Import Question JSON

Current Question (ID: 10500)

Question:

$\text{A soap bubble is blown with the help of a mechanical pump at the mouth of a tube. The pump produces a certain increase per minute in the volume of the bubble, irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time t will be:}$

Options:

1. $\text{Graph showing pressure P decreasing exponentially with time t (curved downward)}$
2. $\text{Graph showing pressure P increasing linearly with time t (straight line upward)}$
3. $\text{Graph showing pressure P decreasing linearly with time t (straight line downward)}$
4. $\text{Graph showing pressure P increasing exponentially with time t (curved upward)}$

Solution:

\text{For a soap bubble: } \Delta P = \frac{4T}{r} \therefore \Delta P \propto \frac{1}{r} \text{Given that the increase in volume is directly proportional to time, the radius of the soap bubble increases with time.} \text{Since volume } V = \frac{4}{3}\pi r^3 \text{ and } V \propto t, \text{ we have } r \propto t^{1/3} \text{Therefore: } \Delta P \propto \frac{1}{r} \propto \frac{1}{t^{1/3}} = t^{-1/3} \text{This means pressure decreases with time following a power law relationship, which appears as a curved exponential-like decay.} \text{The correct graph shows pressure decreasing in a curved manner (not linearly) with time.}

Import JSON File

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
}