Import Question JSON

Current Question (ID: 12526)

Question:

Which one is true for spongy parenchyma

Options:

1. abaxially placed to palisade parenchyma
2. oval or round & loosely arranged cells
3. extends to the abaxial epidermis
4. All of the above

Solution:

Refer NCERT, heading 6.3.5 para 1 page 93 The following options are true for spongy parenchyma: Abaxially placed to palisade parenchyma: Spongy parenchyma is a type of ground tissue found in the leaves of plants. It is located on the abaxial (lower) side of the leaf and is separated from the palisade parenchyma (located on the adaxial side of the leaf) by the mesophyll. Oval or round and loosely arranged cells: The cells of spongy parenchyma are usually oval or round in shape and are loosely arranged, with large intercellular spaces between them. This arrangement allows for the diffusion of gases, such as oxygen and carbon dioxide, during photosynthesis and respiration. Extends to the abaxial epidermis: Spongy parenchyma extends from the palisade parenchyma to the abaxial epidermis of the leaf. It is responsible for gas exchange, as well as some of the storage and support functions of the leaf.

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
}