Import Question JSON

Current Question (ID: 12499)

Question:
In dicot roots, the initiation of the lateral roots and the vascular cambium during the secondary growth takes place in:
Options:
  • 1. Pericycle
  • 2. Endodermis
  • 3. Conjunctive tissue
  • 4. Epidermis
Solution:
In dicot roots, the initiation of lateral roots and the vascular cambium during secondary growth occurs in the pericycle. The pericycle is a layer of cells located just inside the endodermis, which is the innermost layer of the cortex in the root. The pericycle is capable of cell division and can give rise to lateral roots, as well as contribute to the formation of the vascular cambium, which is responsible for secondary growth in roots. Statement 2 is incorrect, as the endodermis is not involved in the initiation of lateral roots or vascular cambium during secondary growth in dicot roots. The endodermis is a specialized layer of cells responsible for regulating the movement of water and nutrients into the vascular tissue of the root. Statement 3 is incorrect, as conjunctive tissue refers to a type of tissue that connects or binds different types of tissues in plants, and it is not directly involved in the initiation of lateral roots or vascular cambium. Statement 4 is incorrect, as the epidermis is the outermost layer of cells in the root and does not play a role in the initiation of lateral roots or vascular cambium during secondary growth.

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
}