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Current Question (ID: 10491)

Question:

$\text{A small tiny water droplet is falling towards earth at a uniform speed of 1 cm/s. When 27 such identical droplets combine together to form a bigger drop, then with what uniform velocity will this bigger drop fall?}$

Options:

1. $27 \text{ cm/s}$
2. $9 \text{ cm/s}$
3. $3 \text{ cm/s}$
4. $81 \text{ cm/s}$

Solution:

\text{Hint: Droplets will fall with terminal velocity.} \text{Step 1: Terminal velocity of the droplets is given as:} v_T = \frac{2r^2g(\sigma - \rho)}{9\eta} \text{Here, } v_T \propto r^2 \text{Step 2: Find the relation between r and R.} \text{As the volume is same:} n \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \text{So, } R = n^{1/3}r R = 3r \text{Step 3: Calculate } v_2 \text{Hence, } \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 \Rightarrow v_2 = 9v_1 = 9 \text{ cm/s}

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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
}