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Current Question (ID: 20544)

Question:

$\text{A transmitting station releases waves of wavelength } 960 \text{ m. A capacitor of } 2.56 \ \mu\text{F} \text{ is used in the resonant circuit. The self-inductance of the coil necessary for resonance is:}$

Options:

1. $10 \times 10^{-8} \ \text{H}$
2. $5 \times 10^{-8} \ \text{H}$
3. $10 \times 10^{-5} \ \text{H}$
4. $5 \times 10^{-4} \ \text{H}$

Solution:

$\text{Hint: } \omega_0 = \frac{1}{\sqrt{LC}}$ $\lambda = 960 \text{ m}$ $C = 2.56 \ \mu\text{F} = 2.56 \times 10^{-6} \ \text{F}$ $c = 3 \times 10^8 \ \text{m/s}$ $L = ?$ $\text{Now at resonance, } \omega_0 = \frac{1}{\sqrt{LC}}$ $\text{[Resonant frequency]}$ $2\pi f_0 = \frac{1}{\sqrt{LC}}$ $\text{On substituting } f_0 = \frac{c}{\lambda}, \text{ we have } 2\pi \frac{c}{\lambda} = \frac{1}{\sqrt{LC}}$ $\text{Squaring both sides: } 4\pi^2 \frac{c^2}{\lambda^2} = \frac{1}{LC}$ $= \frac{4 \times 10 \times (3 \times 10^8)^2}{(960)^2} = \frac{1}{L \times 2.56 \times 10^{-6}}$ $\Rightarrow \frac{1}{L} = \frac{4 \times 10 \times 9 \times 10^{16} \times 2.56 \times 10^{-6}}{960 \times 960}$ $\Rightarrow L = 10 \times 10^{-8} \ \text{H}$

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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
}