Import Question JSON

Current Question (ID: 20652)

Question:

$\text{An EM wave propagating in } x\text{-direction has a wavelength of } 8 \text{ mm.}$ $\text{The electric field vibrating } y\text{-direction has maximum magnitude of } 60 \text{ Vm}^{-1}. \text{ Choose the correct equations for electric and magnetic fields if the EM wave is propagating in a vacuum:}$

Options:

1. $E_y = 60 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\jmath} \text{ Vm}^{-1}$ $B_z = 2 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \text{ T}$
2. $E_y = 60 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\jmath} \text{ Vm}^{-1}$ $B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \text{ T}$
3. $E_y = 2 \times 10^{-7} \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{\jmath} \text{ Vm}^{-1}$ $B_z = 60 \sin \left[ \frac{\pi}{4} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \text{ T}$
4. $E_y = 2 \times 10^{-7} \sin \left[ \frac{\pi}{4} \times 10^4 \left( x - 4 \times 10^8 t \right) \right] \hat{\jmath} \text{ Vm}^{-1}$ $B_z = 60 \sin \left[ \frac{\pi}{4} \times 10^4 \left( x - 4 \times 10^8 t \right) \right] \hat{k} \text{ T}$

Solution:

$\text{Hint: } E_0 = B_0 c$

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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
}