Import Question JSON

Current Question (ID: 19596)

Question:
$\text{Six identical small liquid drops, each falling with a terminal velocity of } 10 \text{ m/s, coalesce to form a single larger spherical drop. Assuming the drops fall through the same viscous medium under gravity and obey Stokes' law, what will be the terminal velocity of the larger drop?}$
Options:
  • 1. $10 \times (6)^{1/3} \text{ m/s}$
  • 2. $10 \times (6)^{2/3} \text{ m/s}$
  • 3. $5 \times (3)^{2/3} \text{ m/s}$
  • 4. $10 \times (6)^3 \text{ m/s}$
Solution:
$\text{Hint: } v_T \propto (\text{radius})^2$ $\text{Step 1: Find the radius of the bigger drop.}$ $\text{The volume of the water remains the same.}$ $\frac{4}{3} \pi R^3 = 6 \left( \frac{4}{3} \pi r^3 \right)$ $\Rightarrow R = 6^{1/3}(r)$ $\text{Step 2: Find the terminal velocity of a bigger drop.}$ $v_T \propto (\text{radius})^2$ $\frac{v_b}{v_s} = \frac{R^2}{r^2}$ $\Rightarrow v_b = 10 \times (6)^{2/3} \text{ m/s}$ $\text{Hence, option (2) is the correct answer.}$

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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
}