Import Question JSON

Current Question (ID: 15710)

Question:

Select the option which correctly fills up the blanks in the following statements. A. Destruction of embryo or foetus in the uterus is called ____________. B. Government of India legalized MTP in the year ____________. C. Natural family planning method is also called ____________. D. ____________ is a method in which the male partner withdraws his penis from vagina just before ejaculation. E. ____________ is the copper releasing and is a hormone releasing intra uterine devices.

Options:

1. A - Foeticide, B - 1961, C - Rhythm method, D - Coitus interruptus, E - Multiload 375, LNG-20
2. A - Foeticide, B - 1971, C - Rhythm method, D - Coitus interruptus, E - Multiload 375, LNG-20
3. A - Foeticide, B - 1965, C - Rhythm method, D - Coitus interruptus, E - Multiload 375, LNG-20
4. A - Foeticide, B -1982, C - Rhythm method, D - Coitus interruptus, E - Multiload 375, LNG-20

Solution:

A. Destruction of embryo or foetus in the uterus is called foeticide. B. Government of India legalized MTP in the year 1971. C. Natural family planning method is also called Rhythm method. D. Coitus interruptus is a method in which the male partner withdraws his penis from vagina just before ejaculation. E. Multiload-375 is the copper releasing and LNG-20 is a hormone releasing intra uterine devices.

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
}