Import Question JSON

Current Question (ID: 14700)

Question:
How many times can human urine be concentrated in counter current mechanism?
Options:
  • 1. Four times
  • 2. Five times
  • 3. Three times
  • 4. Two times
Solution:
Human urine can be concentrated up to four times in the counter current mechanism. This is achieved through the countercurrent multiplier system that involves the loop of Henle and vasa recta. The descending limb of the loop of Henle is permeable to water but not to ions, which allows water to be reabsorbed into the bloodstream, resulting in an increase in solute concentration in the tubular fluid. The ascending limb of the loop of Henle, on the other hand, is permeable to ions but not to water, which allows for the active transport of ions out of the tubular fluid, further increasing the solute concentration. The vasa recta, a specialized capillary network that surrounds the loop of Henle, helps to maintain the concentration gradient by absorbing solutes and water as they leave the loop of Henle, and returning them to the bloodstream. This countercurrent mechanism allows the kidney to produce highly concentrated urine, which is important for maintaining water balance in the body.

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
}