Import Question JSON

Current Question (ID: 14653)

Question:

Urinary excretion is equivalent to: GF = Glomerular filtration. TR = Tubular reabsorption. TS = Tubular secretion.

Options:

1. GF + TR + TS
2. GF - (TR + TS)
3. GF - TR + TS
4. GF - TS + TR

Solution:

Glomerular filtration (GF) is the process by which water and small solutes are filtered from the blood in the glomerulus into the Bowman's capsule. Tubular reabsorption (TR) is the process by which water and solutes are reabsorbed from the tubules back into the bloodstream. Tubular secretion (TS) is the process by which substances are actively transported from the blood into the tubules for excretion in the urine. Therefore, the net amount of a substance excreted in the urine is equal to the amount filtered (GF) minus the amount that is reabsorbed (TR) plus the amount that is secreted (TS), which is represented by the equation: Urinary excretion = GF - TR + TS.

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
}