Import Question JSON

Current Question (ID: 10432)

Question:

A mercury-filled U-tube arrangement is connected to a bulb containing a gas. Atmospheric pressure is $1.012 \times 10^5 \text{ Pa}$ and $H = 0.05 \text{ m}$. Then:

Options:

1. gauge pressure at $\text{R}$ is nil.
2. gauge pressure at $\text{R}$ is $6.56 \times 10^3 \text{ Pa}$.
3. gauge pressure at $\text{R}$ is $1.08 \times 10^5 \text{ Pa}$.
4. pressure at $\text{R}$, $\text{Q}$, and inside the bulb is the same.

Solution:

\text{The gauge pressure is the amount by which the pressure measured in a fluid exceeds that of the atmosphere.} \text{Gauge pressure} = \text{Absolute pressure} - \text{Atmospheric pressure} \text{Step: Find the gauge pressure at different points.} \text{At point R, the mercury surface is open to the atmosphere. Therefore, the absolute pressure at R is equal to the atmospheric pressure.} \text{Pressure at R} = \text{Atmospheric pressure} = 1.012 \times 10^5 \text{ Pa} \text{Gauge pressure at R} = \text{Absolute pressure at R} - \text{Atmospheric pressure} \text{Gauge pressure at R} = (1.012 \times 10^5 \text{ Pa}) - (1.012 \times 10^5 \text{ Pa}) = 0 \text{So, the gauge pressure at R is nil.} \text{The pressure at R is equal to the atmospheric pressure as R lies at the surface of the liquid and the surface of the liquid is open to the air.}

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
}