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Current Question (ID: 17346)

Question:

$\text{A series } LCR \text{ circuit containing } 5.0 \text{ H inductor, } 80 \, \mu\text{F capacitor and } 40 \, \Omega \text{ resistor is connected to } 230 \, \text{V variable frequency AC source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:}$

Options:

1. $46 \, \text{rad/s and } 54 \, \text{rad/s}$
2. $42 \, \text{rad/s and } 58 \, \text{rad/s}$
3. $25 \, \text{rad/s and } 75 \, \text{rad/s}$
4. $50 \, \text{rad/s and } 25 \, \text{rad/s}$

Solution:

$\text{Hint: The power transferred to the circuit is half the power at the resonant frequency when the current amplitude is } \frac{1}{\sqrt{2}} \text{ times its maximum value.}$ $\text{Step 1: Find the resonant frequency.}$ $\text{The resonant frequency, } \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = 50 \, \text{rad/sec}$ $\text{Step 2: Find the bandwidth of the circuit}$ $\text{The bandwidth of the circuit, } 2 \, \Delta \omega = 2 \times \frac{R}{2L} = 2 \times \frac{40}{2 \times 5} = 8 \, \text{rad/sec}$ $\text{Step 3: Find the required angular frequencies.}$ $\omega_1 = \omega_0 + \Delta \omega = 50 + 4 = 54 \, \text{rad/sec}$ $\omega_2 = \omega_0 - \Delta \omega = 50 - 4 = 46 \, \text{rad/sec}$

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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
}