Import Question JSON

Current Question (ID: 11798)

Question:

Practical purpose of taxonomy or classification is to:

Options:

1. facilitate the identification of unknown species
2. explain the origin of organisms
3. know the evolutionary history
4. identify medicinal plants

Solution:

The practical purpose of taxonomy or classification is primarily to facilitate the identification of unknown species. Taxonomy provides a systematic framework for organizing and categorizing living organisms based on their shared characteristics, such as morphology, genetics, and ecology. This organized classification system allows scientists and researchers to identify and name new species by comparing their characteristics with existing species. Identification of unknown species is an essential step in various fields of biology, including biodiversity conservation, ecological research, and agricultural sciences. While taxonomy can also provide insights into the evolutionary history and origin of organisms (option 2), and help in the identification of medicinal plants (option 4), these are secondary purposes and not the primary focus of taxonomy. Taxonomy primarily serves as a tool for identifying and naming species, which is crucial for understanding and managing the diversity of life on Earth. Explanation of the origin of organisms and knowledge of evolutionary history may be inferred from taxonomic relationships, but they are not the main practical purpose of taxonomy.

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
}