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Current Question (ID: 19573)

Question:

$\text{A capillary tube made of glass of radius } 0.15 \text{ mm is dipped vertically in a beaker filled with methylene iodide (surface tension } T = 0.05 \text{ Nm}^{-1}, \text{ density } \rho = 667 \text{ kg m}^{-3}) \text{ which rises to height } h \text{ in the tube.}$ $\text{It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of } 60^\circ \text{ with one another.}$ $\text{Then } h \text{ is close to:}$ $\text{(take } g = 10 \text{ m/s}^2)$

Options:

1. $0.137 \text{ m}$
2. $0.172 \text{ m}$
3. $0.087 \text{ m}$
4. $0.049 \text{ m}$

Solution:

$\text{Hint: } h = \frac{2T}{\rho g R}$ $\text{Step: Find the height } (h) \text{ in the capillary rise.}$ $\text{The methylene iodide rises in the capillary tube is shown in the figure below;}$ $\text{From the figure we know that;}$ $\Rightarrow \frac{r}{R} = \cos \theta$ $\Rightarrow \frac{r}{R} = \cos 30^\circ$ $\Rightarrow R = \frac{2r}{\sqrt{3}} = \frac{2 \times 0.15 \times 10^{-3}}{\sqrt{3}} = \frac{0.3}{\sqrt{3}} \times 10^{-3} \text{ m}$ $\text{The height of the capillary is given by;}$ $\Rightarrow h = \frac{2T}{\rho g R}$ $\Rightarrow h = \frac{2 \times 0.05}{667 \times 10 \times \frac{0.3 \times 10^{-3}}{\sqrt{3}}}$ $\Rightarrow h = 0.087 \text{ m}$ $\text{Hence, option } (3) \text{ is the correct answer.}$

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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
}