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Current Question (ID: 12265)

Question:

The cells proximal to the merismatic zone in root tip represents

Options:

1. Zone of cell elongation
2. Zone of deacceleration
3. Zone of differentiation
4. Zone of maturation

Solution:

The cells proximal (towards the root tip) to the meristematic zone in the root tip represent the zone of cell elongation. The root tip is composed of several distinct zones, including the meristematic zone, zone of cell elongation, zone of differentiation, and zone of maturation. These zones are responsible for different processes involved in root growth and development. The zone of cell elongation is located just above the meristematic zone and is characterized by active cell elongation, where the cells increase in size and push the root tip further into the soil. This zone is responsible for the actual growth in length of the root, as the cells undergo elongation and contribute to the overall increase in root length. The zone of deacceleration is not a recognized term in root anatomy and growth zones. The zone of differentiation is located above the zone of cell elongation and is responsible for the differentiation of cells into specialized cell types, such as root hairs, cortex cells, and endodermis cells, which perform specific functions in root development. The zone of maturation is located above the zone of differentiation and is characterized by the maturation and specialization of root tissues, including the development of root hairs, root cap, and the organization of the mature root structure.

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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
}