular frequency}$ $\omega = 2\pi f$ $= 2\pi \times 50$ $= 100\pi$ $\text{Step 2: Find inductive reactance}$ $X_L = L\omega$ $= 10^{-3} (100\pi) = \pi 10 \Omega$ $\text{Step 3: Find capacitive reactance}$ $X_C = \frac{1}{C\omega} = \frac{1}{3 \times 10^{-7} \times (100\pi)}$ $= \frac{10^5}{3\pi}$ $\text{Step 4: Find impedance of the circuit}$ $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\text{When } R = 0$ $Z = X_L - X_C$ $= \frac{10^5}{3\pi} - \frac{\pi}{10}$

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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
}