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Current Question (ID: 14933)

Question:

During the propagation of a nerve impulse, the action potential results from the movement of

Options:

1. K⁺ ions from extracellular fluid to intracellular fluid
2. Na⁺ ions from intracellular fluid to extracellular fluid
3. K⁺ ions from intracellular fluid to extracellular fluid
4. Na⁺ ions from extracellular fluid to intracellular fluid

Solution:

During the propagation of nerve impulse when a stimulus of adequate strength is applied to a polarised membrane, the permeability of the membrane to Na⁺ is greatly increased at the point of stimulation. As a result the sodium ion channels permit the influx of Na⁺ by diffusion. Since, there are more Na⁺ ions entering than leaving, the electrical potential of the membrane changes from -70 mV towards zero. At 0 mV the membrane is said to be depolarised. While the resting potential is determined largely by K⁺ ions, the action potential is determined largely by Na⁺ ions. Action potential is another name of nerve impulse. The stimulated negatively charged point on the outside of the membrane sends out an electrical current to the positive point adjacent to it. This local current causes the adjacent inner part of the membrane to reverse its potential from -70 mV to +30 mV. NCERT XI, Page 318

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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
}