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Current Question (ID: 19565)

Question:
$\text{Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of- (density of water = 1000 kg/m}^3\text{, coefficient of viscosity of water = 1 m Pa s)}$
Options:
  • 1. $10^4$
  • 2. $10^3$
  • 3. $10^2$
  • 4. $10$
Solution:
$\text{Hint: Volume rate} = v \times \pi r^2$ $\text{Step 1: Find volume flow rate in SI units}$ $\frac{\Delta V}{\Delta t} = \frac{100 \times 10^{-3}}{60} \text{ m}^3/\text{s}$ $\text{Step 2: Find the velocity of flow}$ $\frac{\Delta V}{\Delta t} = Av$ $\text{Step 3: Find Reynold's number}$ $R_e = \frac{\rho v d}{\eta}$ $= \frac{10^3 \times 10^{-1} \times 10^{-1}}{10^{-3} \times 60 \times \pi (25 \times 10^{-4})}$ $\approx 2 \times 10^4$

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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
}