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Current Question (ID: 19607)

Question:

$\text{A water drop of radius } 1 \, \mu m \text{ falls in a situation where the effect of buoyant force is negligible.}$ $\text{Co-efficient of viscosity of air is } 1.8 \times 10^{-5} \, \text{Nsm}^{-2} \text{ and its density is negligible as compared to that of water } 10^6 \, \text{gm}^{-3}. \text{ Terminal velocity of the water drop is:}$ $\text{(Take acceleration due to gravity } = 10 \, \text{ms}^{-2})$

Options:

1. $145.4 \times 10^{-6} \, \text{ms}^{-1}$
2. $118.0 \times 10^{-6} \, \text{ms}^{-1}$
3. $132.6 \times 10^{-6} \, \text{ms}^{-1}$
4. $123.4 \times 10^{-6} \, \text{ms}^{-1}$

Solution:

$\text{The terminal velocity } v_t \text{ is given by Stokes' law:}$ $v_t = \frac{2}{9} \cdot \frac{r^2 \cdot g \cdot (\rho - \sigma)}{\eta}$ $\text{where } r = 1 \times 10^{-6} \, \text{m}, \ g = 10 \, \text{ms}^{-2}, \ \rho = 10^6 \, \text{kgm}^{-3}, \ \sigma \approx 0, \ \eta = 1.8 \times 10^{-5} \, \text{Nsm}^{-2}$ $v_t = \frac{2}{9} \cdot \frac{(1 \times 10^{-6})^2 \cdot 10 \cdot 10^6}{1.8 \times 10^{-5}}$ $v_t = 123.4 \times 10^{-6} \, \text{ms}^{-1}$

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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
}