Import Question JSON

Current Question (ID: 13634)

Question:

All cyclostomes are __________ on some fishes; paired fins are ________, cranium is _______ and circulation is ______ type.

Options:

1. Ectoparasites, Absent, Cartilaginous, Open
2. Endoparasites, Present, Bony, Closed
3. Ectoparasites, Absent, Cartilaginous, Closed
4. Ectoparasites, Present, Cartilaginous, Open

Solution:

Cyclostomes are a group of jawless fish that includes lampreys and hagfish. The given statements about cyclostomes are as follows: Cyclostomes are ectoparasites on some fishes: Lampreys, which are a type of cyclostome, are known to be ectoparasites that attach themselves to the bodies of other fishes and feed on their blood. Paired fins are absent: Cyclostomes, including lampreys, do not possess paired fins like other fishes. They have a unique, elongated body shape without any paired appendages. Cranium is cartilaginous: Cyclostomes have a cartilaginous cranium, which is the part of the skull that encloses and protects the brain. Unlike other fishes, they lack a bony cranium. Circulation is closed type: Cyclostomes, including lampreys, have a closed circulatory system where blood is enclosed within vessels and pumped by a heart. This is in contrast to an open circulatory system, where blood is not enclosed in vessels and circulates more freely in the body cavity.

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
}