Import Question JSON

Current Question (ID: 14614)

Question:

Which one of the following statements is incorrect?

Options:

1. The medullary zone of the kidney is divided into a few conical masses called medullary pyramids projecting into calyces.
2. Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis.
3. glomerulus along with Bowman's capsule is called the renal corpuscle.
4. Renal corpuscle, proximal convoluted tubule (PCT), and distal convoluted tubule (DCT) of the nephron are situated in the cortical region of the kidney.

Solution:

The renal pelvis is not part of the cortical region of the kidney. It is a funnel-shaped structure that collects urine from the calyces (funnel-shaped chambers that receive urine from the medullary pyramids) and funnels it into the ureter for elimination from the body. Option 1 is correct, as the medullary zone of the kidney contains several conical masses called medullary pyramids, which project into the calyces. Option 3 is correct, as the glomerulus along with Bowman's capsule forms the renal corpuscle, which is the site of ultrafiltration in the kidney. Option 4 is also correct, as the renal corpuscle, PCT, and DCT are all located in the cortical region of the kidney.

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
}