Import Question JSON

Current Question (ID: 7476)

Question:

$\text{Match the following electromagnetic radiations with their corresponding frequencies:}$ $\text{Column I:}$ $\text{A. X-rays}$ $\text{B. Ultraviolet wave (UV)}$ $\text{C. Long radio waves}$ $\text{D. Microwave}$ $\text{Column II:}$ $\text{1. } \nu = 10^0 - 10^4 \text{ Hz}$ $\text{2. } \nu = 10^{10} \text{ Hz}$ $\text{3. } \nu = 10^{16} \text{ Hz}$ $\text{4. } \nu = 10^{18} \text{ Hz}$

Options:

1. $\text{A-4, B-1, C-3, D-2}$
2. $\text{A-1, B-4, C-2, D-3}$
3. $\text{A-1, B-4, C-3, D-2}$
4. $\text{A-4, B-3, C-1, D-2}$

Solution:

$\text{To solve this matching problem, we need to compare the frequency ranges of each electromagnetic radiation type with the given options in Column II.}$ $\text{From the given data:}$ $\text{A. X-rays: } 2 \times 10^{16} - 3 \times 10^{19} \text{ Hz}$ $\text{B. Ultraviolet wave (UV): } 7.9 \times 10^{14} - 2 \times 10^{16} \text{ Hz}$ $\text{C. Long radio waves: } 10^0 - 10^4 \text{ Hz}$ $\text{D. Microwave: } 1 \times 10^9 - 5 \times 10^{11} \text{ Hz}$ $\text{Matching with Column II:}$ $\text{A. X-rays (} 2 \times 10^{16} - 3 \times 10^{19} \text{ Hz) matches with option 4 (} \nu = 10^{18} \text{ Hz) as } 10^{18} \text{ falls within the X-ray frequency range.}$ $\text{B. Ultraviolet wave (} 7.9 \times 10^{14} - 2 \times 10^{16} \text{ Hz) matches with option 3 (} \nu = 10^{16} \text{ Hz) as } 10^{16} \text{ is at the upper limit of UV range.}$ $\text{C. Long radio waves (} 10^0 - 10^4 \text{ Hz) matches exactly with option 1 (} \nu = 10^0 - 10^4 \text{ Hz).}$ $\text{D. Microwave (} 1 \times 10^9 - 5 \times 10^{11} \text{ Hz) matches with option 2 (} \nu = 10^{10} \text{ Hz) as } 10^{10} \text{ falls within the microwave frequency range.}$ $\text{Therefore, the correct matching is A-4, B-3, C-1, D-2.}$

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
}