Import Question JSON

Current Question (ID: 13796)

Question:

Which type of tissue is correctly matched with its location?

Options:

1. Areolar tissue - Tendons
2. Transitional epithelium - Tip of nose
3. Cuboidal epithelium - Lining of stomach
4. Smooth muscle - Wall of intestine

Solution:

The cells of smooth muscles are elongated, spindle-shaped, broad from the middle and have tapered ends. Smooth muscles line the hollow organs and are involuntary e.g., posterior region of oesophagus, stomach, intestine, lungs, urinary bladder, urinogenital tract. Columnar epithelium consists of a single layer of tall and slender cells, lying on a basement membrane. It is found in the lining of the stomach and intestine. Tendons are cord-like, strong, inelastic structures that join skeletal muscle to bone. They consist of Dense regular connective tissue. Transitional Epithelium lines the inner surface of the urinary bladder and uterus. The nasal cavity is divided into two segments- the respiratory segment and the olfactory segment. The respiratory segment comprises most of each nasal fossa and is lined with ciliated pseudostratified columnar epithelium (also called respiratory epithelium). The conchae, or turbinates, are located in this region.

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
}