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Current Question (ID: 19619)

Question:
$\text{The excess pressure inside a soap bubble is thrice the excess}$ $\text{pressure inside a second soap bubble. The ratio between the volume}$ $\text{of the first and the second bubble is:}$
Options:
  • 1. $1 : 81$
  • 2. $1 : 9$
  • 3. $1 : 27$
  • 4. $1 : 3$
Solution:
$\text{Let the excess pressure in the first bubble be } P_1 \text{ and in the second}$ $\text{bubble be } P_2. \text{ Given } P_1 = 3P_2.$ $\text{The excess pressure in a soap bubble is given by } P = \frac{4T}{r},$ $\text{where } T \text{ is the surface tension and } r \text{ is the radius.}$ $\text{Thus, } \frac{4T}{r_1} = 3 \times \frac{4T}{r_2} \Rightarrow r_2 = 3r_1.$ $\text{The volume of a bubble is given by } V = \frac{4}{3}\pi r^3.$ $\text{Therefore, the ratio of the volumes is } \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

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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
}