Import Question JSON

Current Question (ID: 12546)

Question:
In which character an isobilateral leaf differs from the dorsiventral leaf?
Options:
  • 1. Scattered vascular bundles
  • 2. Undifferentiated mesophylls
  • 3. Absence of stomata with guard cells
  • 4. Conjoint, collateral, and closed vascular bundle
Solution:
An isobilateral leaf differs from the dorsiventral leaf in the presence of undifferentiated mesophylls. Dorsiventral leaves, also known as bifacial leaves, have a distinct upper and lower surface with different structures and functions. The upper surface of the dorsiventral leaf is called the adaxial surface, which is exposed to the sun and is covered with a waxy cuticle to reduce water loss. The lower surface of the leaf is called the abaxial surface, which is shaded and contains stomata for gas exchange. Dorsiventral leaves have a well-differentiated mesophyll, which is divided into two distinct layers, the palisade mesophyll and the spongy mesophyll. Isobilateral leaves, also known as equifacial leaves, have a symmetrical structure with the same anatomy on both sides. Both surfaces of the leaf have undifferentiated mesophylls, and the stomata are distributed on both sides. Isobilateral leaves lack a distinct upper and lower surface, and there is no division of mesophyll into palisade and spongy layers. Isobilateral leaves have scattered vascular bundles, which are conjoint, collateral, and closed, similar to the dorsiventral leaf.

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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
}