Import Question JSON

Current Question (ID: 17463)

Question:

$\text{Which physical quantity does not change in a vacuum for } X\text{-rays?}$ $1. \text{ Speed of light}$ $2. \text{ Wavelength}$ $3. \text{ Frequency}$ $4. \text{ None of these}$

Options:

1. $\text{Speed of light}$
2. $\text{Wavelength}$
3. $\text{Frequency}$
4. $\text{None of these}$

Solution:

$\text{Hint: Frequency and wavelength depend on the source.}$ $\text{Explanation: In a vacuum, the speed of light is a constant, as it is the fundamental speed limit of the universe.}$ $\text{The speed of light in a vacuum is approximately } 3 \times 10^8 \text{ m/s, and it remains unchanged for } X\text{-rays.}$ $\text{However, wavelength and frequency can change when the } X\text{-rays pass from one medium to another (like from air to glass),}$ $\text{but in a vacuum, the speed of light is constant, and wavelength and frequency are related by the equation:}$ $c = \lambda \nu$ $\text{Since the speed of light is constant in a vacuum, both the wavelength and frequency of } X\text{-rays remain linked,}$ $\text{but the wavelength and frequency can be adjusted depending on the source or if the } X\text{-rays interact with different media.}$ $\text{Hence, option (1) is the correct answer.}$

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
}