Import Question JSON

Current Question (ID: 12553)

Question:

The tissue belonging to bark but not to periderm is

Options:

1. Vascular cambium
2. Secondary phloem
3. Phellem
4. Secondary cortex

Solution:

Secondary phloem is a tissue that belongs to the bark of a plant, which is the outer protective covering of stems and roots. Bark typically includes multiple layers, including the periderm (which consists of the cork cambium, cork cells or phellem, and sometimes phelloderm), as well as the secondary phloem (inner bark) and secondary xylem (wood). Vascular cambium, on the other hand, is a type of meristematic tissue that is responsible for producing secondary growth in plants, which leads to the formation of secondary xylem (wood) towards the inside and secondary phloem (inner bark) towards the outside. Vascular cambium is a component of the periderm, which is part of the bark. Phellem, also known as cork cells, is a type of tissue that is produced by the cork cambium and forms the outermost protective layer of the periderm. Phelloderm is another layer of the periderm, located towards the inside of the cork cambium, and it consists of living parenchyma cells. Both phellem and phelloderm are part of the periderm and belong to the bark. Secondary cortex, also known as secondary phloem parenchyma, is a tissue that is formed towards the inside of the secondary phloem and is not part of the periderm. It consists of living parenchyma cells and is involved in storage and other metabolic functions.

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
}