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Current Question (ID: 15969)

Question:

$\text{A sound wave is passed through a chamber. If the rms speed of molecules in a gas is } v_1 \text{ and the speed of sound is } v_2 \text{ in the gas, then:}$ $1.\ v_1 = v_2$ $2.\ v_1 > v_2$ $3.\ v_1 < v_2$ $4.\ v_1 \leq v_2$

Options:

1. $v_1 = v_2$
2. $v_1 > v_2$
3. $v_1 < v_2$
4. $v_1 \leq v_2$

Solution:

$\text{Hint: Use the formula of RMS speed and speed of sound in any medium.}$ $\text{Step 1: Write the formula of RMS speed.}$ $v_{\text{rms}} = \sqrt{\frac{3P}{\rho}} = v_1$ $\text{Step 2: Write the formula of the speed of sound in any medium and compare both speeds.}$ $v_{\text{sound}} = \sqrt{\frac{\gamma P}{\rho}} = v_2$ $\Rightarrow v_1 > v_2 \ (\text{as } \gamma < 3)$ $\text{Hence, option (2) is the correct answer.}$

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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
}