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Current Question (ID: 14946)

Question:
Five events in the transmission of nerve impulse across the synapse are given– A. Opening of specific ion channels allows the entry of ions, a new action potential is generated in the post synaptic neuron. B. Neurotransmitter binds to the receptor on post synaptic membrane C. Synaptic vesicle fuses with pre-synaptic membrane, neurotransmitter releases into synaptic cleft D. Depolarization of pre-synaptic membrane E. Arrival of action potential at axon terminal. In which sequence do these events occur?
Options:
  • 1. E → D → C → B → A
  • 2. A → B → C → D → E
  • 3. A → B → D → C → E
  • 4. E → D → C → A → B
Solution:
Page 319-320, XI NCERT

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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
}