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Current Question (ID: 19586)

Question:

$\text{A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be:}$ $1.\ 2.8 \times 10^{-4} \text{ J}$ $2.\ 1.5 \times 10^{-3} \text{ J}$ $3.\ 1.9 \times 10^{-4} \text{ J}$ $4.\ 9.4 \times 10^{-5} \text{ J}$

Options:

1. $2.8 \times 10^{-4} \text{ J}$
2. $1.5 \times 10^{-3} \text{ J}$
3. $1.9 \times 10^{-4} \text{ J}$
4. $9.4 \times 10^{-5} \text{ J}$

Solution:

$d = 2 \text{ cm}$ $r = 1 \text{ cm}$ $T = 0.075$ $\Delta SE = T \Delta A$ $= 0.075 \left( A_f - A_i \right)$ $A_i = 4 \pi r^2$ $A_f = 4 \pi r_0^2 \times 64$ $\text{By volume conservation}$ $\frac{4}{3} \pi r^3 = 64 \cdot \frac{4}{3} \pi r_0^3$ $r_0 = \frac{r}{4}$ $A_f = 4 \pi \left( \frac{r}{4} \right)^2 \cdot 64 = 16 \pi r^2$ $\Delta SE = 0.075 \left( 16 \pi r^2 - 4 \pi r^2 \right)$ $= 0.075 \left( 12 \pi (0.01)^2 \right)$ $= 2.8 \times 10^{-4} \text{ J}$

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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
}