Import Question JSON

Current Question (ID: 12827)

Question:
Fruit and leaf drop at early stages can be prevented by the application of
Options:
  • 1. Cytokinins
  • 2. ethylene
  • 3. auxins
  • 4. gibberellic acid
Solution:
Auxin delay abscission of leaves and fruits at early stages. Whenever leaf or fruit fall occurs, the organ concerned stops producing auxin. However, it promotes abscission of older, mature leaves and fruits. Auxins are a powerful growth hormone produced naturally by plants. They are found in the shoot and root tips and promote cell division, stem and root growth. They can also drastically affect plant orientation by promoting cell division to one side of the plant in response to sunlight and gravity. Auxins in the flower promote maturation of the ovary wall and promote steps in the full development of the fruit. Fruit and leaf drop at early stages can be prevented by the application of auxin. Ethylene is also an important natural plant hormone and is used in agriculture to force the ripening of fruits. Gibberellic acid is a hormone found in plants and is a simple gibberellin promoting growth and elongation of cells. Cytokinins are a class of phytohormones that promote cell division in plant roots and shoots. They are involved primarily in cell growth and differentiation, but also affect apical dominance, axillary bud growth, and leaf senescence.

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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
}