Import Question JSON

Current Question (ID: 12324)

Question:

Match the Column-I with Column-II: Column-I (Aestivation in Corolla) A. Valvate B. Twisted C. Imbricate D. Vexillary (papilionaceous) Column-II (Examples) I. China rose, Lady's finger, cotton II. Calotropis III. Cassia, Gulmohur IV. Pea, Beans

Options:

1. A - II, B - I, C - III, D - IV
2. A - II, B - I, C - IV, D - III
3. A - I, B - II, C - III, D - IV
4. A - II, B - IV, C - I, D - III

Solution:

The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation. 1. Valvate: When sepals or petals in a whorl just touch one another at the margin, without overlapping, as in Calotropis, it is said to be valvate. 2. Twisted: If one margin of the appendage overlaps that of the next one and so on as in china rose, lady's finger and cotton, it is called twisted. 3. Imbricate: If the margins of sepals or petals overlap one another but not in any particular direction as in Cassia and gulmohur, the aestivation is called imbricate. 4. Vexillary: In pea and bean flowers, there are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel); this type of aestivation is known as vexillary or papilionaceous.

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
}