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Current Question (ID: 10528)

Question:
$\text{An incompressible fluid flows steadily through a cylindrical pipe which has a radius } 2r \text{ at the point A and a radius } r \text{ at the point B further along the flow direction. If the velocity at the point A is } v\text{, its velocity at the point B is:}$
Options:
  • 1. $2v$
  • 2. $v$
  • 3. $\frac{v}{2}$
  • 4. $4v$
Solution:
\text{Using the equation of continuity for incompressible flow:} \text{Area} \times \text{velocity} = \text{constant} \pi r^2 \times v = \text{constant} \text{Therefore: } r_1^2 v_1 = r_2^2 v_2 \text{Given: } r_1 = 2r, r_2 = r, \text{ and } v_1 = v \text{Substituting: } (2r)^2 \times v = r^2 \times v_2 4r^2 \times v = r^2 \times v_2 \text{Therefore: } v_2 = 4v

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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
}