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Current Question (ID: 16015)

Question:
$\text{A sufficiently long-closed organ pipe has a small hole at its bottom. Initially, the pipe is empty. Water is poured into the pipe at a constant rate. The fundamental frequency of the air column in the pipe:}$
Options:
  • 1. $\text{Constantly increases.}$
  • 2. $\text{Increases at first, then becomes constant.}$
  • 3. $\text{Constantly decreases.}$
  • 4. $\text{Decreases at first, then becomes constant.}$
Solution:
$\text{Hint: Change in frequency depends on the change in length of the air column.}$ $\text{Step 1: Find the change in the length of the air column.}$ $\text{Therefore, length } l \text{ first decreases and then becomes constant.}$ $\text{Step 2: Find the change in fundamental frequency.}$ $f = \frac{v}{4l}$ $\text{The fundamental frequency of the column is inversely proportional to the length of the air column i.e., } f \propto \frac{1}{l}$ $\text{Therefore, the fundamental frequency of the air column in the pipe increases at first, then becomes constant.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}