Import Question JSON

Current Question (ID: 11151)

Question:

$\text{The curve between absolute temperature and } v_{\text{rms}}^2 \text{ is:}$

Options:

1. $\text{Curved line starting from origin, increasing at decreasing rate}$
2. $\text{Straight line passing through origin}$ (Correct)
3. $\text{Curved line starting from origin, increasing at increasing rate}$
4. $\text{Horizontal line}$

Solution:

$\text{Hint: } v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ $\text{Explanation: The root mean square speed } (v_{\text{rms}}) \text{ of gas molecules is given by the formula:}$ $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ $\text{From the equation } v_{\text{rms}}^2 = \frac{3RT}{M}, \text{ we can see that } v_{\text{rms}}^2 \text{ is directly proportional to } T. \text{ This can be expressed as:}$ $v_{\text{rms}}^2 \propto T$ $\text{Since } v_{\text{rms}}^2 \text{ is directly proportional to } T, \text{ we can conclude that if we plot } v_{\text{rms}}^2 \text{ against } T, \text{ the graph will be a straight line passing through the origin. This indicates a linear relationship.}$ $\text{Therefore, the curve between absolute temperature } T \text{ and } v_{\text{rms}}^2 \text{ is a straight line.}$ $\text{Hence, option (2) is the correct answer.}$

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
}