Import Question JSON

Current Question (ID: 17423)

Question:

$\text{The electric field part of an electromagnetic wave in a medium is represented by:}$ $E_x = 0;$ $E_y = \left(2.5 \frac{\text{N}}{\text{C}}\right) \cos \left[ \left(2\pi \times 10^6 \frac{\text{rad}}{\text{s}}\right) t - \left(\pi \times 10^{-2} \frac{\text{rad}}{\text{m}}\right) x \right];$ $E_z = 0.$ $\text{The wave is:}$

Options:

1. $\text{Moving along } y\text{-direction with frequency } 2\pi \times 10^6 \text{ Hz and wavelength 200 m.}$
2. $\text{Moving along } +x\text{-direction with frequency } 10^6 \text{ Hz and wavelength 100 m.}$
3. $\text{Moving along } +x\text{-direction with frequency } 10^6 \text{ Hz and wavelength 200 m.}$
4. $\text{Moving along } -x\text{-direction with frequency } 10^6 \text{ Hz and wavelength 200 nm.}$

Solution:

$\text{(3) Comparing the given equation:}$ $E_y = 2.5 \frac{\text{N}}{\text{C}} \cos \left[ \left(2\pi \times 10^6 \frac{\text{rad}}{\text{s}}\right) t - \left(\pi \times 10^{-2} \frac{\text{rad}}{\text{m}}\right) x \right]$ $\text{With the standard equation}$ $E_y = E_0 \cos(\omega t - kx)$ $\text{we get}$ $\omega = 2\pi f = 2\pi \times 10^6$ $f = 10^6 \text{ Hz}$ $\text{Moreover, we know that}$ $\frac{2\pi}{\lambda} = k = \pi \times 10^{-2} \text{ m}^{-1}$ $\Rightarrow \lambda = 200 \text{ m}$ $\text{Hence, the wave is moving along positive } x\text{-direction with frequency } 10^6 \text{ Hz and wavelength of 200 m.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}