Import Question JSON

Current Question (ID: 20640)

Question:
$\text{A plane electromagnetic wave, has frequency of } 2.0 \times 10^{10} \text{ Hz and its energy density is } 1.02 \times 10^{-8} \text{ J/m}^3 \text{ in vacuum. The amplitude of the magnetic field of the wave is close to:}$ $\left( \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \text{ Nm}^2 \text{ C}^{-2} \text{ and speed of light } = 3 \times 10^8 \text{ ms}^{-1} \right)$
Options:
  • 1. $180 \text{ nT}$
  • 2. $160 \text{ nT}$
  • 3. $150 \text{ nT}$
  • 4. $190 \text{ nT}$
Solution:
$\text{Hint: Energy density } = \frac{B_0^2}{2\mu_0}$ $\text{Energy density } \frac{dU}{dV} = \frac{B_0^2}{2\mu_0}$ $1.02 \times 10^{-8} = \frac{B_0^2}{2 \times 4\pi \times 10^{-7}}$ $B_0^2 = (1.02 \times 10^{-8}) \times (8\pi \times 10^{-7})$ $B_0 = 16 \times 10^{-8} \text{ T} = 160 \text{ nT}$

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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
}