Import Question JSON

Current Question (ID: 10804)

Question:

$\text{Air is a bad conductor of heat or partly conducts heat. Still, a vacuum is to be placed between the walls of the thermos flask because:}$

Options:

1. $\text{it is difficult to fill the air between the walls of the thermos flask.}$
2. $\text{due to more pressure of air, the thermos can get cracks.}$
3. $\text{by convection, heat can flow through the air.}$ (Correct)
4. $\text{on filling the air, there is no advantage.}$

Solution:

$\text{The correct answer is option 3: by convection, heat can flow through the air.}$ $\text{Explanation:}$ $\text{While air is indeed a poor conductor of heat (has low thermal conductivity), it can still transfer heat through convection. In convection, heat transfer occurs due to the movement of air molecules. When air between the walls of a thermos flask gets heated, it becomes less dense and rises, while cooler air sinks down, creating convection currents.}$ $\text{A vacuum eliminates this problem because there are no air molecules present to create convection currents. In a vacuum, heat transfer can only occur through radiation, which is minimal in a well-designed thermos flask with reflective surfaces.}$ $\text{Therefore, a vacuum is used between the walls of a thermos flask to prevent heat transfer by convection, making the thermos much more effective at maintaining temperature.}$

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
}