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Current Question (ID: 19588)

Question:
$\text{A water drop of a radius 1 cm is broken into 729 equal droplets. If the surface tension of water is 75 dyne/cm, then the gain in surface energy up to first decimal place will be:}$ $\text{[Given } \pi = 3.14]$
Options:
  • 1. $8.5 \times 10^{-4} \text{ J}$
  • 2. $8.2 \times 10^{-4} \text{ J}$
  • 3. $7.5 \times 10^{-4} \text{ J}$
  • 4. $5.3 \times 10^{-4} \text{ J}$
Solution:
$\text{Hint: } W = T \Delta A$ $\text{Step 1: Find the radius of the small water droplets.}$ $\text{By conserving the volume of water:}$ $\frac{4}{3} \pi R^3 = 729 \left( \frac{4}{3} \pi r^3 \right)$ $r = \frac{R}{(729)^{1/3}} = \frac{R}{9} = \frac{1}{9} \text{ cm}$ $\text{Step 2: Find the gain in surface energy.}$ $\Delta U = S \Delta A$ $= S \left( 4 \pi r^2 n - 4 \pi R^2 \right)$ $= 75 \times 10^{-5} \left( 4 \pi \left[ 729 \times 10^{-4} - 81 \times 10^{-4} \right] \right)$ $= 8 \times 942 \times 10^{-7} \text{ J}$ $= 7536 \times 10^{-7}$ $= 7.5 \times 10^{-4} \text{ J}$

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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
}