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Current Question (ID: 17426)

Question:
$\text{The electric field part of an electromagnetic wave in a vacuum is;}$ $\vec{E} = (3.1 \text{ N/C}) \cos[(1.8 \text{ rad/m}) y + (5.4 \times 10^8 \text{ rad/s}) t] \hat{i}.$ $\text{What is the frequency of the wave?}$
Options:
  • 1. $5.7 \times 10^7 \text{ Hz}$
  • 2. $9.3 \times 10^7 \text{ Hz}$
  • 3. $8.6 \times 10^7 \text{ Hz}$
  • 4. $7.5 \times 10^7 \text{ Hz}$
Solution:
$\text{Hint: Compare the given equation with the standard equation.}$ $\text{Step: Find the frequency of the wave.}$ $\text{The standard equation of the electric field in the electromagnetic wave is given by;} \ E = E_0 \cos[ky + \omega t]$ $\text{On comparing with the given equation we get;} \ \omega = 5.4 \times 10^8$ $\text{We know that;} \ \omega = 2\pi f$ $\Rightarrow f = \frac{\omega}{2\pi}$ $\text{Substitute the given values we get;} \ f = \frac{5.4 \times 10^8}{2\pi}$ $\Rightarrow f = 8.6 \times 10^7 \text{ Hz}$ $\text{Hence, option (3) is the correct answer.}$

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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
}