Import Question JSON

Current Question (ID: 12087)

Question:

Which one is wrongly matched? 1. Uniflagellate gametes - Polysiphonia 2. Biflagellate zoospores - Brown algae 3. Gemma cups - Marchantia 4. Unicellular organism - Chlorella

Options:

1. Uniflagellate gametes - Polysiphonia
2. Biflagellate zoospores - Brown algae
3. Gemma cups - Marchantia
4. Unicellular organism - Chlorella

Solution:

Red algae belong to classes of Rhodophyceae and are ancient algae. Example: Polysiphonia • Red algae are multicellular. The life cycle of Polysiphonia is diplobiontic and there is no motile stage is found in the life cycle of red algae. • Phaeophyceae or brown algae, the vegetative reproduction takes place by fragmentation. Asexual reproduction in most brown algae is by biflagellate zoospores that are pear-shaped and have two unequal laterally attached flagella. • Gemmae is a widespread means of asexual reproduction in both liverworts and mosses. In liverworts such as Marchantia, the flattened plant body or thallus is a haploid gametophyte with gemma cups scattered about its upper surface. The gemma cups are cup-like structures containing gemmae. • Most living organisms consist of many cells, a single cell of Chlorella is an independent life form and is called an unicellular organism. PAGE 33-34, XI NCERT

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
}