Import Question JSON

Current Question (ID: 19976)

Question:

$\text{In a resonance tube experiment, a tuning fork resonates with the air column when the water level is at } 17.0 \text{ cm from the bottom.}$ $\text{The next resonance with the same tuning fork occurs when the water level is at } 24.5 \text{ cm.}$ $\text{Given that the speed of sound in air is } 330 \text{ m/s, the frequency of the tuning fork is:}$

Options:

1. $1100 \text{ Hz}$
2. $3300 \text{ Hz}$
3. $2200 \text{ Hz}$
4. $550 \text{ Hz}$

Solution:

$\text{Hint: } v = f \lambda$ $\text{Step 1: Find the wavelength of sound wave produced by the tuning fork.}$ $\text{In a closed organ pipe, resonance occurs at odd multiples of the quarter wavelength:}$ $L_n = (2n - 1) \frac{\lambda}{4}$ $\text{So, the difference between two successive resonant lengths in a closed pipe is:}$ $\Delta L = \frac{3\lambda}{4} - \frac{\lambda}{4}$ $\Rightarrow \Delta L = \frac{\lambda}{2}$ $\Rightarrow 24.5 - 17 = \frac{\lambda}{2}$ $\Rightarrow \lambda = 15 \text{ cm} = 0.15 \text{ m}$ $\text{Step 2: Find the frequency of the tuning fork.}$ $v = f \lambda$ $\Rightarrow f = \frac{v}{\lambda}$ $\Rightarrow f = \frac{330}{0.15}$ $\Rightarrow f = 2200 \text{ Hz}$ $\text{Hence, option (3) is the correct answer.}$

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
}