Import Question JSON

Current Question (ID: 12434)

Question:

Collenchyma is characterised by

Options:

1. Absence of protoplast at maturity
2. Supporting young stem and leaf petiole
3. Lignocellulosic thickening on secondary walls
4. The common presence of intracellular spaces

Solution:

Collenchyma is a type of ground tissue in plants that provides mechanical support to young stems, leaf petioles, and other growing plant parts. Unlike sclerenchyma, another type of supportive tissue in plants, collenchyma cells remain alive at maturity and have a relatively thin primary cell wall that can undergo expansion during growth. Collenchyma cells often have a variable shape, ranging from elongated to polygonal, and can have unevenly thickened cell walls that provide additional structural support. Collenchyma cells do not have lignocellulosic thickening on their secondary walls, and while they may have intracellular spaces, this is not a defining characteristic of the tissue. Option 1 is incorrect because collenchyma cells are alive at maturity and have a protoplast. Option 3 is incorrect because collenchyma cells do not have lignocellulosic thickening on their secondary walls. Option 4 is incorrect because the presence of intracellular spaces is not a defining characteristic of collenchyma.

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
}