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Current Question (ID: 15967)

Question:
$\text{Sound waves travel at } 350 \text{ m/s through warm air and at } 3500 \text{ m/s through brass.}$ $\text{The wavelength of a } 700 \text{ Hz acoustic wave as it enters brass from warm air:}$
Options:
  • 1. $\text{increase by a factor of 20.}$
  • 2. $\text{increase by a factor of 10.}$
  • 3. $\text{decrease by a factor of 20.}$
  • 4. $\text{decrease by a factor of 10.}$
Solution:
$\text{Hint: } \nu = \frac{v}{\lambda}$ $\text{Step: Find the wavelength of the acoustic wave.}$ $\text{Given,}$ $\text{Speed of sound in warmer air } = 350 \text{ m/s}$ $\text{Speed of sound in brass } = 3500 \text{ m/s}$ $\text{Frequency of wave } = 700 \text{ Hz}$ $\text{We know,}$ $\text{The velocity of sound is given by; } v = \nu \lambda$ $\text{Here } \nu = \text{frequency of the sound wave}$ $\frac{v_1}{v_2} = \frac{\nu_1 \lambda_1}{\nu_2 \lambda_2} \quad (\text{but } \nu_1 = \nu_2)$ $\lambda_2 = \lambda_1 \frac{v_2}{v_1} = \lambda_1 \times 10$ $\lambda_2 = 10 \lambda_1$ $\text{Therefore, the wavelength of the acoustic wave increases by a factor of 10.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}