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Current Question (ID: 19610)

Question:

$\text{The diameter of an air bubble which was initially } 2 \text{ mm, rises steadily through a solution of density } 1750 \text{ kgm}^{-3} \text{ at the rate of } 0.35 \text{ cms}^{-1}. \text{ The coefficient of viscosity of the solution is (in the nearest integer):}$ $\text{(The density of air is negligible).}$

Options:

1. $15 \text{ poise}$
2. $9 \text{ poise}$
3. $18 \text{ poise}$
4. $11 \text{ poise}$

Solution:

$\text{Hint: The buoyance force acting on the air bubble equals the viscous force offered by the liquid.}$ $\text{Step: Find the coefficient of viscosity of the solution.}$ $v_T = \frac{2gr^2}{9\eta}(\sigma - \rho_{\text{air}})\eta = \frac{2 \times 10 \times (1 \times 10^{-3})^2 \times 1750}{9 \times 0.35 \times 10^{-2}} = \frac{350}{315} \approx 1.11 \text{ Ns/m}^2 = 11 \text{ poise.}$

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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
}