Import Question JSON

Current Question (ID: 10511)

Question:

$\text{A capillary tube of radius } r \text{ is immersed in water and water rises in it to a height } h\text{. The mass of the water in the capillary is 5 g. Another capillary tube of radius } 2r \text{ is immersed in water. The mass of water that will rise in this tube is:}$

Options:

1. $\text{5.0 g}$
2. $\text{10.0 g}$
3. $\text{20.0 g}$
4. $\text{2.5 g}$

Solution:

$\text{As } h \alpha \frac{1}{r}$ $\Rightarrow \frac{h}{h_2} = \frac{2r}{r}$ $\Rightarrow h_2 = \frac{h}{2}$ $\text{Now, } m = \pi r^2 h \rho$ $\Rightarrow \frac{m_1}{m_2} = \frac{\pi r_1^2 h_1 \rho}{\pi r_2^2 h_2 \rho} = \left(\frac{r_1}{2r_1}\right)^2 \times \frac{h}{h/2}$ $\Rightarrow m_2 = 5 \times 2 = 10 \text{ g}$

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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
}