Import Question JSON

Current Question (ID: 19591)

Question:
$\text{A drop of water of } 10 \text{ mm radius is divided into } 1000 \text{ droplets. If the surface tension of the water surface is equal to } 0.073 \text{ J/m}^2 \text{ then increment in surface energy while breaking down the bigger drop in small droplets as mentioned is equal to:}$
Options:
  • 1. $8.25 \times 10^{-5} \text{ J}$
  • 2. $9.17 \times 10^{-4} \text{ J}$
  • 3. $9.17 \times 10^{-5} \text{ J}$
  • 4. $8.25 \times 10^{-4} \text{ J}$
Solution:
$\text{Let the radius of one small droplet is } r \text{ then:}$ $1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (10)^3$ $\Rightarrow \ r = 1 \text{ mm}$ $v_f = 1000 \times 4 \pi r^2 T = 1000 \times 4 \pi \times 10^{-6} \times 0.073$ $v_f = 9.17 \times 10^{-4} \text{ J}$ $v_i = 4 \times \pi \times (10^{-2})^2 T = 9.17 \times 10^{-5} \text{ J}$ $\text{So,}$ $\Delta U = 8.25 \times 10^{-4} \text{ J}$

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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
}